AJAX update database - php

Sorry for not stated my problem. Actually I want to update my data in database.
But the problem now is even i tried to choose approve or reject the ajax still won't update.
I am new in ajax and try search around net but my code still got problem
here is my php page
<?php
$querysel = "SELECT * FROM tblinternapplication WHERE course_code = '{$course_codeapp}' ORDER BY student_id, 1 DESC " ;
$resultsel = mysql_query($querysel, $connection);
echo "<h2><div class=\"h_title\">Status still in pending</div></h2>";
echo "<table>";
echo "<thead>";
echo "<tr>";
echo "<th scope=\"col\">Matric ID</th>";
echo "<th scope=\"col\">Company name</th>";
echo "<th scope=\"col\" width = \"200\">Job Scope</th>";
echo "<th scope=\"col\">Status</th>";
echo "<th scope=\"col\">Action</th>";
echo "</tr>";
echo "</thead>";
while($rowsel = mysql_fetch_array($resultsel)){
if($rowsel['status_approval'] == NULL){
$id = $rowsel['id'];
echo "<tr>";
echo "<tr>"."<td class=\"align-center\">".$rowsel['student_id']."</td>";
echo "<td class=\"align-center\">".$rowsel['company_name']."</td>";
echo "<td class=\"align-center\" width = \"200\">".$rowsel['job_scope']."</td>";
echo "<td class=\"align-center\">";
if($rowsel['status_approval'] != NULL){
if( $rowsel['status_approval'] == 0)
{
echo "Reject";
}
else
{
echo "Approve";
}
}
else
{ echo "Pending";
}
echo "</td>";
echo "<td class=\"align-center\"><select name=\"approve\"
onchange=\"getstatus(this.value)\">";
echo "<option value=\"\">Select status:</option>";
echo "<option value=\"1\">Approve</option>";
echo "<option value=\"0\">Reject</option>";
echo "</select>";
echo "</td>";
echo "</tr>";
}
}
echo "</table>";
here is my jscript page
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script type="text/javascript">
function getstatus(id, approve)
{
if (approve=="")
{
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.open("GET","updatestatus.php?id=" + id + "&status=" + approve,true);
xmlhttp.send();
}
</script>
then here is my updatestatus.php
<?php require_once("../includes/session.php"); ?>
<?php require_once("sessioncourse.php"); ?>
<?php $course_codeapp = $_SESSION['course_code'] ; ?>
<?php confirm_logged_in(); ?>
<?php require_once("../includes/connection.php") ?>
<?php require_once("../includes/functions.php") ?>
<?php
$id = $_GET['id'];
$status =$_GET['status'];
$sql="UPDATE tblinternapplication set status_approval = $status WHERE id = $id ";
$result = mysql_query($sql);
?>
I work for few days but problem still cannot solve. Hope someone can help me. I will appreciate your help!

For preapre result for JS in php use json_encode() function. Make your update script somthing like this:
<?php require_once("../includes/session.php");
require_once("sessioncourse.php");
$course_codeapp = $_SESSION['course_code'] ;
confirm_logged_in();
require_once("../includes/connection.php");
require_once("../includes/functions.php");
$id = $_GET['id'];
$status =$_GET['status'];
$sql="UPDATE tblinternapplication set status_approval = $status WHERE id = $id ";
$result = mysql_query($sql);
$json = array();
while ($row = mysql_fetch_assoc($result)) {
$json[] = $row;
}
echo json_encode($json);
IMPORTANT
Dont close php tag or you may add extra space chars
AJAX
In your case using jQuery ajax was good practic. Mkae you code like this:
<script type="text/javascript">
function getstatus(id, approve)
{
$.ajax({
'url': 'updatestatus.php',
'data': {"id": id, "status": approve},
'success': function (response) {
console.log(response);
//TODO: use server response
}
});
}
</script>

echo "<td class=\"align-center\"><select name=\"approve\"
onchange=\"getstatus(this.value)\">";
In that line above, you are passing only one value. At the same time your JS function waits for 2 parameters. Currently, you have id always equal to 0 or 1 and status is always undefined.
Suppose you just need to change that line like:
echo "<td class=\"align-center\"><select name=\"approve\"
onchange=\"getstatus("+ $id +", this.value)\">";
One note:
$sql="UPDATE tblinternapplication set status_approval  = $status WHERE id = $id ";
Except that you should not use mysql_* functions anymore, code above is opened to sql injections. To avoid sql injections with mysql_ extension you should do something like this:
$sql="UPDATE tblinternapplication set status_approval  = ".mysql_real_escape_string($status) ." WHERE id = " .mysql_real_escape_string($id);
See docs for mysql_real_escape_string here. Also, read a warning message on that page - it tells you what you should use in replacement for mysql extension

