onclick get the project id and send it to back.php

Question

I'm trying to display the details of projects in database and and allow admins to them delete them.

Here's what I have so far:

<?php
    include 'dbc.php';
    $query=mysql_query("select * from pro1");
    while($result=mysql_fetch_array($query)){
        echo '<span>'.$result['name'].'</span>'.'<a href="#" id='.$result['pro_id'].' onclick="delet(this.id);">delet</a>'.'<br/>';
    }
?>       
<html>
    <head>
    </head>
    <body>
        <span>test10</span><a href="#" id=10 onclick="delet(this.id);">delet</a>    

        <script src="http://code.jquery.com/jquery-1.10.1.min.js"></script>
        <script type="text/javascript">
        $(document).ready(function(){
            function delete(proid){
                //alert(proid);
                $.post("back.php",{
                    proid:proid
                },function(data){
                    alert(data);
                });
        });//ready func end
        </script>
    </body>
</html>

I'm getting the proid from db... now I need to send the id with ajax. The loop output will be something like this:

<span>test10</span>
 <a href="#" id=10 onclick="delete(this.id);">delete</a>

score 0 · Answer 1 · answered Jun 11 '13 at 18:54

Move the delet function out of the document.ready wrapper. It can't be accessed because it's out of scope:

            // no $(document).ready()
            function delet(proid){
                $.post("back.php",{
                    proid:proid
                },function(data){
                    alert(data);
                }); // <-- this was missing
            }

score 0 · Accepted Answer · edited Jun 20 '20 at 09:12

Explanation

There is missing one closing bracket for the function.

Note: The function must not be wrapped to be executed on load, otherwise it won't be accessible !

Solution

(JSFiddle)

Javascript/jQuery

    function delet(proid){
        //alert(proid);
        $.post("back.php",{
            proid:proid
        },function(data){
            alert(data);
        });
    } /*This one here */

score 0 · Answer 3 · answered Jun 11 '13 at 19:29

0

Perhaps you want like this

JQuery :-

 $(document).ready(function(){
    function delete(proid)
    {
       $.ajax({
           url: 'back.php',
           type: "get",
           data:'id='+ proid,
           success: function (response) {
              alert(data);
           }

        });      
    }
});

PHP:-

<?php
    include 'dbc.php';

    $id=$_GET['id'];
    $sql=mysql_query("DELETE FROM pro1 WHERE pro_id='$id'");

    if($sql)
       echo "Deleted Succesfully<br>";

    else
       echo "Problem to delete :(<br>";

    $query=mysql_query("select * from pro1");
    while($result=mysql_fetch_array($query)){
        echo '<span>'.$result['name'].'</span>'.'<a href="#" id='.$result['pro_id'].' onclick="delet(this.id);">delet</a>'.'<br/>';
    }
?>

answered Jun 11 '13 at 19:29

sharif2008

2,716
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I may be wrong as i am not an expert in PHP or JQUERY. just let me know plz. – sharif2008 Jun 11 '13 at 19:33
This code is vulnerable to sql injection attacks. Please check out this question for advice on how to make your php code more secure: http://stackoverflow.com/questions/60174/how-to-prevent-sql-injection-in-php – Allan Nienhuis Jun 11 '13 at 19:35
@sharif Also note that `mysql` is *deprecated* in latest PHP versions. You'll want to use `mysqli` or `PDO` statements for regular `MySQL` queries. – Jeff Noel Jun 14 '13 at 11:16

onclick get the project id and send it to back.php

3 Answers3

Explanation

Solution

Javascript/jQuery