Ajax call with javascript not working

Question

I want to try to get some value from page cart.php from and ajax call, but its not working. Ajax code is below:

function temp_sell(id) {
    //var p_id=document.getElementById('temp').value;
    alert(p_id);
    $.ajax({
        type: "POST",
        url: "temp_sell.php",
        data: "value=" + p_id,
        success: function (message) {
            alert(message);
            $("#your_cart").html(message);
        }
    });
}

temp_sell.php is below where I want to show some products details

<?php 
include("connection.php");
echo $p_id = $_POST['value'];
$qty=1;
$query="SELECT * FROM product WHERE id='$p_id'";
    $result=mysql_query($query);
    while($data=mysql_fetch_array($result))
    {
    ?>
            <form>
            <table>
                <tr>
                    <img src="<?php echo "img/".$data['image'];?>"/>
                    <strong><?php echo $data['pro_name'];?></strong>
                    <strong><?php echo $data['price'];?></strong>
                    <input type="text" value="<?php echo $qty;?>"/>
                </tr>
            </table>
            </form>
    <?php
    }                   
?>

p_id is undefined, you have commented the p_id value or change temp_sell(p_id) — Krish R
– Krish R, Commented Jan 10, 2014 at 9:35

Community · Accepted Answer · 2017-05-23 12:13:07Z

1

p_id is undefined, you have commented the p_id value or change temp_sell(p_id).

function temp_sell(p_id)

instead of

function temp_sell(id)

Reference comments : Ajax call with javascript not working

edited May 23, 2017 at 12:13

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answered Jan 10, 2014 at 9:38

Krish R

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3 Comments

Forhad Sikder Over a year ago

yes i correct it but still not working i can't get anything from temp_sell.php

Elias Van Ootegem Over a year ago

@ForhadSikder: because, as the 2 other answers explain, you're not fetching the resultset as an associative array (while($data=mysql_fetch_array($result)) should be while($data=mysql_fetch_assoc($result)))

Krish R Over a year ago

did you get the alert when you call that function. Make sure your jquery libaray path is correct. Checck your browser console, it maythrough some errors if any

Alireza Fallah · Accepted Answer · 2014-01-10 09:59:39Z

0

Do this :

function temp_sell(id){
var p_id=document.getElementById('temp').value; // <--- uncomment this line
alert(p_id);
$.ajax({
         type: "POST", 
         url: "temp_sell.php",
         data: {value:p_id}, // <--- change this
         success: function(message){ 
         alert(message);
         $("#your_cart").html(message);
        }       
       });
   }

edited Jan 10, 2014 at 9:59

answered Jan 10, 2014 at 9:41

Alireza Fallah

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5 Comments

Krish R Over a year ago

mysql_fetch_array — Fetch a result row as an associative array, a numeric array, or both

Alireza Fallah Over a year ago

thank you, I did not know it, but user is still wrong, because MYSQL_ASSOC) should be as the second argument of mysql_fetch_array, to return a associative array

Krish R Over a year ago

By default, it fetches both, but you can pass second parametr based on your need.

Elias Van Ootegem Over a year ago

@KrishR: either way, sure the fetching works using the default behaviour of mysql_fetch_array, but it won't work for long... once the ancient extension is finally removed, the code won't work anyway. In fact, it will already issue E_DEPRECATED notices on up-to-date systems

Alireza Fallah Over a year ago

I updated my answer, thank you @KrishR, after 3 years of php programming, I didn't know that :)

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