这个javascript回调函数如何工作？
[英]How does this javascript callback function work?
问题描述
can anyone explain line by line, I don't get how this call back and prototype works especially the function(callback) in js file 
任何人都可以逐行解释，我不知道此回调和原型如何工作，尤其是js文件中的函数（回调）
user.getUsers(function (theUsers) {
$('#users-table-wrapper').html(user.getATable(theUsers));
});
this part in HTML HTML中的这一部分
Js File Js文件
function User () {
}
User.prototype.getUsers = function (callback) {
$.ajax({
url: 'posting.php',
data: {
request:'get-users'
},
type:'post',
dataType: 'json',
success: function(users){
// callback(users);
if (callback) { callback(users); }
}
});
}
Here is my index.html 这是我的index.html
theUser is not declared but still works. 没有声明theUser，但仍然可以使用。 when I type funcion (theUser) as far as I know theUser is a argument or a parameter. 据我所知，当我输入funcion（theUser）时，User是一个参数或参数。 It has to be declared somewhere. 必须在某处声明它。
It seems it is neither of them.... how does this work? 似乎他们俩都不是...。这是怎么工作的？
<!DOCTYPE html>
<html>
<head>
<title>Users</title>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script src="User.js"></script>
<script>
$(function () {
var user = new User();
user.getUsers(function (theUsers) {
$('#users-table-wrapper').html(user.getATable(theUsers));
});
});
</script>
</head>
<body>
<div class='main-wrapper'>
<h3>Users</h3>
<div id="users-table-wrapper">
</div>
</div>
</body>
</html>
2 个解决方案
解决方案1
1 2013-09-18 19:25:33
theUsers is a parameter to the anonymous function you provide as a callback: theUsers是您作为回调提供的匿名函数的参数：
function (theUsers) {
$('#users-table-wrapper').html(user.getATable(theUsers));
});
In the success method of User.getUsers , you'll see it works like this: 在User.getUsers的成功方法中，您将看到它的工作方式如下：
success: function(users){
if (callback) { callback(users); }
}
Thus, if the AJAX call succeeds, and a callback is defined, the users parameter containing the data successfully retrieved is passed as the first argument to the callback function. 因此，如果AJAX调用成功，并且定义了回调，则包含成功检索到的数据的users参数将作为第一个参数传递给回调函数。 Since the first argument is named theUsers in your anonymous callback definition, that's how it appears inside the method, making itself available for user.getATable(theUsers) . 由于第一个参数在您的匿名回调定义中被命名为theUsers ，因此它就出现在方法中，从而可用于user.getATable(theUsers) 。
解决方案2
0 2013-09-18 19:21:04
theUser is a named argument to your function. theUser是函数的命名参数。
When calling the function, it gets the value of the first parameter passed. 调用该函数时，它将获取传递的第一个参数的值。
You're calling the function here: 您在这里调用该函数：
callback(users);
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