Ajax 从 php 获取返回值？
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Ajax get a return value from php?
提问by user1854438
I want to alert the return value from a php method, but nothing happens. Here is the ajax and php methods. Can anyone see what I am doing wrong?
我想提醒 php 方法的返回值，但没有任何反应。这里是ajax和php方法。谁能看到我做错了什么？
--------------------------------------… Ajax script
--------------------------------------... Ajax 脚本
$.ajax({
type: 'get',
url: '/donation/junk/4',
data: datastring,
success: function(data) {
alert(data');
}
});
--------------------------------------… php method
-------------------------------... php 方法
function junk($id)
{
return "works11";
}
回答by Kristian
in PHP, you can't simply return your value and have it show up in the ajax response. you need to printor echoyour final values. (there are other ways too, but that's getting off topic).
在 PHP 中，您不能简单地返回您的值并将其显示在 ajax 响应中。您需要print或echo您的最终值。（也有其他方法，但这已经离题了）。
also, you have a trailing apostrophe in your alert()call that will cause an error and should be removed.
此外，您的alert()调用中有一个尾随撇号，这会导致错误，应该将其删除。
Fixed:
固定的：
$.ajax({
type: 'get',
url: '/donation/junk/4',
data: datastring,
success: function(data) {
alert(data);
}
});
PHP:
PHP：
function junk($id)
{
print "works11";
}
回答by zgr024
You have an extra ' in there on the alert(data') line
您在警报（数据）行上有一个额外的 '
This should work
这应该工作
$.ajax({
type: 'get',
url: '/donation/junk/4',
data: datastring,
success: function(data) {
alert(data);
}
});
And your PHP code should call the method also and echo the value
并且您的 PHP 代码也应该调用该方法并回显该值
function junk($id) {
return 'works11';
}
exit(junk(4));
All you're doing currently is creating the method
您当前所做的就是创建方法
回答by Peter
ajax returns text, it does not communicate with php via methods. It requests a php page and the return of the ajax request is whatever the we babe would have showed if opened in a browser.
ajax 返回文本，它不通过方法与 php 通信。它请求一个 php 页面，而 ajax 请求的返回是我们宝贝在浏览器中打开时会显示的任何内容。
