未捕获的TypeError: string不是函数 - Uncaught TypeError: string is not a function

Uncaught TypeError: string is not a function

本文关键字：string 函数 TypeError | 更新日期: 2023-09-27

我试图创建一个登录部分，它应该采取电子邮件id/电话号码和密码，并检查数据库。就像它应该的那样简单，我在-

行上得到一个TypeError，声明"string is not a function"。

if ($e_check!=0) {

  <?php
if (isset($_POST['username']) && isset($_POST['password'])) {
    include_once("db_conx.php");
    $u = $_POST["username"];
    $p1 = $_POST["password"];
    $p = md5($p1);
    if ($u!=""){
        if (is_numeric($u)) {
            $sql = "SELECT * FROM users WHERE phone='$u' AND password='$p' LIMIT 1"; //check against phone numbers
            $query = mysqli_query($con, $sql);
             $result = mysqli_result($query);
        $e_check = mysqli_num_rows($result);

            //var_dump($e_check);
             if ($e_check != 0) {
                echo "Success!";
                exit();
            }
            else {
                echo "Password is incorrect";
                exit();
            } 
        }
        else {
            $sql = "SELECT * FROM users WHERE email='$u' AND password='$p' LIMIT 1"; //check against email
            $query = mysqli_query($con, $sql);
             $result = mysqli_result($query);
        $e_check = mysqli_num_rows($result);
            var_dump($e_check);
             if ($e_check != 0) {
                echo "Success!";
                exit();
            }
            else {
                echo "Password is incorrect";
                exit();
            } 
        }
    }
}
?>
<html>
<head>
<meta charset="UTF-8">
<title>Sign In</title>
<script src="ajax.js"></script>
<script type='text/javascript'>
function login(){
    var username = document.getElementById("username").value();
    var password = document.getElementById("pwd").value();
    var status = document.getElementById("status");
    if (username=="" || password == "") {
        status.innerHTML = "Email/Phone or Password is missing";
    }
    else {
        status.innerHTML = "Please wait...";
        var ajax = ajaxObj("POST", "index2.php");
        ajax.onreadystatechange = function() {
            if(ajaxReturn(ajax) == true) {
                if(ajax.responseText != "Success!"){ ///returned from php file
                    status.innerHTML = ajax.responseText;
                }
                else {
                    //alert("Logged in");
                    //do nothing
                }
            }
        }
       ajax.send("username="+username+"&password="+password); //shoots variable to php 
    }
}
</script>
</head>
<body>
<div id="pageRight">
    <form id="sign-up" onSubmit="return false;">
        Email/Phone: <input type="text" id="username" >
        Password: <input type="password" id="pwd">
        <button id="signin" onClick="login()" style="background-color:#9C092B; color: white; padding: 9px 15px">Sign In </button><br>
        <span id="status" style="color: red"></span>
    </form>
</div>
</body>
</html>

我知道变量e_check没有检索到正确的返回类型，但我不明白为什么会发生这种情况。是否有更好的方法来检查查询是否成功，或者您可以建议对现有代码进行任何修改，以消除此错误?我是PHP的新手，如果这是一个愚蠢的问题，我很抱歉。提前感谢您的帮助!:)

var username = document.getElementById("username").value;
var password = document.getElementById("pwd").value;

不

var username = document.getElementById("username").value();
var password = document.getElementById("pwd").value();

您需要一个结果，以获得num_row计数。使用:

        $sql = "SELECT * FROM users WHERE phone='$u' AND password='$p' LIMIT 1"; //check against phone numbers
        $query = mysqli_query($con, $sql);
        $result = mysqli_result($query);
        $e_check = mysqli_num_rows($result);

您需要在语句中使用空格(您在if语句的两个条件中都有空格)，这只是糟糕的形式-

if ($e_check != 0) {

然而，这并不像其他人指出的那样是你的问题。TypeError问题是JavaScript的问题，不是PHP的问题。您创建的语句将在PHP中工作。