I would suggest to use the developer tools like F12 in IE or chrome or the fiddler to check the request you are sending to the PHP code . In case you are getting any error you can easily see in the response

Related

Can't get Post value ID from row table to PHP using Jquery

I know that my question may be duplicate, but I've looked through a ton of questions with the same problem, but none of the solutions worked for me.
It's pretty simple, get the item-id value and POST it to del.php.
I can't get the POST value in the php after the click.
<?php
include '../include/db.php';
$result = $mysqli->query("SELECT * FROM usuario WHERE nivel = '1'");
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td class='item-id'>" . $row['id'] . "</td>";
echo "<td>" . $row['nome']. "</td>";
echo "<td>" . $row['email']. "</td>";
echo "</tr>";
}
?>
<script type="text/javascript">
$("tr").click(function() {
var id = $(this).find('.item-id').text();
$.post('del.php',{id: id},function(json){
window.location.href = "http://localhost/unesc/3/admin/del.php";
}, "json")
});
</script>
del.php
<?php
if (isset($_POST['id'])){
$id = $_POST['id'];
echo $id;
} else {
echo "erro";
}
?>
The del.php just get the $_post['id'] and echo it.
The window.location is there so i won't have to enter the address manually.
The trigger may not be necessarily a Click, it can be a Button, woks just fine for me too.
EDIT: got the POST running with $.ajax, but now, with firebug i noticed that the post in del.php keeps returning empty.
EDIT2: got the del.php post response, but as an error, it says that 'id' don't exist.
If THIS is what you need here is the code of both pages I used
del2.php
<script type="text/javascript" language="javascript" src="jquery.min.js"></script>
<?php
//include '../include/db.php';
include 'includeconnect.php';
//$result = $mysqli->query("SELECT * FROM usuario WHERE nivel = '1'");
$result = mysql_query("SELECT * FROM usuario WHERE nivel = '1'");
while($row = mysql_fetch_array($result))
{
echo "
<table border='1'>
<tr>";
echo "<td class='item-id'>" . $row['id'] . "</td>";
echo "<td>" . $row['nome']. "</td>";
echo "<td>" . $row['email']. "</td>";
echo "</tr>
</table>
</br>";
}
echo"<div id='show_responce'>I am a text on page del2.php and I am pretty much useless apart from showing you where the responce will display</div>"
?>
<script type="text/javascript">
$("tr").click(function() {
var rep = $("#show_responce");
var id = $(this).find('.item-id').text();
var dataString = 'id=' + id;
$.ajax({
type: 'post',
url: 'del.php',
data: dataString,
success: function(html) {
rep.empty().append(html);
rep.fadeIn("slow")
}
});
});
</script>
Here is your modified del.php
<?php
if (isset($_POST['id'])){
$id = $_POST['id'];
echo "I am coming from del.php and the ID you clicked on is $id";
} else {
echo "error";
}
?>
By the way because of my account's low reputation I could not ask you what was you trying to do with this window.location but whatever. Instead of what you asked for which was basically to make sure that del.php gets the value of id as you will see I displayed it on del2.php which is the page I am making the request from.
$( "table > td" ).find( ".item-id" ).val();
or
<?php
while($row = mysqli_fetch_array($result))
{
echo "<tr id='row'>";
echo "<td class='item-id'>" . $row['id'] . "</td>";
echo "<td>" . $row['nome']. "</td>";
echo "<td>" . $row['email']. "</td>";
echo "</tr>";
}
?>
<script>
var id = [];
var values = [];
$("#row > .item-id").each(function(index){
id.push($(this).attr("id")); // [a,b,c,....]
values.push($(this).text()); //[ Dummy1, Dummy2, Dummy3, Dummy4,..]
});
</script>
Possible errors:
}, "json") here last argument means, that function expects you return valid JSON string from del.php file.
After window.localtion.href you can't access previous data, because it redirects to another page, and thus you loose all your previous data.

