$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
如果连接sql server时出现问题,我不希望显示警告。我只想要显示自己的错误。
答案 0 :(得分:9)
2种可能的选择:
@
符号,这会发出警告消息答案 1 :(得分:0)
在每个函数隐藏错误之前放置@符号
$conn = @mysqli_connect($mysqlServer, $mysqlUser, $mysqlPass) or die("failed to connect to db");
答案 2 :(得分:0)
试试这个:
$conn = mysqli_connect($mysqlServer, $mysqlUser, $mysqlPass, $mysqlDb);
使用connect作为第四个参数传递数据库名称。