Passing values from a php page to a modal from other page - php

I'm trying to edit a post using this snippiet of from my query.php page:
echo '<button type="submit" onclick="openModal(this)" id="btn-edit"
name="edit" value='.$postID.'></button>';
function openModal(id)
{
var editpost = id.value;
$.ajax({
url:"query.php",
method:"POST",
data:{ editpost : editpost },
success:function(data)
{ //
},
error: function () {//
}
});
$('#edit').modal('show');
}
The function to get the data from the post selected is on the same php page. This is the code:
if(isset($_POST['editpost']))
{
session_start();
$editpostid = $_POST['editpost'];
if($editpostid != "")
{
$sql = "SELECT * FROM post WHERE PostId = '" . $editpostid . "'";
}
if($sql != "")
{
$qry = mysqli_query($connection, $sql);
if (mysqli_num_rows($qry) > 0)
{
foreach($qry as $row)
{
$_SESSION['editpostdesc'] = $row['PostDesc'];
$_SESSION['editpostfile'] = $row['PostFile'];
$_SESSION['editpostid'] = $row['PostId'];
}
}
}
}
But the modal is on the other page (main page). I want to get the data of the post and display it on the modal, so I tried using $_SESSION. Yes, I got the data and was able to display it on the modal, but the problem is when I try to edit another post, the first value assigned to the session cannot be replaced. Is there is any other way that I can pass the values without using session? I'm really running out of ideas, I'm just starting my first web project.

if you get multiple row means this methods will worked...
if(isset($_POST['editpost']))
{
session_start();
$editpostid = $_POST['editpost'];
if($editpostid != "")
{
$sql = "SELECT * FROM post WHERE PostId = '" . $editpostid . "'";
}
if($sql != "")
{
$qry = mysqli_query($connection, $sql);
if (mysqli_num_rows($qry) > 0)
{
foreach($qry as $row)
{
$res[] = array(
'editpostdesc' => $row['PostDesc'],
'editpostfile' => $row['PostFile'],
'editpostid' => $row['PostId']
);
}
}
}
}

Related

How do I display different table from database on one page using scroll function

Right now, I am working on a project where I need to display data from different tables on one page, one after another using scroll function
For this task I am using ajax, php, mysql.
So here is what I have managed so far:
PHP code:
<?php
$conn = mysqli_connect("127.0.0.1", "Got", "nokia", "myddb");
$feedb = mysqli_query($conn,"SELECT * FROM feedb");
$feedb_count=mysqli_num_rows($feedb);
$photo = mysqli_query($conn,"SELECT * FROM photob");
$photo_count=mysqli_num_rows($photo);
$limito=$_POST["limit"];
$pagefeed=$_POST["start"];
$pagephoto=$_POST["startphoto"];
$pagetexto=$_POST["start3"];
$tumb=$_POST["change"];
if($tumb==0)
{
$query = "SELECT * FROM feedb ORDER BY cid DESC LIMIT $pagefeed , $limito ";
$result = mysqli_query($conn, $query);
while($row = mysqli_fetch_array($result))
{
echo "
<div class=\"block\">
<img class=\"imago\"src=\"img/$row[cphoto].jpg\">
<span class=\"namo\">$row[cname]</span><br>
<span class=\"texto\">$row[ctext]</span>
</div>"."Page".$pagefeed."Func".$tumb;
}
}
if($tumb==1)
//echo "<br>"."Page".$pagefeed."Func".$tumb;
{
$query = "SELECT * FROM textb ORDER BY cid DESC LIMIT $pagephoto , $limito ";
$result = mysqli_query($conn, $query);
while($row = mysqli_fetch_array($result))
{
echo "
<div class=\"block\">
<span class=\"namo\">$row[cname]</span><br>
<span class=\"texto\">$row[ctext]</span>
</div>";
}
}
Javascript code:
<script>
$(document).ready(function(){
var limit = 7;
var start = 35;
var start2 = 0;
var start3 = 0;
var action = 'inactive';
var change = 0;
var ncount =0;
if(active='inactive'){
action='active';
load_country_data();
}
function load_country_data()
{
$.ajax({
url:"resout.php",
method:"POST",
data:{limit:limit,start:start,change:change,startphoto:start2},
cache:false,
success:function(dataz)
{
$('#load_data').append(dataz);
if(dataz == '')
{
action = 'inactive';
change++;
if(change==2){action = 'active'; $('#load_data_message').html('Больше нету данных'); }
}
else
{
$('#load_data_message').html("<button type='button' class='btn btn-warning'>Please Wait....</button>");
action = "inactive";
}
}
});
}
$(window).scroll(function(){
if(($(window).scrollTop() + $(window).height())==$(document).height() && action == 'inactive' && change==0)
{
action = 'active';
start=start+limit; setTimeout(function(){load_country_data();}, 300);
}
else if(($(window).scrollTop() + $(window).height())==$(document).height() && action == 'inactive' && change==1)
{
action = 'active';
setTimeout(function(){load_country_data();}, 300);
start2=((ncount-1)*limit);ncount++;
}
});
});
</script>
but after displaying one base, when it start to dislay second it has an error
mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given
right after error it displays the second base properly.
I think this problem is because of a method that I chose for displaying second base "star2= (ncount-1)*limit ;ncount++", but I didn't see a different solution.
How can I fix this problem or how can I display different
table in different way?

