我把PHP代码放在我的文件之上,虽然路径是正确的,但是当我运行它时,它说没有找到对象 [英] I put a PHP code on top of my file, though the path is right but when I run it, it said object not found
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问题描述
这是我的代码
文件名是admin2.php,目前在grapesjs文件夹中
<?php require../login/loginheader.php;
需要../login/dbconf.php;
$ dbconn = mysqli_connect($ host,$ username,$ password,$ db_name);
函数清理($ str){
global $ dbconn;
$ str = trim($ str);
$ str = stripslashes($ str);
$ str = htmlentities($ str,ENT_QUOTES);
$ str = mysqli_real_escape_string($ dbconn,$ str);
返回$ str;
}
if(isset($ _SESSION ['username'])){
$ thisUser = clean($ _ SESSION ['username']);
}
if(isset($ _ POST ['action'])){
$ content = mysqli_real_escape_string($ dbconn,$ _ POST ['content'] );
$ sql1 =UPDATE`member` SET`content` ='。$ content。'WHERE`username` ='。$ thisUser。';
$ query1 = mysqli_query($ dbconn,$ sql1);
}
?>
和这里我的文件夹里面是什么
grapesjs
> admin2.php
登录
> loginheader.php
> dbconf.php
我尝试过:
i改变了路径
<?php require./ login / loginheader.php;
需要./login/dbconf.php;
$ dbconn = mysqli_connect($ host,$ username,$ password,$ db_name);
解决方案
dbconn = mysqli_connect(
宿主
用户名,
here is my code
the file name is admin2.php, currently in grapesjs folder
<?php require "../login/loginheader.php"; require "../login/dbconf.php"; $dbconn = mysqli_connect($host, $username, $password, $db_name); function clean($str){ global $dbconn; $str = trim($str); $str = stripslashes($str); $str = htmlentities($str, ENT_QUOTES); $str = mysqli_real_escape_string($dbconn, $str); return $str; } if( isset( $_SESSION['username'])){ $thisUser = clean($_SESSION['username']); } if( isset($_POST['action'])){ $content = mysqli_real_escape_string($dbconn,$_POST['content']); $sql1 = "UPDATE `members` SET `content`= '".$content."' WHERE `username` ='".$thisUser."'"; $query1 = mysqli_query($dbconn, $sql1); } ?>
and here whats inside my folder
grapesjs
>admin2.php
login
>loginheader.php
>dbconf.php
What I have tried:
i changed the path to
<?php require "./login/loginheader.php"; require "./login/dbconf.php"; $dbconn = mysqli_connect($host, $username, $password, $db_name);
解决方案
dbconn = mysqli_connect(
host,
username,
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