Question

我试图通过函数将变量返回true或false。我想这样做，所以我只能调用一次ajax，并在一个函数中接收两个变量。但是，我在返回变量时遇到了麻烦。这是我的代码：

var emailExists = false;
var userExists = false;

function checkExisting(emailExists,userExists) {
    var emailExists = true;
return emailExists;
}
alert(emailExists);

我想不出的是为什么警报会让我虚假，当我认为它会让我真实时。这个设置有什么问题？非常感谢你！

Answer 1

你有3个版本的“emailExists”变量：全局一个，checkExisting（）的参数，checkExisting（）中的本地参数！摆脱除第一个之外的所有。此外，您永远不会调用checkExisting（）。

var emailExists = false;

function checkExisting() {
    emailExists = true;
}
checkExisting();
alert(emailExists);

或

var emailExists = false;

function checkExisting() {
    return true;
}
emailExists = checkExisting();
alert(emailExists);

Answer 2

var emailExists = false;
var userExists = false;

function checkExisting(emailExists,userExists) {
    emailExists = true;
return emailExists;
}
checkExisting(false,true); //FOR EXAMPLE !
alert(emailExists);

你应该调用checkExisting函数，而不需要从var使用函数体，因为它是在页面上定义的。

Answer 3

总之......一切。

我认为你是javascript和编程新手？您需要进行大量阅读，以便了解对象范围以及javascript的工作原理。我会快速介绍一下你所写的内容，以便你能够学到一些东西。

// Here you're declared two objects. 'emailExists' and 'userExists'.
// These Boolean objects, since they are not wrapped in a closure are now global
// (you can reference them anywhere) in your script.   
var emailExists = false;
var userExists = false;


// This function never gets called. If it did, it would always return true 
// since you have created a new 'emailExists' Boolean object in your function 
// and would return that each time.
function checkExisting(emailExists,userExists) {
    // This whilst only available within the function closure, is a no, no.
    // You're just confusing things by creating objects with the same name 
    // as global ones.
    var emailExists = true;

    // I'm returning true.
    return emailExists;
}

// Here you are returning your first declared Boolean (the one at the top)
// this will always return false.
alert(emailExists);

难以通过函数返回变量

3 个答案: