我在PHP中有一个cookie,如果它存在,页面应该转到用户页面而不要求登录,但它不起作用。它给了我警告:
The page isn't redirecting properly
Firefox has detected that the server is redirecting the request for this address in a way that will never complete.
This problem can sometimes be caused by disabling or refusing to accept cookies.
允许使用cookie,我的cookie确实存在,我检查了它。
这是我的登录代码(它在cookie之前工作):
<?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 = '{$pass}'");
if(isset($_POST['mantemLog']) && $_POST['mantemLog'] == "1"){
setcookie("mantemUsr",$_POST["con1"],time()+60*60*24*30);
}
if($usuario != null || isset($_COOKIE['mantemUsr'])){
$_SESSION["sucesso"] = "Usuario logado com sucesso";
loggingUsr($mail);
header("Location: slides.php");
}else{
$_SESSION["deny"] = "Usuario ou senha invalidos!";
header("Location: view.php");
}
die();
?>
类usuarios中的函数:
function buscarUsers($query){
$conexao = mysql_connect($this->host, $this->usuario, $this->senha);
mysql_select_db($this->banco, $conexao);
$result = mysql_query($query, $conexao);
$usr = mysql_fetch_assoc($result);
return $usr;
mysql_close($conexao);
}
html:
<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" />
<label for="chk">
<input type="checkbox" name="mantemLog" id="mantemLog" value="1" />Manter logado</label>
</form>
还有index.php:
<?php
require_once("acessa.php");//calls the login code
require_once("view.php");//calls the html
答案 0 :(得分:0)
你有2个问题。第一个是您无法访问与您设置的完全相同的HTTP请求的cookie,必须重新加载该页面才能再次访问它。
第二个问题是您无法在不启动会话的情况下设置会话。你想要使用会话的每一个页面,你必须开始会话,这就像把它放在php脚本的顶部一样简单:
session_start();
另一个错误,您的函数buscarUsers()
执行以下操作:
return $usr;
mysql_close($conexao);
现在mysql_close($conexao);
永远不会运行,因为您之前会返回$usr
。
您应该在创建脚本时打开错误报告,以帮助调试您的内容:
ini_set('display_errors', 1);
error_reporting(-1); // or E_ALL