Passing onclick Jquery variable issue in php

Question

I'm working on Core php, i'm on click of div data is coming and i store a variable and i want pass check with database query it it is matching will come all the records how to do this please help any one.

Html code:

<div class="col-lg-3 col-md-6  test" data-idtest="Diabetes">
    <a href="" class="box_cat_home">
        <i class="icon-info-4"></i>
        <img src="assets/img/icon_cat_3.svg" width="60" height="60" alt="">
        <h3>Diabetes</h3>
        <ul class="clearfix">
            <li><strong>124</strong>Doctors</li>
            <!-- <li><strong>60</strong>Clinics</li> -->
        </ul>
    </a>
</div>

Jquery code:

<script type="text/javascript">
    $(document).ready(function() { 
        $('.test').click(function (){
            var id = $(this).data('idtest') ; 
            //alert(id);
    $.ajax({
         url : "getting.php",
         type: 'POST',
         data    : {id:id },
          }).done(function(response) {
        });
      })
    });
</script>

Php code:

<?php
echo "ejejejej";
$dbHost = 'localhost';
$dbUsername = 'root';
$dbPassword = 'mysql';
$dbName = 'fre';

$id =$id;
$db = new mysqli($dbHost,$dbUsername,$dbPassword,$dbName);

      $men ="select * from tbl_users where doctor_speciality = $id";
      $men_result=$db->query($men);
      $projects=array();
      while($row=mysqli_fetch_assoc($men_result)){ 
      $projects[] = $row;

    }
?>

Also please look into [prepared statements](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php) to make your queries secure. — Jeto, Sep 28 '18 at 12:12

Shubham Azad · Answer 1 · 2018-09-28T12:54:08.997

0

Use this this code to get result in your ajax success.

<div class="col-lg-3 col-md-6  test" data-idtest="Diabetes">
    <a href="javascript:void(0);" class="box_cat_home">
        <i class="icon-info-4"></i>
        <img src="assets/img/icon_cat_3.svg" width="60" height="60" alt="">
        <h3>Diabetes</h3>
        <ul class="clearfix">
            <li><strong>124</strong>Doctors</li>
            <!-- <li><strong>60</strong>Clinics</li> -->
        </ul>
    </a>
</div> 


<?php

$dbHost = 'localhost';
$dbUsername = 'root';
$dbPassword = 'mysql';
$dbName = 'fre';

$id =$_REQUEST['id'];
$db = new mysqli($dbHost,$dbUsername,$dbPassword,$dbName);

      $men ="select * from tbl_users where doctor_speciality = $id";
      $men_result=$db->query($men);
      $projects=array();
      while($row=mysqli_fetch_assoc($men_result)){ 
      $projects[] = $row;

    }
    if($projects){
       $result=array(
       "responseCode" =>1,
       "message" =>"Data found",
       "data" => $projects
       );
    }else{
       $result=array(
       "responseCode" =>0,
       "message" =>"No data found regarding this id"
       );
    }
echo json_encode($result);die();
?>
<script type="text/javascript">
    $(document).ready(function() { 
        $('.test').click(function (){
            var id = $(this).data('idtest') ; 
            //alert(id);
        $.ajax({
        type: "POST",
         url : "getting.php",
        data    : {id:id },
        cache: false,
        success: function(data){
           console.log(data);
        }
      });
      })
    });
</script>

edited Sep 28 '18 at 12:54

answered Sep 28 '18 at 12:18

Shubham Azad

677
8
22

Hi Shubham Azad,, in php file echo is not coming sir. – rajkumar Sep 28 '18 at 12:19
what do you mean , echo is not coming? – Shubham Azad Sep 28 '18 at 12:22
is it null response you are getting? – Shubham Azad Sep 28 '18 at 12:23
No sir nothing i'm getting. – rajkumar Sep 28 '18 at 12:29
Yes sir it is reloading sir – rajkumar Sep 28 '18 at 12:33
what i have to do this – rajkumar Sep 28 '18 at 12:34
Yes sir i check above you code it's not reloading but it is not coming any response like alert also – rajkumar Sep 28 '18 at 12:38
Please see agian my php code it will return a proper response whenever data is present or not... – Shubham Azad Sep 28 '18 at 12:44
Thanks for your replaying, still not responding sir, t think my ajax is going to wrong ? – rajkumar Sep 28 '18 at 12:49
please see i ihave added ajax code.. and see result in browser network or console... – Shubham Azad Sep 28 '18 at 12:54
i checked sir in console like this is coming : {"responseCode":0,"message":"No data found regarding this id"} – rajkumar Sep 28 '18 at 12:59
have you got the solution? – Shubham Azad Sep 28 '18 at 13:05
Hi Shubham Azad, All the data is coming but it's not redirecting to the listing page. – rajkumar Oct 03 '18 at 11:28
what you getting in console? – Shubham Azad Oct 03 '18 at 12:55
have you got the solution or not? – Shubham Azad Oct 04 '18 at 12:33

Hary · Answer 2 · 2018-09-28T12:17:07.270

-1

You should $_POST to get the posted parameters in PHP

use

$id =  $_POST["id"]

Also, use PreparedStatement to avoid SQL Injection

$db = new mysqli($dbHost,$dbUsername,$dbPassword,$dbName);
$stmt = $db->prepare("select * from tbl_users where doctor_speciality = ?");
$stmt->bind_param('s', $id));
$stmt->execute();

edited Sep 28 '18 at 12:17

answered Sep 28 '18 at 12:10

Hary

5,208
6
31
67

Hi mbharanidharan88,,, This is not html form correct, if html form post method means will work your. – rajkumar Sep 28 '18 at 12:17
@rajkumar, What you meant? You're confused it seems. With `$.ajax` and setting `type:POST` you're already posting values to PHP page and you can it read it back only with `$_POST` or `$_GET` if it is of `type:GET ` – Hary Sep 28 '18 at 12:20
You should check [this](https://stackoverflow.com/questions/6009206/what-is-ajax-and-how-does-it-work) – Hary Sep 28 '18 at 12:22

Passing onclick Jquery variable issue in php

2 Answers2