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I have a 2 ajax call. First ajax call populates the dropdown box which works correctly, bu the second ajax call uses change event function. Whenever the dynamically populated item from the dropdown is clicked it lists the products details with image.

In below code the change function triggers becos it alerts the data item selected. but the problem starts inside the ajax i guess. so i am not getting the result(product details)

Problem: it donot list the products

also how to use var_dump($_POST) to check name1 is passed correctly

second ajax code:

<script src="https://code.jquery.com/jquery-3.2.1.min.js"></script>
<script>
$("#name").on('change',function (e) { 
    var name1 = this.value;
    alert("name1 = " + name1);
    $.ajax ({
        data:{name1: name1},
        type: 'POST',
        url: 'dataprod.php',
        success: function (response) {
            console.log(response);

            $('.products-wrp').html('');
            $('.products-wrp').hide();
            $('.products-wrp').html(response);
            $('.products-wrp').show();            
        },
    });
});
</script>

dataprod.php

<?php
$name1 = $_POST['name1'];
$results = $mysqli_conn->query("SELECT product_name, product_desc, product_code,  
product_image, product_price FROM products_list where product_name='$name1'");

$products_list =  '<ul id ="products_list" class="products-wrp">';
while($row = $results->fetch_assoc()) {
$products_list .= <<<EOT
<li>
<form class="form-item">
<h4>{$row["product_name"]}</h4>
<div>
<img src="images/{$row["product_image"]}" height="62" width="62">
</div>
<div>Price : {$currency} {$row["product_price"]}<div>
</form>
</li>
EOT;
}
$products_list .= '</ul></div>';
echo $products_list;
?>
Lilly Meow
  • 45
  • 6
  • Your code is vulnerable to [**SQL injection**](https://en.wikipedia.org/wiki/SQL_injection) attacks. You should use [**mysqli**](https://secure.php.net/manual/en/mysqli.prepare.php) or [**PDO**](https://secure.php.net/manual/en/pdo.prepared-statements.php) prepared statements with bound parameters as described in [**this post**](https://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php). – Alex Howansky Jun 02 '17 at 16:50
  • 2
    Didn't you post this already? https://stackoverflow.com/questions/44327551/change-event-not-firing-on-ajax-loaded-elements – Funk Forty Niner Jun 02 '17 at 16:51
  • You should add `error: function(e) { alert(e) }` to your ajax call. – Chris G Jun 02 '17 at 16:51
  • It's ironic that the "Try this" first posted answer got an upvote, and the other "try this" with more code got downvoted (none of them were mine, I might add) and the other one also contains `'/dataprod.php'` with the slash. – Funk Forty Niner Jun 02 '17 at 16:52

1 Answers1

-1

Try this code:

...
url: '/dataprod.php'
...
Nikita Goncharov
  • 196
  • 6