AJAX script not parsing variable to php correctly

UPDATE 19/01/2017 - PROBLEM RESOLVED - Removed 'intval' on get request within get.php file and the value q is now capturing the correct data.
What i'm trying to achieve can be summed up in 2 points:
1.) A dropdown box which dynamically populates according to a MySql table
2.) Depending on whats chosen in the dropdown the textfield below will display a corresponding id via a SQL query.
I've been trying to follow W3Schools guide and modifying it accordingly to match my needs, but i'm getting stuck.
Here is what I currently have:
PHP -
<?php
mysql_connect('localhost', 'root', '');
mysql_select_db('core_log');
$sql = "SELECT local_authority FROM ons_code_tbl";
$result = mysql_query($sql);
echo "<tr>";
echo "<td> Local Authority: </td>";
echo "<td>";
echo "<select name='local_authority'onchange='showLocal_authority(this.value)'>";
echo '<option value=""></option>';
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['local_authority'] . "'>" . $row['local_authority'] . "</option>";
}
echo "</select>";
echo "</td>";
echo "</tr>";
?>
Script -
<script>
function showLocal_authority(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("txtHint").innerHTML = this.responseText;
}
};
xmlhttp.open("GET","get.php?q="+str,true);
xmlhttp.send();
}
}
</script>
get.php -
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('localhost','root','','core_log');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"ons_code_tbl");
$sql="SELECT ons_code FROM ons_code_tbl WHERE local_authority = '".$q."'";
$result = mysqli_query($con,$sql);
$row = mysqli_fetch_array($result);
$name = $row['ons_code'];
echo "<td>" . $q . "</td>";
mysqli_close($con);
?>
I did a test and simply echoed the value q to see what its outputting, it should be the value of whatever they pick in the dropdown, but instead its displaying 0?
In the screenshot below, it should display 'Amber Valley' with the correct ONS code pulled from the MySQL table.
If anyone can tell me why the value q is not storing 'Amber Valley' for example, that would be great, thanks!
Replace this line:
$q = intval($_GET['q']);
for
$q = $_GET['q'];
Then make a query with prepared statement, to avoid SQL Injection:
mysqli_select_db($con,"ons_code_tbl");
$sql="SELECT ons_code FROM ons_code_tbl WHERE local_authority = ?";
$stmt = mysqli_prepare($con, $sql);
mysqli_stmt_bind_param($stmt, "s", $q);
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $ons_code);
mysqli_stmt_fetch($stmt);
Just replace echo "<td>" . $q . "</td>"; for echo "<td>" . $ons_code . "</td>";

