ajax not returning value from the mysql database

Question

I am hoping to be able to use .post method to send variables to a php file where I could be able to retrieve data from database with mysql query. So far I can send information about the variable using ajax but not able to retrieve output from the php script using ajax. I am very immature to ajax, hence learning through errors...Thanks for looking into my query.

How can I get to display php output within the div tag of index.php

test.php

   <?php require('../config/connection.php'); ?>
        <?php
        $value = $_POST['value'];
        $query = mysqli_query($dbc, "SELECT DISTINCT class,product FROM prdct_categories WHERE class = '$value'");
        while($list = mysqli_fetch_assoc($query)){
        $prdct = $list['product'];
        echo $prdct;    
        }
   ?>

ajax code (index.php)

    <div class="col-md-2" >

         <?php

        $test  = $_GET['product'];
        $q = "SELECT * FROM prdct_categories WHERE product = '$test' ";
        $r = mysqli_query($dbc, $q);
        $path_info = get_path();
        $test1 = $path_info['call_parts'][1];

       While($list = mysqli_fetch_assoc($r)) {?>
      <li class="nav" <?php if($test1==$list['slugs']){echo'id="actives"';} ?>>

<a href="<?php echo $test;?>/<?php echo $list['slugs'];?> ">

<?php echo $list['subgroup']."(".$list['contains'].")".'<br/>';?></a></li>

                 <?php }?>   
                 </div>

                <div class="col-md-4" id="testing">

                </div>

                <script>
                    $(document).ready(function(){

                        $(".nav").click(function(){

                        $.post("test.php", {value:$(this).text()}, function(data) {$("#testing").text(data)} );

                        event.preventDefault();
                        });
                    });
                </script>

Your script is at risk for SQL Injection Attacks. Learn about prepared statements for MySQLi. — Jay Blanchard
– Jay Blanchard, Commented Feb 15, 2016 at 14:27
Have you watched the request / response in your browser's developer tools? — Jay Blanchard
– Jay Blanchard, Commented Feb 15, 2016 at 14:28
php.net/manual/en/function.error-reporting.php - php.net/manual/en/mysqli.error.php and make sure you've loaded the jQuery library. — Funk Forty Niner
– Funk Forty Niner, Commented Feb 15, 2016 at 14:31

PeterKA · Accepted Answer · 2016-02-15 15:19:17Z

1

Please use this code instead. You do have to prevent the default action of the links that you're clicking within the .nav elements:

    $(".nav > a").click(function(e){ // see change here
            e.preventDefault();   // added here e must match var e above.
            $.post("test.php", {value:$(this).text()}, function(data) {$("#testing").text(data)} );
    });

And your .nav script should be:

<li class="nav<?php if($pageid==$nav['product']){echo ' active';}?>"> 
....

Your current script does not create any elements with class="nav" and therefore, the selector .nav returns nothing. Your current script also creates several elements with the same id, id="nav", which makes your HTML invalid.. Make the above changes and

edited Feb 15, 2016 at 15:19

answered Feb 15, 2016 at 15:12

PeterKA

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2 Comments

Gautam P Behera Over a year ago

I am getting this notice but not the output from database also apology about the confusion. Please check my updated nav bit... Notice: Array to string conversion in C:\xampp\htdocs\series\dynamics\admin\CleanURL\test.php on line 11 Array

PeterKA Over a year ago

In effect you're confirming that the above changes indeed do work, in that you're now getting a response from test.php, only that the php script needs to be fixed, right? If this is so, would it be better to ask a new PHP question?

Maxi Schvindt · Accepted Answer · 2016-02-15 17:03:16Z

0

Form to return data from ajax.

<?php require('../config/connection.php'); 
      header('Content-Type: application/json');

        $value = $_POST['value'];
        $prdct = array();
        $query = mysqli_query($dbc, "SELECT DISTINCT product FROM prdct_categories WHERE class LIKE '{$value}'");
        while($list = mysqli_fetch_assoc($query)){
        $prdct[] = $list['product'];
        }
        $return = json_encode($prdct);
        echo $return;
        die;
   ?>

In javascript replace:

$.post("test.php", {value:$(this).text()}, function(data) {$("#testing").text(data)} );

to

$.post("test.php", {value:$(this).text()}, function(data){
    for(var k in data) {
    $("#testing").append(data[k]+"<br />");
    }
});

edited Feb 15, 2016 at 17:03

answered Feb 15, 2016 at 14:47

Maxi Schvindt

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6 Comments

Maxi Schvindt Over a year ago

In your javascript, data is a list to results. Use console.log(data) and view in console firebug or console chrome to see the elements.

Gautam P Behera Over a year ago

this is the output [], with above changes

Maxi Schvindt Over a year ago

So, your query return empty.

Maxi Schvindt Over a year ago

try : "SELECT DISTINCT product FROM prdct_categories WHERE class LIKE '{$value}'"

Maxi Schvindt Over a year ago

And in your javascript, to set text o code in de tag, use html for example. I edit my answer.

|

Nana Partykar · Accepted Answer · 2016-02-15 17:06:48Z

0

Use $("#testing").html(data); instead of $("#testing").text(data);

<script>
    $(document).ready(function(){
        $(".nav").click(function(){
              $.post("test.php", {value:$(this).text()}, function(data) {
                        $("#testing").html(data);
                    });
              event.preventDefault();
        });
    });
</script>

answered Feb 15, 2016 at 17:06

Nana Partykar

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1 Comment

Gautam P Behera Over a year ago

using .tml doesn't make any difference to the output

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