问题描述
我正在这样做(是的，我使用了错误的连接数据，这是为了强制连接错误)
try {
$connection = new mysqli('localhost', 'my_user', 'my_password', 'my_db') ;
} catch (Exception $e ) {
echo "Service unavailable";
exit (3);
}
但是 PHP 正在这样做 php_warning:
But PHP is doing this php_warning:
mysqli::mysqli(): (28000/1045): 用户 'my_user'@'localhost' 访问被拒绝(使用密码:YES)
mysqli::mysqli(): (28000/1045): Access denied for user 'my_user'@'localhost' (using password: YES)
在示例中，我使用错误的连接数据来强制连接错误，但在现实世界中，数据库可能已关闭，或者网络可能已关闭......等等.
In the example I'm using wrong connection data to force a connection error, but in the real world the database could be down, or the network could be down... etc..
问题:有没有办法在不抑制警告的情况下拦截数据库连接问题?
Question: Is there a way, without suppressing warnings, to intercept a problem with the database connection ?
推荐答案
你需要告诉mysqli抛出异常:
You need to tell mysqli to throw exceptions:
mysqli_report(MYSQLI_REPORT_STRICT);
try {
$connection = new mysqli('localhost', 'my_user', 'my_password', 'my_db') ;
} catch (Exception $e ) {
echo "Service unavailable";
echo "message: " . $e->message; // not in live code obviously...
exit;
}
现在您将捕获异常并从那里获取它.
Now you will catch the exception and you can take it from there.
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