AJAX/PHP/Jquery Issue with callback
I can't seem to figure out why this won't work. I have stripped all elements out and have striped this done to a base example. can someone please tell me what I am doing wrong. When I click on the button on the html page I get my initial alert, but never get an alert from my call backs.
Here is the code for my html file. test.html
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="开发者_如何学Ctext/html; charset=utf-8" />
<title>Untitled Document</title>
<script src="http://code.jquery.com/jquery-1.5.js"></script>
<script language="javascript" type="text/javascript">
$(document).ready(function()
{
$("#test").click(function()
{
alert("!st");
$.ajax({
type : 'POST',
url : 'test_ajax.php',
dataType : 'json',
data: {
email : 'Jeremy'
},
success : function(data){
alert(data.msg);
},
error : function(XMLHttpRequest, textStatus, errorThrown) {
alert("error");
}
});
return false;
});
});
</script>
</head>
<body>
<input name="test" id="test" type="button" value="Click Me"/>
<div id="results">
Hello
</div>
</body>
</html>
Now for the php: test_ajax.php
<?php
if (empty($_POST['email'])) {
$return['error'] = true;
$return['msg'] = 'You did not enter you email.';
}
else {
$return['error'] = false;
$return['msg'] = 'You\'ve entered: ' . $_POST['email'] . '.';
}
echo json_encode($return);
?>
Please any assistance would be much appreciated.
Thanks Jeremy
The code looks clean. Try to use php buffering. Put ob_start() at the very beginning of PHP file and in the code
// cleans the buffer
ob_clean();
echo json_encode($return);
die();
I've just tried the exact code and it works ok.
You've missed the opening php brackets (<?php) from the code listed above, but was that just a mistake?
I think you missed the opening of the php file and the declaration of $return as array. I tried this and it works fine:
<?
$return = array();
if (empty($_POST['email'])) {
$return['error'] = true;
$return['msg'] = 'You did not enter you email.';
}
else {
$return['error'] = false;
$return['msg'] = 'You\'ve entered: ' . $_POST['email'] . '.';
}
echo json_encode($return);
?>
加载中,请稍侯......
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