PHP不显示来自AJAX的变量
[英]php not showing variable from AJAX
问题描述
I am trying to post a variable from javascript to the same php page. 
我正在尝试将变量从javascript发布到同一php页面。 Here is the code: 这是代码：
<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. 我也尝试过使用POST，但没有在php中接听。 Also tried having a separate php file and using include_once() and still doesn't work. 还尝试过使用单独的php文件并使用include_once（）仍然无法正常工作。
Am I missing a file to load in? 我是否缺少要加载的文件？
2 个解决方案
解决方案1
7 2015-04-17 09:56:41
The problem is that you are sending a string that is not a key-value pair. 问题在于您要发送的字符串不是键值对。 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. 您可以发送一个字符串（无论如何在屏幕后面都会发生），但是它必须是一个有效的查询字符串，其中包含正确编码的键-值对。
It is easier to send an object and let jQuery take care of the encoding: 发送对象更容易，让jQuery处理编码：
$('#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: 会变成这样：
$('#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: 编辑：要使您的php脚本的结果返回页面，您需要使用传递给success函数的变量：
...
success: function(output_from_php_script) {
$('#showresult').text(output_from_php_script);
}
Also note that posting to the original page that generated the page in an ajax call, is not very convenient: It will return you loads of html (the whole page...) that you don't need. 还要注意，在ajax调用中发布到生成该页面的原始页面并不是很方便：它将返回您不需要的html负载（整个页面...）。
You'd better write a separate script to process your ajax calls and have that return ( echo out...) only what you need. 您最好编写一个单独的脚本来处理您的ajax调用，并仅根据需要返回（ echo ...）。
解决方案2
0 2015-04-17 09:57:54
- Even when you happen to succeed with it, you won't really know whether the output was a success since the ajax request will be successful, but the outcome of your PHP might not be. 即使您碰巧成功，由于ajax请求将成功，您也不会真正知道输出是否成功，但是PHP的结果可能不会成功。 This is probably not what you wanted. 这可能不是您想要的。
- You don't need to use absolute URL, simply use
url: "page.php"您不需要使用绝对URL，只需使用url: "page.php"
What I would suggest you to do is to create a separate php page that only contains <?php echo 'Result is '; if (isset($_GET['testing'])) { echo $_GET['testing']; } ?> 我建议您做的是创建一个单独的php页面，该页面仅包含<?php echo 'Result is '; if (isset($_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 $_GET['testing']; } ?> and the call this page using ajax. <?php echo 'Result is '; if (isset($_GET['testing'])) { echo $_GET['testing']; } ?>然后使用ajax调用此页面。
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