Ajax refine search

I have a table for a sports day where there are 4 columns name, house, event, result. I have no problem creating and displaying the database but i want to be able to search in a bar and to use AJAX to automatically search all 4 columns for whats in the search bar. I am using PHPmyadmin to store the database with mySQLI. i am able to display the database on the page that i want. I also want when the page starts for the whole table to be displayed and then when you start typing it just removes any items that do not match the search. I have never used Ajax before so sorry for my bad code as it is all from w3schools site. the DB is called sports_day and the table is called full_results. here is my current code.
<script>
function showUser(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("txtHint").innerHTML = this.responseText;
}
};
xmlhttp.open("GET","results_query.php?q="+str,true);
xmlhttp.send();
}
}
</script>
<form>
search for pupil
<input type="text" size="30" name="user" onkeyup="showUser(this.value)">
<div id="livesearch"></div>
<br>
</form>
<div class="col-sm-12">
<div id="txtHint"><b> pupil's info will be listed here</b></div>
</div>
and on a page called results_query.php is this code
<body>
<?php
$q = intval($_GET['q']);
$con = mysqli_connect("localhost","root","","sports_day");
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"sports_day");
$sql="SELECT * FROM full_results WHERE id = '".$q."'";
$result = mysqli_query($con,$sql);
echo '<tr>';
echo '<th>NAME</th>';
echo '<th>HOUSE</th>';
echo '<th>EVENT</th>';
echo '<th>RESULT</th>';
echo ' </tr>';
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['NAME'] . "</td>";
echo "<td>" . $row['HOUSE'] . "</td>";
echo "<td>" . $row['EVENT'] . "</td>";
echo "<td>" . $row['RESULT'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
</body>
at the moment what happens is none of the table is shown and when i type anything in the search box the whole table appears along with in plain text at the bottom the title and all the contents of the table in a long line.
any suggestion to get my code to work would be greatly appreciated!
thanks!
The solution would be like this:
Keep your HTML search form as it is.
<form>
search for pupil
<input type="text" size="30" name="user" onkeyup="showUser(this.value)">
<div id="livesearch"></div>
<br>
</form>
... I also want when the page starts for the whole table to be displayed and then when you start typing it just removes any items that do not match the search.
See this <div> section here,
<div class="col-sm-12">
...
</div>
You didn't put anything in this <div> section. First of all, you have to display your entire table in this section, which you can later filter out using the AJAX request. Also, assign an id to this <div> section so that it could be easier for you put the AJAX response in this <div> section. So the code for this <div> section would be like this:
<div class="col-sm-12" id="pupil-info">
<?php
$con = mysqli_connect("localhost","root","","sports_day");
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"sports_day");
$sql = "SELECT * FROM full_results";
$result = mysqli_query($con,$sql);
echo '<table>';
echo '<tr>';
echo '<th>NAME</th>';
echo '<th>HOUSE</th>';
echo '<th>EVENT</th>';
echo '<th>RESULT</th>';
echo ' </tr>';
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['NAME'] . "</td>";
echo "<td>" . $row['HOUSE'] . "</td>";
echo "<td>" . $row['EVENT'] . "</td>";
echo "<td>" . $row['RESULT'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
</div>
Change your Javascript/AJAX code in the following way,
<script>
function showUser(str){
var str = str.trim();
var xmlhttp;
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("pupil-info").innerHTML = this.responseText;
}
};
xmlhttp.open("GET","results_query.php?q="+encodeURIComponent(str),true);
xmlhttp.send();
}
</script>
Please note that you should encode the user inputted str value using encodeURIComponent() function before passing it to the results_query.php page.
Finally, on results_query.php page process your AJAX request like this:
<?php
