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How to pass PHP variable to AJAX URL – Jquery ajax

Hello guys I am new to javascript
I am trying to send php variable to AJAX url file but i was unable to do it. I don’t know where the problem actually arise. your help will be highly appreciated

Here i want to send the below PHP Variable "$cheque" to the AJAX URL Page cheque_select.php

<php? $cheque = '78964Y' ?>
$(document).ready(function(){  
    
    function fetch_data()
    {
        $.ajax({
            url:"cheque_select.php",
            method:"POST",
            dataType:"json",
            
            success:function(data)
            {

                var html = '';
                for(var count = 0; count < data.length; count++)
                {
                    html += '<tr>';
                    html += '<td><input type="checkbox" id="'+data[count].id+'" data-cheque_no="'+data[count].cheque_no+'" data-id="'+data[count].id+'" data-name="'+data[count].name+'" data-sum="'+data[count].sum+'" data-account="'+data[count].account+'" class="check_box"  /></td>';
                    html += '<td>'+data[count].cheque_no+'</td>';
                    html += '<td>'+data[count].id+'</td>';
                    html += '<td>'+data[count].name+'</td>';
                    html += '<td>'+data[count].sum+'</td>';
                    html += '<td>'+data[count].account+'</td></tr>';
                }
                $('tbody').html(html);
            }
        });
    }

    fetch_data();

This is my cheque_select.php i want to fetch data from mysql by the above variable

<?php


include('connection.php');


 


$query = "select   * FROM entry where entry.bank= $cheque";

$statement = $connect->prepare($query);

    
    
    
if($statement->execute())
{
 while($row = $statement->fetch(PDO::FETCH_ASSOC))
 {
  $data[] = $row;
 }

 echo json_encode($data);
}

?>
Tags:

2

Answers


  1. 0

    Add data in your AJAX request like –

    <?php $cheque = '78964Y' ?>
$(document).ready(function(){  

 function fetch_data()
 {
    $.ajax({
        url:"cheque_select.php",
        method:"POST",
        dataType:"json",
        data:{'cheque': "<?php echo $cheque; ?>"},
        success:function(data)
        {

            var html = '';
            for(var count = 0; count < data.length; count++)
            {
                html += '<tr>';
                html += '<td><input type="checkbox" id="'+data[count].id+'" data-cheque_no="'+data[count].cheque_no+'" data-id="'+data[count].id+'" data-name="'+data[count].name+'" data-sum="'+data[count].sum+'" data-account="'+data[count].account+'" class="check_box"  /></td>';
                html += '<td>'+data[count].cheque_no+'</td>';
                html += '<td>'+data[count].id+'</td>';
                html += '<td>'+data[count].name+'</td>';
                html += '<td>'+data[count].sum+'</td>';
                html += '<td>'+data[count].account+'</td></tr>';
            }
            $('tbody').html(html);
        }
    });
}

fetch_data();

    It will send the cheque value to the server end.

    And get it at the PHP end just before the query by adding –

    <?php
include 'connection.php';
$cheque = $_POST['cheque'];
$query = "SELECT * FROM entry WHERE entry.bank = ?"; 
$statement = $connect->prepare($query);
$statement->execute([$cheque]);
$data = $statement->fetchAll(PDO::FETCH_ASSOC);
echo json_encode($data);
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  2. 0

    Simply add a "data" value in the ajax/jquery request you are firing. This will send data in the form of a POST value to the page that is receiving your ajax request. This is the revised "fetch_data" function after adding the data you wish to send:

    function fetch_data()
    {
        $.ajax({
            url:"cheque_select.php",
            method:"POST",
            dataType:"json",
            data:{"cheque":"<?= $cheque ?>"},
            success:function(data)
            {

                var html = '';
                for(var count = 0; count < data.length; count++)
                {
                    html += '<tr>';
                    html += '<td><input type="checkbox" id="'+data[count].id+'" data-cheque_no="'+data[count].cheque_no+'" data-id="'+data[count].id+'" data-name="'+data[count].name+'" data-sum="'+data[count].sum+'" data-account="'+data[count].account+'" class="check_box"  /></td>';
                    html += '<td>'+data[count].cheque_no+'</td>';
                    html += '<td>'+data[count].id+'</td>';
                    html += '<td>'+data[count].name+'</td>';
                    html += '<td>'+data[count].sum+'</td>';
                    html += '<td>'+data[count].account+'</td></tr>';
                }
                $('tbody').html(html);
            }
        });
    }

    Then on the page that is receiving the request, you would get the cheque value by saying php $_POST['cheque'] . Also one mistake I noticed is you have mistyped your opening PHP tag. You have wrote <php? . The correct way is <?php .

    One more thing – you haven’t prepared your statement correctly in the PHP page. This is how you should prepare it:

    $pre_stmt = "SELECT * FROM entry WHERE entry.bank=?";
$stmt = $conn->prepare($pre_stmt);
$stmt->bind_param(
    "i", # if cheque is a numerical value keep it "i" otherwise change it to "s" if it is a string
    $cheque
);
$stmt->execute();
#do whatever success condition

    Preparing statements like this prevents SQL injection

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