2017-05-04 67 views
-3

我想將我的MySQL代碼轉換爲MYSQLI,但它在第21行發生錯誤。我無法確定我必須糾正第21行錯誤。誰能幫幫我嗎?從mysql轉換mysqli錯誤

$conn = mysqli_connect($host,$username,$password) or die(mysqli_error()); 
mysqli_select_db($conn,$dbname) or die(mysqli_error()); 
mysqli_set_charset($conn , "utf8"); 






mysqli_query($conn,"CREATE TABLE IF NOT EXISTS `cookies` (
`ip` varchar(32) NOT NULL DEFAULT '', 
`time` varchar(32) DEFAULT NULL, 
`waktu` varchar(255) DEFAULT NULL, 
PRIMARY KEY (`ip`) 
ENGINE=MyISAM DEFAULT CHARSET=utf8; 
) 
"); 
+0

這是行21 ** mysqli_select_db($康恩,$ DBNAME)或死亡(mysqli_error()); ** –

+0

什麼錯誤? – Swellar

+0

這是錯誤** PHP警告:mysqli_error()期望只有一個參數,0在第21行/home/flexytk1/public_html/liker.php中給出** –

回答

0

您的mysqli_error應該有一個參數。 mysqli_error($connection)

編輯代碼在這裏:

$conn = mysqli_connect($host,$username,$password); 
mysqli_select_db($conn,$dbname) or die(mysqli_error($conn)); 
mysqli_set_charset($conn , "utf8"); 

mysqli_query($conn,"CREATE TABLE IF NOT EXISTS `cookies` (
     `ip` varchar(32) NOT NULL DEFAULT '', 
     `time` varchar(32) DEFAULT NULL, 
     `waktu` varchar(255) DEFAULT NULL, 
     PRIMARY KEY (`ip`) 
     ENGINE=MyISAM DEFAULT CHARSET=utf8; 
     ) 
");