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Javascript 如何使此功能正常工作?_Javascript_Jquery - Fatal编程技术网

Javascript 如何使此功能正常工作?

Javascript 如何使此功能正常工作?,javascript,jquery,Javascript,Jquery,当我运行此命令时,我得到一个错误:getInfo未定义。那么,我该如何做到这一点呢 (我知道这不是一个安全登录,我现在不担心) 这个怎么样: function getInfo(){ var user = document.getElementById('username'); var username = user.value; var pass = document.getElementById('password'); var password = pass.

当我运行此命令时,我得到一个错误:getInfo未定义。那么,我该如何做到这一点呢

(我知道这不是一个安全登录,我现在不担心)

这个怎么样:

function getInfo(){
    var user = document.getElementById('username');
    var username = user.value;
    var pass = document.getElementById('password');
    var password = pass.value;
    return [username, password];
 });

 document.getElementById("submit").addEventListener("click", function (){ getInfo();});
 //or
 //document.getElementById("submit").addEventListener("click", getInfo);

 var userinfo = getInfo();
 var username = userinfo[0];
 var password = userinfo[1];

 console.log(username, password);

document.getElementById(“提交”).addEventListener(“单击”,getInfo,false)
function getInfo(){
    var user = document.getElementById('username');
    var username = user.value;
    var pass = document.getElementById('password');
    var password = pass.value;
    return [username, password];
 });

 document.getElementById("submit").addEventListener("click", function (){ getInfo();});
 //or
 //document.getElementById("submit").addEventListener("click", getInfo);

 var userinfo = getInfo();
 var username = userinfo[0];
 var password = userinfo[1];

 console.log(username, password);