$con = mysqli_connect("localhost","root","","sports_day");
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"sports_day");
$sql = "SELECT * FROM full_results";
if(isset($_GET['q']) && !empty($_GET['q'])){
$sql .= " WHERE CONCAT(id, NAME, HOUSE, EVENT, RESULT) LIKE '%".$_GET['q']."%'";
}
$result = mysqli_query($con,$sql);
echo '<table>';
echo '<tr>';
echo '<th>NAME</th>';
echo '<th>HOUSE</th>';
echo '<th>EVENT</th>';
echo '<th>RESULT</th>';
echo ' </tr>';
if(mysqli_num_rows($result)){
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['NAME'] . "</td>";
echo "<td>" . $row['HOUSE'] . "</td>";
echo "<td>" . $row['EVENT'] . "</td>";
echo "<td>" . $row['RESULT'] . "</td>";
echo "</tr>";
}
}else{
echo "<tr>";
echo "<td colspan='4' style='text-align:center;'>No records found</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
Sidenote: Learn about prepared statement because right now your query is susceptible to SQL injection. Also see how you can prevent SQL injection in PHP.
If you use your 'results_query.php' file only for getting the data from database, then you don't need to create a <body> tag. If you use only PHP then you can easily skip any plane HTML. That's just a digression :)
But to the point.
You can change the way you return your data from database. I think, instead of doing a lot of echo's it is better to add result to the variable and echoing the variable at the end.
$data = '<tr>' . '<th>NAME</th>' . '<th>HOUSE</th>' . '<th>EVENT</th>' . '<th>RESULT</th>' . '</tr>';
while($row = mysqli_fetch_array($result)) {
$data .= '<tr>';
$data .= '<td>' . $row['NAME'] . '</td>';
$data .= '<td>' . $row['HOUSE'] . '</td>';
$data .= '<td>' . $row['EVENT'] . '</td>';
$data .= '<td>' . $row['RESULT'] . '</td>';
$data .= '</tr>';
}
$data .= '</table>';
mysqli_close($con);
echo $data;
See if this changes something.
What about showing entire table after the page's loaded, you will have to change both PHP and JavaScript code a little bit.
You can change your JS so it gets everything from your full_results table after page is loaded.
There are several ways to do this and you can read about them here:
pure JavaScript equivalent to jQuery's $.ready() how to call a function when the page/dom is ready for it
The easiest way would be to do this this way:
<script>
function showUser(str) {
var url;
var xmlhttp;
if (str == "") { //if empty string - get all data
url = "results_query.php";
} else { //get particular data otherwise
url = "results_query.php?q="+str;
}
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("txtHint").innerHTML = this.responseText;
}
}
xmlhttp.open("GET", url, true);
xmlhttp.send();
}
</script>
<form>
search for pupil
<input type="text" size="30" name="user" onkeyup="showUser(this.value)">
<div id="livesearch"></div>
<br>
</form>
<div class="col-sm-12">
<div id="txtHint"><b> pupil's info will be listed here</b></div>
</div>
<script>
//calling your function with empty string because we want to get all data
showUser("");
</script>
and in the PHP file you can do something like this:
<?php
$q = 0;
//check if 'q' parameter was passed
if(isset($_GET['q'])) {
$q = intval($_GET['q']);
}
$con = mysqli_connect("localhost","root","","sports_day");
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"sports_day");
$sql = ($q) ? "SELECT * FROM full_results WHERE id = '".$q."'" : "SELECT * FROM full_results";
Now your JavaScript function will be called after loading your page.
It will call your PHP script with AJAX and this script should return all data from your table.
In line ($q) ? "SELECT * FROM full_results WHERE id = '".$q."'" : "SELECT * FROM full_results"; there is a simple check if $q is different from 0. Our variable will be set to 0 if no argument was passed, so whenever $q is equal to '0', we just want to get all the data from full_results and specific data otherwise.
I also added var xmlhttp because it is only local variable.
You can read more about that in here:
https://stackoverflow.com/a/1471738/7301294
I hope it will help you.
Let me know if you have any other problems and never be afraid to ask.
Good luck!

Post array from form using xmlhttprequest

I have an php generated table/form with checkboxes like this:
if ($query) {
if (!mysqli_num_rows($query)) {
//Empty storage
echo "There are no items in '$storage'.";
exit();
} else {
//form
echo "<form name='send_parts_form' action='../includes/generatetab.php' method='post'>";
//Table header
echo "
<table>
<tr>
<th>Part</th>
<th>PN</th>
<th>Manufactured</th>
<th>Serial</th>
<th>Site</th>
<th>Date replaced</th>
<th>By user</th>
<th>Faulty</th>
<th>Send</th>
<th>Select</th>
</tr>";
//Retrieved data
while ($row = mysqli_fetch_array($query)) {
echo "<tr>";
echo "<td>" . $row['part_type'] . "</td>";
echo "<td>" . $row['pn'] . "</td>";
echo "<td>" . $row['manufactured'] . "</td>";
echo "<td>" . $row['serial'] . "</td>";
echo "<td>" . $row['site_id'] . "</td>";
echo "<td>" . $row['date'] . "</td>";
echo "<td>" . $row['user'] . "</td>";
echo "<td>" . $row['faulty'] . "</td>";
echo "<td><input type='checkbox' name='send_parts[]' class='checkclass' value=" . $row['id'] . "></td>";
echo "</tr>";};
echo "</table>";
echo "</br>";
echo "</br>";
echo "<input type='button' onclick='sendToZG()' value='Send'/>";
echo "</br>";
echo "<input type='submit' name='submit' value='Generate tab' />";
echo "</form>";
exit();
}
} else {
die("Query failed");
}
User then checks option they want and upon submiting (Generate tab) they get tab delimited text with values they selected.
I now want when they click "Send" to have values posted to another php page and results returned on the same page (under SentList div). I have js like this:
//Browser Support Code
function sendToZG(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
}catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
}catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4 && ajaxRequest.status == 200){
var ajaxDisplay = document.getElementById('SentList');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
// Now get the value from page and pass it to server script.
var formData = new FormData( document.getElementsByName("send_parts")[0] );
ajaxRequest.open('POST', "../includes/sendtozg.php", true);
ajaxRequest.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
ajaxRequest.send("send_parts=" + formData);
}
Edited: ajaxRequest.send("send_part=" + formData); to ajaxRequest.send("send_parts=" + formData);
Now it returns:
Invalid argument supplied for foreach() on line 53 (That is where I fetch my data in sendtozg.php).
I'll add sendtozg.php at the end of the post.
If instead of:
<form name='send_parts_form' action='../includes/generatetab.php' method='post'>
I echo:
<form name='send_parts_form' action='../includes/sendtozg.php' method='post'>
Upon submit, script sendtozg.php gets executed fine but on a different page.
So basically what I'm trying to do is to have 2 options for the php generated form:
Generate tab delimited txt file
Execute sendtozg.php and return results on same page
I already have both scripts (generatetab.php and sendtozg.php) and they work fine.
sendtozg.php:
if (!empty($_POST['send_parts'])){
foreach ($_POST['send_parts'] as $send_parts){
$getchecked = mysqli_query($con, "SELECT * FROM $storage WHERE id=$send_parts");
while ($row = mysqli_fetch_array($getchecked)) {
$copypart = mysqli_query($con, "INSERT INTO sent_parts (part_type, pn, manufactured, serial, site_id, date, user, faulty, log)
SELECT part_type, pn, manufactured, serial, site_id, date, user, faulty, log
FROM $storage WHERE id=$send_parts");
// check to see if it copied
$getserial = mysqli_query($con, "SELECT serial FROM $storage WHERE id=$send_parts");
$getserial_row = mysqli_fetch_array($getserial, MYSQLI_NUM);
$foundserial = $getserial_row[0];
$checkcopy = mysqli_query($con, "SELECT id FROM sent_parts WHERE serial = '$foundserial'");
// add user info and date
$addinfo = mysqli_query($con, "UPDATE sent_parts SET sent2zg='$user', date2zg='$strdate' WHERE serial = '$foundserial'");
if (!$check1_res) {
printf("Error: %s\n", mysqli_error($con));
exit();
};
//delete from storage
if($checkcopy > 0) {
$getpart = mysqli_query($con, "SELECT part_type FROM sent_parts WHERE serial='$foundserial'");
$getpart_row = mysqli_fetch_array($getpart, MYSQLI_NUM);
$deletedpart = $getpart_row[0];
$removepart = mysqli_query($con, "DELETE FROM $storage WHERE id = '$send_parts'");
echo "Part " . $deletedpart . " has been transfered";
} else {
echo "Part " . $row['part_type'] . "was NOT transfered";
};
};
} exit ();
} else {
echo "Nothing was selected, please try again!";
}
Your <form> doesn't have an id attribute on it so you'll either need to add id="send_parts" to the <form> or you'll need to change your code from getElementById to getElementsByName like this:
// Now get the value from page and pass it to server script.
// Serialize the data
var formData = new FormData( document.getElementsByName("send_parts")[0] );
ajaxRequest.open('POST', "../includes/sendtozg.php", true);
ajaxRequest.send(formData);
Then inside sendtozg.php you'll need to change the first two lines to:
if (!empty($_POST)){
foreach ($_POST as $send_parts){
This is the final code for the sendtozg.js:
// Get get the value from page and pass it to server script.
var formData = new FormData( document.getElementsByName("send_parts")[0] );
ajaxRequest.open('POST', "../includes/sendtozg.php", true);
ajaxRequest.setRequestHeader("X-Requested-With", "XMLHttpRequest");
ajaxRequest.send(formData);
}
and sendtozg.php should be:
if (!empty($_POST)){
foreach ($_POST['send_parts'] as $send_parts){
$getchecked = mysqli_query($con, "SELECT * FROM $storage WHERE id=$send_parts");
By the way:
print_r ($some_array)
and
if (!$check1_res) {
printf("Error: %s\n", mysqli_error($con));
exit();
};
Are great tools for troubleshooting.

