pass ajax variabel to other php page

Question

This is my ajax.php

I try to pass source1 to other php I got error when I use

$data = json_decode($_POST['source1']); It says Undefined index. What should I do?

<?php 
session_start();
$data = json_decode($_POST['source1']);

 ?> 
<!DOCTYPE html>
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>


  <title></title>
</head>
<!-- <form action="sendinto.php" method="post"><center>

<button id="btnSelectedRows" type="button">test send</button>
</center>
</form> -->

<center>
      <button id="btnSelectedRows" >test</button>
    <!-- input type="submit" name="upload" value="Compare"> -->
  </center>
<body>





</body>
</html>


<script type="text/javascript">

this is my ajax code

$('#btnSelectedRows').on('click', function() {
 $.post({
  type: "POST",
  url: 'sendinto.php',
  datatype : 'text',
  data: {source1 : "heyo"} ,
   // or GET

}).done(function(data) {
    //Redirect to another page if you want...
    window.location.href = "sendinto.php";
   });
  });

this is my sendinto.php

<?php 

  session_start();
  $coba = $_SESSION['source1'];
  echo $coba;

?>

`$coba = $_POST['source1'];` – guradio Nov 26 '19 at 08:23 — guradio, Nov 26 '19 at 08:23

Jatin Parmar · Answer 1 · 2019-11-26T08:42:58.787

0

you are doing it wrong, as per your source code, you are sending source1 to sendinto.php and trying to access in ajax.php. first, remove the following lines from ajax.php

<?php 
 session_start();
 $data = json_decode($_POST['source1']);
?>

and put it in sendinto.php like:

 <?php 
  session_start();
    $data=null;
   //you first need to check it
   if(isset($_POST['source1']))
   $data = json_decode($_POST['source1']);
 ?>

edited Nov 26 '19 at 08:42

answered Nov 26 '19 at 08:26

Jatin Parmar

2,759
5
20
31

it works, but there is error in sendinto.php – Sherlin Tjoa Nov 26 '19 at 08:33
I have updated my answer please check it – Jatin Parmar Nov 26 '19 at 08:34
i already use your code but `$data` return null it mean the ajax not passing the variabel? – Sherlin Tjoa Nov 26 '19 at 08:39

score 0 · Answer 2 · answered Nov 26 '19 at 08:30

0

you can use if condition for stop error

if($_POST['source1']){
   $data = json_decode($_POST['source1']);
}else{
   $data = "";
}

answered Nov 26 '19 at 08:30

Darshan Patani

208
1
9

and what i should write in sendinto.php? – Sherlin Tjoa Nov 26 '19 at 08:32
1

you can simply fetch data in sendinto.php like this $value = $_POST['source1'];echo $value; – Darshan Patani Nov 26 '19 at 08:42
@SherlinTjoa if your problem has been solved, please don't forget to tick my answer as correct answer. – Darshan Patani Nov 26 '19 at 10:55

pass ajax variabel to other php page

2 Answers2