PHP实例化通过AJAX调用打破脚本

I called a PHP script with jQuery but it doesn't return any response if i instantiate a class before echo-ing the response. Meaning that it works if i instantiate the object after.

我用jQuery调用了一个PHP脚本，但是如果我在回显响应之前实例化一个类，它就不会返回任何响应。这意味着如果我实例化之后的对象，它的工作原理。

javascript code :

javascript代码：

$.post('libraries/test.php', {action: 'duplicateForm', service: service_name}, function(response)
{
    console.log(response);
});

php script:

php脚本：

$response = '1';
$f = new usvn_form_insert();
$response = '2';
echo $response;

I don't use the $_POST variables in this test. With this, the response is empty but if I put the echo before $f = new usvn_form_insert(); it prints 1

我在这个测试中不使用$ _POST变量。有了这个，响应是空的，但如果我把回声放在$ f = new usvn_form_insert（）之前;它打印1

1 个解决方案

#1

Put error_reporting(E_ALL) at the top of your php script and see if it returning any error back to the browser.

将error_reporting（E_ALL）放在php脚本的顶部，看看它是否将任何错误返回给浏览器。