Question

我正在尝试在网站上创建管理区域，当我需要用户登录时，它不会去。我的查询总是返回<div id="logFrm"> <h5>Por favor, insira o seu email<br/ > e senha para continuar.</h5> <form action="acessa.php" method="POST"> <label for="con1" class="lblLog">Email</label> <input type="text" name="con1" id="con1" /> <br /><br /> <label for="con2" class="lblLog">Senha</label> <input type="password" name="con2" id="con2" /> <br /><br /> <input type="submit" name="logBtn" id="logBtn" value="Logar" /> </form> </div>，使登录无法进行。我做了所有这样的查询，一切正常，再加上，当我在mysql控制台中进行查询时，一切都很好。有人能帮助我吗？

html表单：

<?php

include_once '../usersDB.php';
include_once '../usersFunctions.php';

$conexao = new usuarios();

$mail = $_POST["con1"];
$pass = $_POST["con2"];

$usuario = $conexao->buscarUsers("select * from users where email = '{$mail}' and senha = '{$pass}'");

//var_dump($usuario); //always return null
//var_dump($mail); //return the correct value
//var_dump($pass); ////return the correct value

    if($usuario == null){
        $_SESSION["deny"] = "Usuario ou senha invalidos!";
        header("Location: index.php");
    }else{
        $_SESSION["sucesso"] = "Usuario logado com sucesso";
        loggingUsr($mail);
        header("Location: slides.php");
    }

    die();
?>

应该验证登录的文件：

<?php class usuarios{

    private $host= "*****";
    private $usuario = "*****";
    private $senha = "*****";
    private $banco = "*****";
    private $conexao;

    function buscarUsers($query){
        $conexao = mysql_connect($this->host, $this->usuario, $this->senha);
        $result = mysql_query($conexao, $query);
        $usr = mysql_fetch_assoc($result);
        return $usr;
        mysql_close($conexao);
    }

}

?>

usersDB.php：

<?php
session_start();

function userLogged(){
    return isset($_SESSION["usuario_logado"]);
}

function logUsr(){
    return $_SESSION["usuario_logado"];
}

function verificaLog(){
    if(!userLogged()){
        $_SESSION["deny"];
        die("Você precisa estar logado para acessar esse recurso.");
    }
}

function loggingUsr($email){
    $_SESSION["usuario_logado"] = $email;
}

function logoutUsr(){
    session_destroy();
    session_start();
}
?>

usersfunctions.php：

Player 1 wins 50% of his games
Player 2 wins 20% of his games
Player 3 wins 90% of his games
Player 4 wins 20% of his games

Answer 1

在mysql_中，之后是连接，而不是第一次。

此mysql_query($conexao, $query）应为mysql_query($query, $conexao)

http://php.net/manual/en/function.mysql-query.php

混合mysql_query（字符串$ query [，resource $ link_identifier = NULL]）

首先使用数据库连接，是mysqli_方法。

http://php.net/manual/en/mysqli.query.php

混合mysqli_query（mysqli $ link，字符串$ query [，int $ resultmode = MYSQLI_STORE_RESULT]）

无法让用户登录adm部分

1 个答案: