$mysqlServer = "***";
$mysqlDb = "***";
$mysqlUser = "***";
$mysqlPass = "***";
$conn = mysqli_connect($mysqlServer, $mysqlUser, $mysqlPass) or die("failed to connect to db");
mysqli_select_db($conn, $mysqlDb) or die("failed to connect select db");我有这个代码,它的工作没有任何问题。但是,如果我尝试输入错误的sql服务器或测试它来执行错误。这将显示:
Warning: mysqli_connect(): (HY000/2002): A connection attempt failed because the connected party did not properly respond after a period of time, or established connection failed because connected host has failed to respond.
failed to connect select db如果在连接sql服务器时出现问题,我不希望显示警告。我只想显示我自己的错误。
发布于 2013-06-20 16:24:31
2种可能的选择:
@ sign,这将抑制警告消息发布于 2013-06-20 16:25:20
将@符号放在每个函数前面会隐藏错误
$conn = @mysqli_connect($mysqlServer, $mysqlUser, $mysqlPass) or die("failed to connect to db");发布于 2013-06-20 16:27:36
试试这个:
$conn = mysqli_connect($mysqlServer, $mysqlUser, $mysqlPass, $mysqlDb);使用connect作为第四个参数传递DB名称。
https://stackoverflow.com/questions/17208678
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