Question

我一直在尝试向我的登录php代码中添加一种方法来验证用户是否被充血，以便能够首先从数据库中激活用户（默认情况下未批准），所以我整天都在工作尝试对该登录进行编码。不幸的是，我相信我破坏了我的代码，因为现在连正常的登录都不能用:( 我得到了错误

PHP语法检查：解析错误：语法错误，意外的“ COUNT” （T_STRING），在第31行的代码中应为'，'或'）'

代码如下：

<?php

session_start(); $servername = ""; $dbusername = ""; $dbpassword = ""; $dbname = ""; $conn = new mysqli($servername, $dbusername, $dbpassword, $dbname); if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error); }

$id="''"; $username = $_POST['username']; $password = md5($_POST['password']); $func = "SELECT contrasena FROM users WHERE username='$username'";

$realpassask = $conn->query($func); $realpassaskres = $realpassask->fetch_assoc(); $realpass= $realpassaskres[contrasena];

$func = "SELECT bloqueado FROM users WHERE username='$username'"; 
$bloquedask = $conn->query($func); 
$bloquedres = $bloquedask->fetch_assoc(); 
$bloqueado = $bloquedres[bloqueado]; 
//Login //die("password = " . $password . "realpass = " . $realpass ); 
if(!empty($username)) {     // Check the email with database 
    $userexists
= $userexists->prepare(“SELECT COUNT(`username`) FROM `users` WHERE `username`= $username LIMIT 1”);    
$userexists->bindParam(':user_email', $username);   
$userexists->execute();
        // Get the result   $userexistsres = $userexists->fetchColumn();
        // Check if result is greater than 0 - user exist   if( $userexistsres == 1)    {
       if ($bloqueado == NO) {
            if ($password != $realpass) {
            die("contrasena incorrecta");
            } Else{
            $_SESSION['loguin']="OK";
            $_SESSION['username']="$username";
            header("Location: ./herramientas.php");
            } 
        } Else{
            die("Tu usuario ha sido bloqueado o todavía no ha sido aceptado por un administrador. Si el problema persiste contacta con contacto@leonmacias.com");
        } Else{
            die("No hay ninguna cuenta con este nombre de usuario");    } } else {
        echo 'El campo usuario esta vacio'; } ?>

我的php代码存在语法错误，我无法修复

0 个答案: