php variable to another php via javascript

Question

I have an image link in PHP that passes a variable to a script in a different file (functions.js), which is:

"img src=\"images/del.jpg\" onclick='delete_user_program(".$row1['program_name'].")' onmouseover=\"this.style.cursor='pointer'\" /"

The script is:

   function delete_user_program(program_name){

    var confirmed = confirm("Are you sure;");
    if (confirmed == true){
        var str="./delete_user_program.php?p1="+program_name;
        window.location=str; 
    }
}

I try to pass "p1".

Then the delete_user_program.php is:

<?php
    session_start();
    include("connect_db.php"); 
    $con = $_SESSION['connection'];


    $select_query ="SELECT * FROM user_program WHERE program_name='".$_GET['p1']."'";
    $result=@mysqli_query($con,$select_query) or die('Error, query failed');
    $num_result=mysqli_num_rows($result);


    if($num_result>0) {

        //some code
    }


 else {
        echo '<html><script language="javascript">alert("Program not exist.");</script></html>';

When calling delete_user_program.php I get the error that p1 is undefined and the message "Program not exist.". Any tips? Thanks in advance.

Possible duplicate of [PHP: "Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset"](http://stackoverflow.com/questions/4261133/php-notice-undefined-variable-notice-undefined-index-and-notice-undef) — Tom Udding, Mar 17 '17 at 16:17
@MarcoSalerno, sorry, that was copy-paste mistake. Ignore it. — Nikos Kiriakakis, Mar 17 '17 at 16:29
@mtizziani It's a big project and I keep all scripts in a .js file — Nikos Kiriakakis, Mar 17 '17 at 16:29

score 0 · Answer 1 · answered Mar 17 '17 at 16:25

0

You should use ajax.

$(document).ready(function(){
    function delete_user_program(program_name){

        $.get("./delete_user_program.php",
        {
          p1: program_name
        },
        function(data,status){
            alert("Data: " + data + "\nStatus: " + status);
        });
    }
});

answered Mar 17 '17 at 16:25

Dawid Karasiewicz

144
5

Print p1 in PHP file and see it's ok. – Dawid Karasiewicz Mar 17 '17 at 16:35
Not ok. Returns "undefined". Script doesn't pass p1.It has to do with the script. – Nikos Kiriakakis Mar 17 '17 at 16:51
Check program_name variable in js. And look into debug console. – Dawid Karasiewicz Mar 17 '17 at 17:27

score 0 · Answer 2 · answered Mar 17 '17 at 16:26

0

<?php
  // file A.php
  echo 'A';
?>


<?php
  // file B.php;
  $a = file_get_contents('A.php');
  // thats synchron 
  echo $a;
?>

answered Mar 17 '17 at 16:26

mtizziani

956
10
23

While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value. – Donald Duck Mar 17 '17 at 18:55

php variable to another php via javascript

2 Answers2