javascript - 这里的数组输出类型是什么

Question

在第 24 行，我测试 test2[1] == "invalid" 是否永远不起作用，它总是转到 else 并带我到下一页。我认为 test2[1] 只是不是一个字符串，但我不知道它还会是什么，请帮忙

function login() {
      var email = document.getElementById("email").value;
      var passW = document.getElementById("password").value;

      if (email == "" || passW == "") {
        alert("Please enter a valid email or password.");
      } else {
        var myXMLRequest = new XMLHttpRequest();
        myXMLRequest.onload = openWorkout;
        var url = "assignment10.php?em=" + email + "&pass=" + passW;
        myXMLRequest.open("POST", url, true);
        myXMLRequest.send();
      }
    }

    function openWorkout() {
      var invalid = "invalid";
      var step = this.responseText;
      var test = step.split(",");
      var test2 = test[0].split(":");
      console.log(step);
      console.log(test);
      console.log(test2[1]);
      if (test2[1] == "invalid") {
        alert("The email or password you entered is invalid. Please try again.");
      } else {
        window.location = "#workoutPage";
      }
    }

    <?php
    //TASK 1: MAKE A CONNECTION TO THE DATABASE, DISPLAY ERROR FOR FAILED CONNECTIONS
    //(FOR GODADDY) NOTE: $mysqli = new mysqli("127.0.0.1", "username", "password", "database", 3306);

    $mysqli = new mysqli("localhost", "User", "1234", "ass10");
    if ($mysqli->connect_errno) {
      echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
    }
    //CHECK IF EMAIL AND ENTERED PASSWORD VALID (LOGIN PAGE [first part of open workout checks login password])
    $entEmail = $_GET['em'];
    $entPassword = $_GET['pass'];
    $sql = "SELECT * FROM membership_table WHERE Email = $entEmail, Password = $entPassword";
    $result = $mysqli->query($sql);
    if($result->num_rows == 0) {
         $data = "invalid";
    } else {
        $data = "valid";
    }


    //Pass to JSON
    $json = array(
        "data" => $data,
        "Email" => $entEmail,
        "Password" => $entPassword

    );

    header("Contenttype:Application/json");
    print(json_encode($json));
    ?> 

这是第 21 - 23 行的 console.log 输出

assignment10.js:21 {"data":"invalid","Email":"q","Password":"1"}
assignment10.js:22 (3) ["{"data":"invalid"", ""Email":"q"", ""Password":"1"}"]
assignment10.js:23 "invalid"

Answer 1

您正在从 PHP 返回 JSON，因此请使用 JSON.parse 对其进行处理。 ，而不是尝试将字符串分开:

var response = JSON.parse(this.responseText);
var test = response.data;
if (test == 'invalid') {
    ...

请注意，当前代码的问题是 test2[1] 是字面上 "invalid"，包括双引号，因此对于您的测试要工作，您需要使用

if (test2[1] == '"invalid"') {

以下代码片段使用您的问题中的 console.log(step) 的输出来演示代码:

const responseText = '{"data":"invalid","Email":"q","Password":"1"}';
var response = JSON.parse(responseText);
var test = response.data;
if (test == 'invalid') {
  console.log('Invalid!');
} else {
  console.log('Valid!');
}