我已经在外部文件中编写了一些字符串,因此我没有到处都有字符串,但它似乎不起作用。
Init.inc.php
$get_user = mysql_query("SELECT * FROM users WHERE username = '".$_SESSION['user']."'");
$user_data = mysql_fetch_array($get_user);
$userid = mysql_real_escape_string($user_data['id']);
$username = mysql_real_escape_string($user_data['username']);
$email = mysql_real_escape_string($user_data['email']);
Session.php
require("includes/config.php");
require("includes/init.inc.php");
echo "Hello", $username, $email;
它应该输出用户的用户名和电子邮件地址。 预先感谢:)
最佳答案
如果您定义为 GLOBAL 则有效
require("includes/config.php");
require("includes/init.inc.php");
global $username, $email;
echo "Hello", $username, $email;
关于PHP:字符串在外部文件中不起作用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20956681/