Send a variable to a PHP page and open the page in new tab

Question

I've this code

$(document).on('click', '#button', function(){
    var my_data = 'Test Test Tes';

    $.ajax({
      type: "POST",
      url: "preview.php",
      data: "my_data=" + my_data,
      success: function() {
        window.open('preview.php');
      },
    });
  });

Above codes open the preview.php but I'm getting this error -

Notice: Undefined index: my_data in C:\\xampp\\htdocs\\tb-builder\\preview.php on line 3

This is what I've in preview.php -

<?php
    $data= $_POST['my_data'];
    echo $data;
?>

Can anybody help me to solve this please? Thanks

Answer 1

Main error is , window.open('preview.php'); loads the page preview.php ,
You are not sending any data with post/get request while loading it using window.open('preview.php'); and you are expecting value at $_POST['my_data']

  $(document).on('click', '#button', function(){
    var my_data = 'Test Test Tes';

    $.ajax({
      type: "POST",
      url: "preview.php",
      data: "my_data=" + my_data,
    });
  });

preview.php

<?php
    if(isset($_POST['my_data']){
    $data= $_POST['my_data'];
    header('location:preview.php?my_data=$data');  //reload preview.php with the `my_data`
   }
   if(isset($_GET['my_data'])){
      $data=$_GET['my_data'];    
    echo $data;             //display `my_data` as you requested
   }
?>

Update

With out using ajax ,to achieve what you required.

$(document).on('click', '#button', function(){
    var my_data = 'Test Test Tes';
        window.open('preview.php?my_data='+my_data);
  });

preview.php

<?php 
 if(isset($_GET['my_data'])){
        $data=$_GET['my_data'];    
        echo $data;          
 }
?>

---

final Update

Actually you don't need ajax at all , because you are going to another page with the data .

to avoid get request

Wrap the button inside a form with hidden input having value my_data and set action="preview.php" like below

<form action="preview.php" method="post">
  <input type="hidden" id="my_data" value="Test Test Tes">
  <button type="submit" id="button">post</button>
</form>

Answer 2

window.open just create a new window object. That window will have no $_POST globals. Use window.open('preview.php?my_data=' + my_data); and use $_GET['my_data'] in that PHP.

The post is related to the ajax not for the new window.

EDIT

Ok guys, I see here is a little confusion.

Above codes open the preview.php but I'm getting this error -

So what happens?

OP is sending a data called my_data to the preview.php . This data is travelling through $_POST to the preview.php . At this point, there is $_POST['my_data'];

When it finish with success, the ajax call, then the window.open opens a new window , what has no $_POST data, this is why OP get this error message.

If he want to use the same data what he posted in my_data then he need to add it to the URL and use $_GET .

I think the misunderstood was what @HoangHieu wrote and me pointed:

yes. in new window – Xahed Kamal

And if you want to use response, then use

success: function(reponse) {
   window.open('preview.php?my_data=' + response);
},

and then use $_GET['my_data'] in new window.

Answer 3

use

if (isset($_POST['my_data']))    
{    
    $data= $_POST['my_data'];
    echo $data; 
}   

Answer 4

Please change your preview.php code with the following :-

<?php
    if (isset($_POST['my_data']))
    {
       $data= $_POST['my_data'];
       echo $data;
    }
?>

It may help you.

Answer 5

It was completed question in

Is your error in new window ?? – HoangHieu 1 min ago

yes. in new window – Xahed Kamal

First Ajax post it completed when success: function called... In new window you doesn't have any post data, I don't know any solution post data by javascript to new window.. only one you can send data by GET request ..

Answer 6

Your construct:

data: "my_data=" + my_data,

Is wrong. It should be:

data: {my_data: $my_data},

And I changed your JS variabale to $my_data , so it's less confusing.