CODEWITHSUNDEEP
phpjqueryajaxpost
Kay
Kay

Reputation: 87

jQuery post not loading from PHP file

Ok so I have a link created with JS:

var content = document.getElementById("Profile");
  var entry = document.createElement('li');
    entry.innerHTML= ' <a href="eventList2.html" onclick ="post()"> Click Me </a>';
    content.appendChild(entry); 

function post(){
$.post( "random.php", { variable: "hello"} );
}

And in my PHP:

random.php

<?php
Header("content-type: application/x-javascript");
error_reporting(E_ERROR | E_WARNING | E_PARSE);

$variable   = $_POST['variable'];   
echo"alert($variable);";

?>

"undefined" gets echoed instead of "hello". Could anybody tell me what I'm doing wrong? I'm not too familiar with AJAX and Jquery, I'm just going by whatever I see in tutorials and examples. Thanks

Upvotes: 0

Views: 92

Answers (2)

busypeoples
busypeoples

Reputation: 737

Why have PHP return an alert? This is a bad approach. Let PHP return the json encoded result and have jquery handle the alert. This is one possible way to achieve this:

$.post( "random.php", { variable: "hello"}, function(data) {
    alert(data);    
});

PHP:

<?php
Header("content-type: application/x-javascript");
error_reporting(E_ERROR | E_WARNING | E_PARSE);

$variable   = $_POST['variable'];   
echo json_encode($variable);

?>

Upvotes: 1

Barmar
Barmar

Reputation: 780724

Change:

echo"alert($variable);";

to:

$js = json_encode($variable);
echo "alert($js);";

This will put the contents of $variable into the proper format for a Javscript literal.

Upvotes: 0

Related Questions