php not showing variable from AJAX
Question
I am trying to post a variable from javascript to the same php page. Here is the code:
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.11.3/jquery-ui.min.js"></script>
</head>
<body>
<?php echo 'Result is '; if (isset($_GET['testing'])) { echo $_GET['testing']; } ?>
<?php $thisPage = "http://$_SERVER[HTTP_HOST]$_SERVER[REQUEST_URI]"; ?>
<script>
$(document).ready(function() {
$('#testing123').on('change',function () {
var testing = "confirmed";
$.ajax({
type: "GET",
url: "<?php echo $thisPage; ?>",
data: testing,
success: function() { $('#showresult').html("success"); }
})
});
});
</script>
<form>
<select id="testing123">
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
</select>
</form>
<div id="showresult"></div>
</body>
</html>
I have also tried using POST but doesn't pick up in php. Also tried having a separate php file and using include_once() and still doesn't work.
Am I missing a file to load in?
2 answers
solution1
7 2015-04-17 09:56:41
The problem is that you are sending a string that is not a key-value pair. You can send a string (that is what happens behind the screens anyway) but it needs to be a valid query string consisting of key - value pairs that are correctly encoded.
It is easier to send an object and let jQuery take care of the encoding:
$('#testing123').on('change',function () {
var testing = "confirmed";
$.ajax({
type: "GET",
url: "<?php echo $thisPage; ?>",
data: testing,
success: function() { $('#showresult').html("success"); }
})
});
Would become something like:
$('#testing123').on('change',function () {
// send key - value pairs
var testing = {'testing': "confirmed"};
$.ajax({
type: "GET",
url: "<?php echo $thisPage; ?>",
data: testing,
success: function() { $('#showresult').html("success"); }
})
});
Edit: To get the results of your php script back on your page, you need to use the variable that is passed to the success function:
...
success: function(output_from_php_script) {
$('#showresult').text(output_from_php_script);
}
Also note that posting to the original page that generated the page in an ajax call, is not very convenient: It will return you loads of html (the whole page...) that you don't need.
You'd better write a separate script to process your ajax calls and have that return ( echo out...) only what you need.
solution2
0 2015-04-17 09:57:54
- Even when you happen to succeed with it, you won't really know whether the output was a success since the ajax request will be successful, but the outcome of your PHP might not be. This is probably not what you wanted.
- You don't need to use absolute URL, simply use
url: "page.php"
What I would suggest you to do is to create a separate php page that only contains <?php echo 'Result is '; if (isset($_GET['testing'])) { echo $_GET['testing']; } ?> <?php echo 'Result is '; if (isset($_GET['testing'])) { echo $_GET['testing']; } ?> <?php echo 'Result is '; if (isset($_GET['testing'])) { echo $_GET['testing']; } ?> and the call this page using ajax.
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