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I'm attempting to post some data back to the same page through ajax. In the example below, the $name variable is not being updated in my page once the button is clicked. However, if I look at the console log of the response from ajax (using firebug) it shows the correct html, with the name inserted (i.e. <div>Matthew</div>) - the page just isn't being updated with this response. Any ideas for how to fix this would be greatly appreciated.
The code I have is:
<!DOCTYPE html>
<html>
<head>
<script src=".11.3/jquery.min.js"></script>
</head>
<body>
<?php
$name = "A name";
if ($_SERVER["REQUEST_METHOD"] == "POST") {
if (empty($_POST["name"])) {
$name = "No name!";
}
else{
$name = $_POST["name"];
}
}
?>
<script>
$(document).ready(function(){
$("button").click(function(){
var mydata = {name: "Matthew"};
$.ajax({
type: 'POST',
data: mydata,
success: function(data){
console.log(data);
}
});
});
});
</script>
<button>Send an HTTP POST request to a page and get the result back</button>
<div id="name">
<?php echo $name;?>
</div>
</body>
</html>
I'm attempting to post some data back to the same page through ajax. In the example below, the $name variable is not being updated in my page once the button is clicked. However, if I look at the console log of the response from ajax (using firebug) it shows the correct html, with the name inserted (i.e. <div>Matthew</div>) - the page just isn't being updated with this response. Any ideas for how to fix this would be greatly appreciated.
The code I have is:
<!DOCTYPE html>
<html>
<head>
<script src="http://ajax.googleapis./ajax/libs/jquery/1.11.3/jquery.min.js"></script>
</head>
<body>
<?php
$name = "A name";
if ($_SERVER["REQUEST_METHOD"] == "POST") {
if (empty($_POST["name"])) {
$name = "No name!";
}
else{
$name = $_POST["name"];
}
}
?>
<script>
$(document).ready(function(){
$("button").click(function(){
var mydata = {name: "Matthew"};
$.ajax({
type: 'POST',
data: mydata,
success: function(data){
console.log(data);
}
});
});
});
</script>
<button>Send an HTTP POST request to a page and get the result back</button>
<div id="name">
<?php echo $name;?>
</div>
</body>
</html>
- The page wont just update its self - thats the whole point of ajax, you need to use the returned data (in the success function) to update it via javascript. – Steve Commented Jun 10, 2015 at 22:21
-
Ajax doesn't work that way. Go to your
success: function, there, you assigndatato thedivtag. – Jose Manuel Abarca Rodríguez Commented Jun 10, 2015 at 22:21 - or just do a regular form post, sans ajax – Steve Commented Jun 10, 2015 at 22:22
- I've just posted a simple test case here, but in my real application I need to use javascript functionality on the output of the form and then resubmit that, so ajax seemed the best option. – Matt Pitkin Commented Jun 10, 2015 at 22:34
1 Answer
Reset to default 3It is because <?php echo $name;?> does not run again when doing the ajax call. You have to replace the content of the div in the success function like this:
success: function(data){
$("div").html(data);
}
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