Question

我有两个php文件对活动目录用户进行身份验证，我希望从中获取属性url并将此变量$data从authenticate.php传递给login.php如果函数返回true在header("Location: *URL*");的位置，怎么办呢？

authenticate.php：

<?php
  // Initialize session
  session_start();

  function authenticate($user, $password) {
  if(empty($user) || empty($password)) return false;

  // Active Directory server
  $ldap_host = "CRAMSDCR01V.cloud4rain.local";

  // connect to active directory
  $ldap = ldap_connect($ldap_host);

  $ldap_dn="OU=by-style,DC=cloud4rain,DC=local";

  // verify user and password
  if($bind = @ldap_bind($ldap, $user, $password)) 
  {
    $result = ldap_search($ldap,$ldap_dn, "(cn=*)") or die ("Error in search query: ".ldap_error($ldap));
    $data = ldap_get_entries($ldap, $result);
    echo $data["url"];
    return true;    
  } 
  else 
  {
    // invalid name or password
    return false;
  }
 }
?>

的login.php：

<?php
include("authenticate.php");

// check to see if user is logging out
if(isset($_GET['out'])) {
// destroy session
session_unset();
$_SESSION = array();
unset($_SESSION['user'],$_SESSION['access']);
session_destroy();
}

// check to see if login form has been submitted
if(isset($_POST['btn-login'])){
// run information through authenticator
if(authenticate($_POST['userLogin'],$_POST['userPassword']))
{
  // authentication passed
  header("Location: authenticate.php?$data");
  die();
 } else {
  // authentication failed
  $error = "Login failed: Incorrect user name, password, or rights<br /-->";
}
}

// output logout success
if(isset($_GET['out'])) echo "Logout successful";
?>

Answer 1

<强>的login.php

<?php
include("authenticate.php");

这实际上就像粘贴 login.php 中的 authenticate.php 的内容一样，所以虽然它在技术上是2个文件，但它就好像它是＆＃39 ;只是一个 - 但$data在authenticate()函数中定义，因此只有scoped within that function。

在 authenticate.php 中 - 从函数

返回数据

// verify user and password
if($bind = @ldap_bind($ldap, $user, $password)) 
{
    $result = ldap_search($ldap,$ldap_dn, "(cn=*)") or die ("Error in search query: ".ldap_error($ldap));
    $data = ldap_get_entries($ldap, $result);
    // echo $data["url"]; // I assume this is just for debugging...

    // return $data from the function which should be "truthy"
    return $data;
} 
else 
{
    // invalid name or password
    return false;
}

在 login.php 中 - 评估authenticate()函数的返回值 - 因为PHP是松散类型的，函数返回的任何（非空）字符串都可以被评估为＆＃ 34; truthy＆＃34; - 你从函数中获得的唯一其他回报是false所以......

// run information through authenticator
if($authData = authenticate($_POST['userLogin'],$_POST['userPassword']))
{
  // authentication passed
  // renamed the variable $authData just for clarity
  header("Location: authenticate.php?$authData"); 
  die();
 } 

 else {
  // authentication failed
  $error = "Login failed: Incorrect user name, password, or rights<br />";
}

Answer 2

不确定为什么你在login.php中有$_SESSION = array();但如果你想将$ data从一个php传递到另一个php，那么只需将它设置为session

$_SESSION['data'] = $data;

在另一个文件中使用

获取它

$data = $_SESSION['data'];

如何在2个php文件之间传递变量

2 个答案: