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!
Related
The code is supposed to fetch the query result and display it according to the number written in the text box, i tried (onkeydown and onkeyup) and both didn't work, i have no idea why it is not working and what is my mistake.
HTML code:
<script>
function showRoom(rid) {
xmlhttp = new XMLHttpRequest();
xmlhttp.open("GET", "getRoomAJAX.php?q="+rid, true);
xmlhttp.send();
xmlhttp.onreadystatechange = function () {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
}
}
}
</script>
<form>
<input type="text" name="rid" onkeydown="showRoom(this.value)">
</form>
<div id="txtHint">Room...</div>
getRoomAJAX.php code:
<?php
session_start();
ob_start();
$q = $_GET["q"];
$db = new Database();
$dbc = $db->getConnection();
if (!$dbc) {
die('Could not connect: ' . mysql_error());
}
$query = "SELECT * FROM indvProj_room WHERE rid = '".$q."'";
$result = mysqli_query($dbc, $query);
//start creating the table HTML
if ($result) {
echo "<table border='1' align='center' cellspacing = '2' cellpadding = '4' width='100%'>
<tr>
<th><b>Room ID</b></th>
<th><b>Room Description</b></th>
</tr>";
while ($row = mysqli_fetch_array($result, MYSQLI_BOTH)) {
echo "<tr>";;
echo "<td>" . $_SESSION['adminRoomChoice'] = $row['rid'] . "</td>";
echo "<td>" . $row['roomDesc'] . "</td>";
echo "</tr>";
}
echo "</table>";
} else {
echo '<p class="error">Sorry, cannot find the room, are you sure of the entered Room ID?</p>';
echo '<p class = "error">' . mysqli_error($dbc) . '</p>';
}
?>
Does showRoom() get called at all? Put a console.log(rid); inside showRoom() to see if that is getting called and whether the value is being passed in correctly.
If you can determine that it's getting into the showRoom(), then follow the code down and into PHP to see where it's failing. Echo some sample text at the top of the PHP file with return; right below it. That will tell you if the error is in the XMLHttpRequest code or somewhere in the PHP file.
Normally, I don't pass any variables/params with onkeydown type of events. Usually, I call a method like this without params and then retrieve the value based on the element's id. See this answer for more detail on that: https://stackoverflow.com/a/54040431/3103434
I have the following drop-down list that fetches a list coalitions from the database (this list is populated as expected)
<form style = "display: inline-block;" >
<select class="form-control" name ="coalition_select" onchange = "showCandidates(this.value)" method="GET">
<option id = "coalition_id" value="coalition">Coalitions</option>
<?php
include_once 'connection.php';
$sql_coalition = mysqli_query($conn, "SELECT coalition FROM candidates");
while ($row = $sql_coalition->fetch_assoc()) {
echo "<option value=\"coalition\">" . $row['coalition'] . "</option>";
}
?>
</select>
</form>
The problem begins here. Here I'm trying to first get the selected value (which is on of the coalitions) from the drop-down list and second use the value to display users with similar coalition attribute.
here is the script to get the value from the drop-down:
<script>
function showCandidates (str) {
if (str.length == "") {
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","includes/admin.users.list.inc.php?q="+str,true);
xmlhttp.send();
}
}
</script>
and here is the admin.users.list.inc.php file
<body>
<?php
$q = $_GET['q'];
$conn = mysqli_connect('localhost','root','','voting');
if (!$conn) {
die('Could not connect: ' . mysqli_error($conn));
}
mysqli_select_db($conn,"osako_Voting");
$sql="SELECT * FROM candidates WHERE coalition = '".$q."'";
$result = mysqli_query($conn,$sql);
echo "<table>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Coalition</th>
<th>Email</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['firstname'] . "</td>";
echo "<td>" . $row['lastname'] . "</td>";
echo "<td>" . $row['coalition'] . "</td>";
echo "<td>" . $row['email'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($conn);
?>
</body>
The problem seem to be that I'm unable to set the variable $q so that it captures the selected value. right now as it is set it seems to capture the index instead of the value itself. How can this be correctly done? If it is of any help, I'm using this tutorial as a guideline
https://www.w3schools.com/php/php_ajax_database.asp
IN SUMMERY:
If we have a drop-down list that has been populated dynamically using php script. How can be get the selected value using ajax and use the said value in another php script.
Thanks
It seems the problem is in this row:
echo "<option value=\"coalition\">" . $row['coalition'] . "</option>";
You're passing the select value to showCandidates(), to be passed via AJAX but you're setting the option value to be always static "coalition" and not the dynamic values you're fetching from DB.
Maybe you should change the row to
echo "<option value=\"" . $row['coalition'] . "\">" . $row['coalition'] . "</option>";
your not setting value to value attribute . your just setting constant string for all option values coalition
while ($row = $sql_coalition->fetch_assoc()) {
echo "<option value=\"$row['coalition']\">" . $row['coalition'] . "</option>";
}
I know you have accepted an answer but i am puzzled as to this;
in your function you have this
document.getElementById("txtHint").innerHTML = "";
But you don't have any form element with the id "txtHint"
So what element is being sort with that ID?
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>";
I'm trying to retrieve data from multiple table. So when the user clicks and select an item php file to retrieve the data from the corresponding table.
this is my first try on Ajax used w3 school code and trying to modify, I guess the way I'm using if condition is the problem? Before I try with the multiple table it worked.
My Script
<script>
function showUser(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML = "Please Select type of users to view";
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 (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
}
}
xmlhttp.open("GET","get_users.php?q="+str,true);
xmlhttp.send();
}
}
</script>
My Form
<form>
<select name="users" onchange="showUser(this.value)">
<option value="">Select a person:</option>
<option value="1">Staff</option>
<option value="2">Admin</option>
<option value="3">Customers</option>
</select>
</form>
My PHP Page
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('localhost','X','X','X');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con);
if ($q='Admin')
{
$sql="SELECT * FROM admin";
$result = mysqli_query($con,$sql);
echo "<table>
<tr>
<th>Email</th>
<th>Name</th>
<th>Mobile</th>
<th>Address</th>
<th>Password</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['email'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['mobile'] . "</td>";
echo "<td>" . $row['address'] . "</td>";
echo "<td>" . $row['password'] . "</td>";
echo "</tr>";
}
echo "</table>";
else if ($q='Staff')
{
echo "This is from Database B
}
else
{
echo "This is from database C";
}
mysqli_close($con);
?>
You're using $q = intval($_GET['q']); with a string value for your assignments.
intval() "Get the integer value of a variable"
<option value="1">Staff</option>
<option value="2">Admin</option>
<option value="3">Customers</option>
and using an assignment rather than a comparison for both if ($q='Admin') and else if ($q='Staff') both being strings, as opposed to integers in your options.
which should read as and with 2 equal signs if ($q=='2') and else if ($q=='1') in order to match your select's numerical value options.
Or, change them accordingly with your options to read as Staff and Admin (and Customers) in the values, while removing the intval() from the GET array; the choice is yours.
You also don't need mysqli_select_db($con); since you've declared your 4 parameters in your initial connection and would fail for another reason; technically you didn't select a database for it.
You're also missing a quote and semi-colon in (a parse syntax error)
echo "This is from Database B
which should read as
echo "This is from Database B";
Footnotes:
The page which I believe you based yourself on, seems to be http://www.w3schools.com/php/php_ajax_database.asp
If so, then I suggest you follow their example completely and modify it slowly to fit your needs. I myself have used that demo (in the distant past) before with success.
Their example uses a WHERE clause also.
$sql="SELECT * FROM user WHERE id = '".$q."'";
Edit: Just for the record, there is a missing brace for this condition:
if ($q='Admin')
{
I have an php generated table/form with checkboxes like this:
if ($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.