使用未定義的常量db_selected - 假定'db_selected

Question

<?php 
//connecting to server 
$dbc = mysql_connect('localhost','root',''); 
if(!$dbc){ 
    die('Not Connected : '.mysql_error()); 
    } 
//connecting to database 
$db_selected = mysql_selectdb("kalhan",$dbc); 
if(!db_selected){ 
    die('Cannot Connect'.mysql_error()); 
    } 
//testing database 
$query = "UPDATE userinformation SET username = 'i am not working properly' where name = 'kalhan'"; 
$result = mysql_query($query); 
?> 

該代碼顯示上述錯誤。我只是在考驗我的數據庫編碼的連接部分後...幫助

Answer 1

您在db_selected

$db_selected = mysql_selectdb("kalhan",$dbc); 

if(!db_selected){ 
// ^-- No dollar sign 

錯誤消息的前面需要一個美元符號是告訴你，PHP假定你的意思是將db_selected放在引號中，以便您的條件變爲if (!'db_selected')這可能不是您想要的，因爲每次都會將該字符串評估爲true。