Question

我想检查一组用户名和密码来验证用户身份。我的代码在数组的每个元素处停止并显示警报。我怎样才能让它遍历整个阵列然后显示警报。

for(i= 0 ; i< user.length; i+=1) {
  if (username === user[i].username && password === user[i].password) {

    alert("Log In Successful");

    document.getElementById("btn").style.display= "none"
    document.getElementById("navbar-user").innerHTML = username;
    modal.style.display= "none";
  } else {
      alert("Username or Password Incorrect")
  }

Answer 1

我假设你想要在找到正确的用户名密码组合时停止。所以你的ELSE打破了循环，因为它在每次运行时执行。

以下是提案：

var success = false;

for(i= 0 ; i< user.length; i+=1) {
  if (username === user[i].username && password === user[i].password) {
     success = true; 
  } 
}

if(success){
    alert("Log In Successful");

    document.getElementById("btn").style.display= "none"
    document.getElementById("navbar-user").innerHTML = username;
    modal.style.display= "none";
}else{
    alert("Username or Password Incorrect"); 
}

现在代码将运行并完成数组中的所有元素，如果找到用户，它会将success设置为true，并更改它，否则默认为false。

Answer 2

这是一种更实用的方法，使用find在继续之前搜索user数组中的匹配项。

let successfulLogin = user.find(u => {
    return u.username === username && u.password === password;
});

if(successfulLogin) {
    alert("Log In Successful");
    document.getElementById("btn").style.display = "none"
    document.getElementById("navbar-user").innerHTML = username;
    modal.style.display = "none";
} else {
    alert("Username or Password Incorrect");
}

Answer 3

使用Array some（）方法：

if (! user.some(u => username === u.username && password === u.password) )
    alert('Username or Password Incorrect');

Answer 4

希望您只是在客户端测试代码，包括密码。

let alertSucc =true;
for(i= 0 ; i< user.length; i+=1) {
  if (username === user[i].username && password === user[i].password) {

      document.getElementById("btn").style.display= "none"
      document.getElementById("navbar-user").innerHTML = username;
      modal.style.display= "none";
  } else {
      alertSucc =false;
      
  } 
}
  alertSucc ? alert("Log In Successful"):alert("Username or Password Incorrect");

Answer 5

使用一对返回并在if语句之外向下移动警报将有所帮助，因为return将退出当前循环的迭代

for(i= 0 ; i< user.length; i+=1) {
  if (username === user[i].username && password === user[i].password) {
    alert("Log In Successful");
    document.getElementById("btn").style.display= "none"
    document.getElementById("navbar-user").innerHTML = username;
    modal.style.display= "none";
    return;
  } else {
    console.log("We haven't proven whether true or false.
    return;
  } 
  alert("We did not find proper credentials that were true, and tested all inputs")
}

Answer 6

目前尚不清楚是否要在所有有效或有效用户上显示SUCCESS？我祈祷逻辑永远不会到达世界！：）

let userGOOD = [{username: "a", password: "b"}, {username: "a", password: "b"}];
let userBAD = [{username: "a", password: "b"}, {username: "a", password: "c"}];
let test = (username, password) => (user) => username === user.username && 
password === user.password;
let testAB = test("a", "b");
let GOOD = userGOOD.reduce((p, c) => p && testAB(c), true);
let BAD = userBAD.reduce((p, c) => p && testAB(c), true);

在显示警报之前循环遍历数组中的所有元素

6 个答案: