2017-05-04 60 views
-3

我想将我的MySQL代码转换为MYSQLI,但它在第21行发生错误。我无法确定我必须纠正第21行错误。谁能帮帮我吗?从mysql转换mysqli错误

$conn = mysqli_connect($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

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

+0

什么错误? – Swellar

+0

这是错误** PHP警告:mysqli_error()期望只有一个参数,0在第21行/home/flexytk1/public_html/liker.php中给出** –

回答

0

您的mysqli_error应该有一个参数。 mysqli_error($connection)

编辑代码在这里:

$conn = mysqli_connect($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; 
     ) 
");