Question

我写了一个从php文件中获取一些数据的函数。当我解析它并使用警告框来显示它时，它工作正常，但当我尝试返回值时，它是未定义的。我不知道为什么会发生这种情况

function getUser() {

    var httpRequest = new createAjaxRequestObject();

    httpRequest.open('GET', 'getUser.php', true);
    var name;

    httpRequest.onreadystatechange = function() {

        if (httpRequest.readyState == 4) {

            if (httpRequest.status == 200) {
                name = JSON.parse(httpRequest.responseText);
                alert(name); //this works just fine and display the name john
            } else {
                alert('Problem with request');
            }
        }
    }

    httpRequest.send();
    return name; //however this is returning null 
}

Answer 1

现在它发送null，因为它会在调用httpRequest.send();时立即获取值。

在这种情况下，您需要将回调传递给将接收返回值的函数

像这样改变，

function foo(callback) {
    httpRequest = new XMLHttpRequest();
    httpRequest.onreadystatechange = function () {
        if (httpRequest.readyState === 4) { // request is done
            if (httpRequest.status === 200) { // successfully
                callback(httpRequest.responseText); // we're calling our method


            }
        }
    };
   httpRequest.open('GET', 'getUser.php', true);
    httpRequest.send();
}

foo(function (result) {
    var name = result;
});

Javascript函数返回null

1 个答案: