如何在函数外部访问JavaScript函数参数? [英] How can I access JavaScript function argument outside the function?

问题描述

我可以在函数外部访问函数参数吗?

Can I access the function arguments outside the function?

这是我的代码:

function viewmessage(username,name) {
        //alert(name + " : " + username);
        $('#heading').html(name);
        $.get('/notification/viewmessage', {user:username}, function(data) {
            $('#messagesfrom').html(data);
            $('#newmessage').slideDown(200);
        });
    }
alert(name + " : " + username);

推荐答案

除非您在函数外部声明了变量，否则您将无法这样做.

You can't, unless you declare the variable outside the function.

您只能在全局范围内使用相同的变量名称:

You can only use the same variable names in the global scope:

function viewmessage(username, name){
    window.username = username;
    window.name = name;
}
alert(window.name + " : " + window.username ); // "undefined : undefined"
alert(name+" : "+username); // ReferenceError: 'name' not defined

在本地范围内，必须使用在函数内部重新声明的变量名称:

In a local scope, you have to use variable names which are re-declared inside the function:

var username2, name2;
function viewmessage(username, name){
    username2 = username; // No "var"!!
    name2 = name;
}
alert(username2 + " : " + name2); // "undefined : undefined"
viewmessage('test', 'test2');
alert(username2 + " : " + name2); // "test : test2"

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