Getting Ajax reponse as null in php

I have a form in that I am trying to do inline editing and adding using AJAX call.
Firstly I am displaying data in HTML table. And then if enter data into text boxes and click on add button record adding displaying data in HTML table. After I click edit button data showing in the textboxes fine.
But I am getting the ajax response as null.
I couldn't figure it out.
This is my AJAX code PHP file:
$(function() {
$(".scrollingTable tbody a").click(function() {
//debugger;
var link = $(this).attr('href');
var arr = link.split('=');
var id = arr[1];
//alert(id);
$.ajax({
url: "insertgr.php",
type: "POST",
data: {
cntid: id
},
success: function(datas) {
var data = $.parseJSON(datas);
$("#num").val(data.id);
$("#namegr").val(data.vndr_cntname);
$("#designation").val(data.designation);
$("#mobilegr").val(data.vndr_cntmobile);
$("#maildgr").val(data.vndr_cntmail);
}
});
});
});
$(function() {
$('.txtcbt a').click(function() {
debugger;
var cntname, designation, mobile, email, vndrid, id, cid;
cid = $("#num").val();
cntname = $("#namegr").val();
designation = $("#designation").val();
mobile = $("#mobilegr").val();
email = $("#maildgr").val();
vndrid = "<?php echo $selectid; ?>";
//alert(cid);
if (cntname == "" || designation == "" || mobile == "" || email == "") {
alert("fields should not be empty");
} else {
$.ajax({
url: "insertgr.php",
type: "POST",
data: {
id: cid,
name: cntname,
dgnation: designation,
mobileno: mobile,
emailid: email,
vid: vndrid
},
success: function(html) {
var dat = $.parseJSON(html);
alert(html);
alert("it came to success");
$("#num").val("");
$('#namegr').val("");
$('#designation').val("");
$('#mobilegr').val("");
$('#maildgr').val("");
}
});
}
});
});
This is file AJAX is calling:
<?php
require('Assests/connection/connection.php');
error_reporting(0);
$vcntlist = "";
if (!empty($_POST['cntid'])) {
$id = $_POST['cntid'];
$result = mysqli_query($conn, "SELECT `id`, `vndr_cntname`, `designation`, `vndr_cntmobile`, `vndr_cntmail`,`vndr_id` FROM `vndr_cntdtls`where id=$id");
$rowcount = mysqli_num_rows($result);
if ($rowcount > 0) {
$row = mysqli_fetch_array($result);
$vcntid = $row['id'];
$cntname = $row['vndr_cntname'];
$cntdesignation = $row['designation'];
$cntmobile = $row['vndr_cntmobile'];
$cntmail = $row['vndr_cntmail'];
}
}
if (!empty($_POST['name']) && !empty($_POST['dgnation']) &&
!empty($_POST['mobileno']) && !empty($_POST['emailid']) &&
!empty($_POST['vid'])) {
$id = $_POST['id'];
$name = $_POST['name'];
$degination = $_POST['dgnation'];
$mobile = $_POST['mobileno'];
$email = $_POST['emailid'];
$vndrid = $_POST['vid'];
if (empty($_POST['id'])) {
$query = mysqli_query($conn, "INSERT INTO `vndr_cntdtls`(`vndr_cntname`, `designation`, `vndr_cntmobile`, `vndr_cntmail`, `vndr_id`) VALUES ('$name','$degination','$mobile','$email',$vndrid)");
} else {
$update1 = mysqli_query($conn, "UPDATE `vndr_cntdtls` SET `vndr_cntname`='$name',`designation`='$degination',`vndr_cntmobile`='$mobile',`vndr_cntmail`='$email' WHERE id=$id") or die(mysqli_error($conn));
}
$result = mysqli_query($conn, "SELECT DISTINCT `id`, `vndr_cntname`, `designation`, `vndr_cntmobile`, `vndr_cntmail`,vc.vndr_id FROM `vndr_cntdtls` vc INNER JOIN vendors v ON vc.vndr_id=$vndrid") or die(mysqli_error($conn));
$rowcount = mysqli_num_rows($result);
if ($rowcount > 0) {
while ($row = mysqli_fetch_array($result)) {
$vcntid = $row['id'];
$cntname = $row['vndr_cntname'];
$cntdesignation = $row['designation'];
$cntmobile = $row['vndr_cntmobile'];
$cntmail = $row['vndr_cntmail'];
}
}
}
echo json_encode($row);
?>
There could be a lot of ways it is not working.
First:
You need to put a application/json in your php so it can be compatible with the browser you are using.
header('Content-Type: application/json');
echo json_encode($row);
Second:
Why are you doing a while loop and assigning it into an unused variable?
You can simplify it by doing:
$row = mysqli_fetch_array($result);
header('Content-Type: application/json');
echo json_encode($row);
Unless it is multiple rows then:
$rowcount = mysqli_num_rows($result);
$rows = array();
if ($rowcount > 0) {
while ($row = mysqli_fetch_array($result)) {
$rows[] = $row;
}
}
header('Content-Type: application/json');
echo json_encode($row);
Third
null values usually appears when a variable you are trying to use is not initialized. By having the error_reporting turned off it does not display the error.
error_reporting(true);
Fourth
Check also the logs for database error, I assume it has something to do with MySQL query not having to reach the $row initialization.
Fifth
I believe you need to have it fetched as associative array for it to be useful
$row = mysqli_fetch_assoc($result);