Improving a star rating system

I made a star rating system for a school project and need some help improving it a bit. It's a fairly simple setup: index.php contains list-items (content is fetched from a DB) with an image and a div holding the stars for rating. Each star is a link that triggers a function which saves the rating in a database.
Here's the link! for starters.
If you click on any star, the first click will result in a green check mark on the first list-item. The idea is that this check mark will appear on each list-item when rated. That's where I need you guys to point me in the right direction. First of all I know I can't echo out multiple divs with the same id, but I had to in order for the ajax function to work (document.getElementById("rated")). Any ideas on how to solve this?
CODE
insert_receive.php:
<?php
session_start();
$sess = session_id();
$mysqli = new mysqli("", "", "", "");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$stmt = $mysqli->prepare("REPLACE INTO Ratings (projId_fkey, sessionId, rvalue) VALUES ('".$_GET['projId']."','".$sess."','".$_GET['rating']."')");
$stmt->execute();
printf("✔");
?>
ajax_framework.js:
function saveClick(rating,projId)
{
var xmlhttp;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("rated").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","insert_receive.php?rating=" + rating + "&projId=" + projId,true);
xmlhttp.send("");
}
index.php: (the part that matters)
<?php
$select = "SELECT id, projName, location FROM Projects";
if($result = $mysqli->query($select))
{
while($row = $result->fetch_assoc())
{
echo '<li id="'.$row['id'].'">';
echo '<h1 class="header">'.$row['projName']."</h1>";
echo '<img src="'.$row['location'].'" alt=""/>';
echo '<div class="rating">';
echo ''.★★★★★."";
echo ''.★★★★."";
echo ''.★★★."";
echo ''.★★."";
echo ''.★."";
echo '<div id="rated">'.""."</div>";
echo "</div>";
echo "</li>";
}
}
?>
You can use an iterator to give each element a different ID, after that, you can indeed dynamically get the element by the correct ID. For example:
echo '<div id="rated'.$row['id'].'">
and then:
document.getElementById("rated"+projId).innerHTML=xmlhttp.responseText;
This will dynamically select the element you want.
Regarding your code in general, I simply must point you in the direction of http://jquery.com/
It will make element selection much simpler, and will also normalize your use of ajax.
Change your php to:
while($row = $result->fetch_assoc())
{
echo '<li id="'.$row['id'].'">';
echo '<h1 class="header">'.$row['projName']."</h1>";
echo '<img src="'.$row['location'].'" alt=""/>';
echo '<div class="rating">';
echo ''.★★★★★."";
echo ''.★★★★."";
echo ''.★★★."";
echo ''.★★."";
echo ''.★."";
echo '<div id="rated'.$row['id'].'">'.""."</div>";//updated this line
echo "</div>";
echo "</li>";
}
And the line in your javascript to:
document.getElementById("rated"+projId).innerHTML=xmlhttp.responseText;
Use a counter$id to increment li id
<?php
$select = "SELECT id, projName, location FROM Projects";
if($result = $mysqli->query($select))
{
$id = 0;
while($row = $result->fetch_assoc())
{
echo '<li id="'.$id.'">';
echo '<h1 class="header">'.$row['projName']."</h1>";
echo '<img src="'.$row['location'].'" alt=""/>';
echo '<div class="rating">';
echo ''.★★★★★."";
echo ''.★★★★."";
echo ''.★★★."";
echo ''.★★."";
echo ''.★."";
echo '<div id="rated">'.""."</div>";
echo "</div>";
echo "</li>";
$id++;
}
}
?>
First, use a class instead, obviously.
echo '<div class="rated">'.""."</div>";
Second, in your JavaScript, instead of this line
document.getElementById("rated").innerHTML=xmlhttp.responseText;
Use this
var proj = document.getElementById(projId);
var rated = proj.getElementsByClassName('rated')[0];
rated.innerHTML=xmlhttp.responseText;
What we're doing here is that we find the element with an id matching projId. Then we find the elements with the class rated INSIDE the projID we found earlier. Now notice that getElementsByClassName returns an array regardless of the number of elements, but since we only have a single one here we just select the item at the first index [0].

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