2014-12-07 20 views
-4

這是代碼:我不能定義變量來更新我的數據庫

<?php 
include "../include/koneksi_db.php"; 
include "../user/link.php";  
$query=mysql_query("SELECT * FROM admin WHERE id=$id", $konek); 
$hasil=mysql_fetch_array($query); 
$id =$hasil['id']; 
$user=$hasil['username']; 
$pass=$hasil['password']; 
$hak =$hasil['hak_akses']; 
?> 

問題: 如何解決這個問題:Nocticce:未定義的變量:用C ID:.... \ ed_user。 5號線上的PHP

+0

什麼?給一個代碼? – 2014-12-07 12:04:40

+0

'$ id = 5; mysql_query(「SELECT * FROM admin WHERE id = $ id」)'你在找那種東西嗎? – Rizier123 2014-12-07 12:05:55

+0

你的問題是什麼?什麼不起作用? – redelschaap 2014-12-07 12:06:09

回答

0

不知道我是否理解正確:這?

<?php 
include "../include/koneksi_db.php"; 
include "../user/link.php"; 

$check_the_query = "SELECT * FROM admin WHERE id='".$id."'"; 
$query=mysql_query(, $konek); 

while($hasil=mysql_fetch_array($query)){ 
    $id =$hasil['id']; 
    $user=$hasil['username']; 
    $pass=$hasil['password']; 
    $hak =$hasil['hak_akses']; 

    echo "id -> ".$id."<br>"; 
    echo "user -> ".$id."<br>"; 
    echo("<br>"); 
} 

?>