How to get a single mysql value and output it to an ajax call?

I'm trying to get a number from a mysql line then outputting it to ajax. the number can't be a string because I will multiply it in ajax. This is what i have so far. I'm not sure what to do from here.
ajax:
$(document).ready(function()
{
$("#btnCalc").click(function()
{
var user = $("#txtUser").val();
var amount = $("#txtAmount").val();
var category = $("txtCat").val();
var number = $("txtNum").val();
var result = '';
$.get("code/value.php",
{
ID:user,
amount:amount,
result:result
},function(query)
{
if ( user > 0 and user < 30 ){
alert(result);
}
else{
alert( 'invalid user ID');
}
});
});
});
php:
<?php
$userID = $_GET["ID"];
$amount = $_GET["amount"];
$category = $_GET["category"];
$num = $_GET["number"];
require "../code/connection.php";
$SQL = "select userAmount from user where userID= '$userID'";
$reply = $mysqli->query($SQL);
while($row = $reply->fetch_array() )
{
}
if($mysqli->affected_rows > 0){
$msg= "query successful";
}
else{
$msg= "error " . $mysqli->error;
}
$mysqli->close();
echo $msg;
?>
Pretty straightforward - you just grab the value from the row and cast it as a float.
while($row = $result->fetch_array() )
{
$msg = floatval($row['userAmount']);
}
if($msg > 0) {
echo $msg;
} else {
echo "error" . $mysqli->error;
}
$mysqli->close();
And one small change in your ajax call:
$.get("code/value.php",
{
ID:user,
amount:amount,
result:result
},function(query)
{
alert(query);
});
});
You need to add echo $row['userAmount']; inside or after your while loop, and drop the second echo. You should be able to take result within your AJAX code and use it as a number directly.
Here function(query), query is the response from the AJAX call. So your alert should be:
alert(query);
result is empty.
You also should be using prepared statements and outputting the value you want.
Something like:
<?php
$userID = $_GET["ID"];
$amount= $_GET["amount"];
require "../code/connect.php";
$SQL = "SELECT userAmount FROM user WHERE userID= ?";
$reply = $mysqli->prepare($SQL);
if($mysqli->execute(array($userID))) {
$row = $reply->fetch_array();
echo $row['amount'];
}
else
{
$msg = "error" . $mysqli->error;
}
$mysqli->close();
?>
Then JS:
$(document).ready(function()
{
$("#btnCalc").click(function()
{
var user = $("#txtUser").val();
var amount = $("#txtAmount").val();
var result = '';
$.get("code/value.php",
{
ID:user,
amount:amount,
result:result
},function(query)
{
alert(query);
});
});
});
You can use https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/parseFloat or https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/parseInt to convert the value to an integer/float in JS.

Query in Jquery IF statement

I have a jquery save script like :
naam = prompt('Give a name for your file.');
if(naam != null)
{
var div_contents = $("#print").html();
$.post("save.php", { 'contents': div_contents,'naam':naam });
alert('Your file is save as : '+ naam);
window.location.replace("index.php?id=latest");
}
else
{
alert('Not saved');
}
I save a div in save.php which creates an new id in the database
What I want to achive is were
window.location.replace("index.php?id=latest");
id=latest must become (id=id from last saved file).
I tried
$q = "select MAX(id) from Moodboards";
$result = mysql_query($q);
$data = mysql_fetch_array($result);
$MBId = $data[0];
window.location.replace("index.php?id="+MBId);
and
var MBID =
<?php
$q = "select MAX(id) from Moodboards";
$result = mysql_query($q);
$data = mysql_fetch_array($result);
$MBId = $data[0];
echo $MBId ?>
window.location.replace("index.php?id="+MBId);
They both failed.
How can I run the query in the if(naam !=null) statement?
At first place you must fix your jQuery POST... You don't use POST respond which is wrong.. You should wait for it and then continue with other actions
naam = prompt('Give a name for your file.');
if(naam != null)
{
var div_contents = $("#print").html();
$.post("save.php", { 'contents': div_contents,'naam':naam }, function(responde){
if(responde.id)
window.location.replace("http://yoururl.com/index.php?id="+responde.id);
else
alert("No responde...");
}, "json");
}
else
{
alert('Not saved');
}
For better results I suggest you to use JSON data in that post/respond..
At your PHP code you have to set:
<?php
$q = "select MAX(id) from Moodboards";
$result = mysql_query($q);
$data = mysql_fetch_array($result);
$MBId = $data[0];
echo json_encode(array('id'=>$MBId));
exit();
?>
P.S. For window.location.replace please set your FULL url: "http://localhost/index.php?id=" OR atleast put slash at start of it "/index.php?id="
Solution
if(naam != null)
{
var div_contents = $("#print").html();
$.post("save.php", { 'contents': div_contents,'naam':naam });
alert('Uw moodboard is opgeslagen als '+ naam);
window.location.replace("index.php?id=<?php $q = "select MAX(id) from Moodboards";
$result = mysql_query($q);
$data = mysql_fetch_array($result);
$MBId = ($data[0] + 1); echo "$MBId";?>");
}
This Works for me , i didnt need to make a jquery var i could echo the variable in php.
And i had to add 1 cause the sql query is loaded when the page is loaded.
So the file isn't saved yet when i get the highest id.

Using jQuery and php to pull data from database

I have a page that is pulling data through jQuery but it is only pulling the return code. Here is my code:
<script type='text/javascript' language='javascript'>
function showCU() {
$.post('getCU.php', {
cuid: $('#cuContact').val()
},
function (response) {
$('#contact').val(response).show();
$('#email').val(response).show();
$('#phone').val(response).show();
})
}
</script>
$select = "SELECT priContact, priEmail, priPhone FROM user WHERE id = '" . $_POST['id'] . "'";
$query = mysql_query($select) or die ("Could not get info: " . mysql_error());
if (mysql_num_rows($query) > 0) {
while ($get = mysql_fetch_array($query)) {
$priContact = $get['priContact'];
echo $priContact;
echo $get['priEmail'] . " | " . $get['priPhone'];
}
} else {
echo "No users";
}
So the call is pulling from getCU.php whenever the onchange event handler is called. That is why this is in a function. What I want to do is every time a user chooses something from the option list the text values change according to what was selected. I have the php page pulling from a db and echoing out the code correctly. jQuery id pulling the data from the php page correctly, but I cannot get the code to place the single details in each of the text boxes.
So what I want to happen is this:
A user selects a name from a drop-down box. Then the mysql data attached to that name would be displayed on the page in form text fields.
Let me know if more information or code is needed.
I think you'll be better off structuring your data. My general recommendation is JSON.
// QUICK WARNING: Don't take unparse GET/POST responses.
// This is asking for trouble from SQL injection.
$select = "SELECT priContact, priEmail, priPhone FROM user WHERE id = '" . mysql_escape_string($_POST['id']) . "'";
$query = mysql_query($select) or die ("Could not get info: " . mysql_error());
$retVal = array();
if (mysql_num_rows($query) > 0) {
$retVal['data'] = array();
while ($get = mysql_fetch_array($query))
{
$retVal['data'][] = $get;
}
} else {
$retVal['error'] = 'No users';
}
header('Content-type: application/json');
echo json_encode($retVal);
Javascript:
<script type="text/javascript">
function showCU() {
$.post('getCU.php', {
cuid: $('#cuContact').val(),
dataType:'json'
},
function (response) {
if (response.error) {
//handle error
}
else
{
$('#contact').val(response.data.priContact).show();
$('#email').val(response.data.priEmail).show();
$('#phone').val(response.data.priPhone).show();
}
})
}
</script>

Categories