new mysqli - 为什么会显示错误消息

Question

I want to check if accessing mysql is possible or not.

I don't wont to have an error message, only true/false if the connection is ready or not.

I tried to do it with this :

$appDB = new mysqli($data[0]['MYSQL']['HOST'], $data[0]['MYSQL']['BENUTZER'], $data[0]['MYSQL']['PW'], $data[0]['MYSQL']['TABELLE']);
  if ($appDB->connect_error) 
  { $res['code']=FALSE; } 
  else 
  { 
    $appDB -> close();
    $res['code']=TRUE; 
  }

But always a error message will come up - how can i prevent this message from showing ?

"<b>Fatal error</b>:  Uncaught exception 'mysqli_sql_exception' with message 'Access denied for user ''
@'localhost' (using password: NO)' in /is/htdocs/...mypath
.php:33
Stack trace:
#0 /is/htdocs...mypath.myfile.php(33): mysqli-&gt;mysqli('', ''
, '', '')
#1 {main}
  thrown in <b>/is/htdocs/wp1076647_373QG1K1B0/butobo/module/4/code/cms_checkcon.php</b> on line <b>33
</b><br />"

Answer 1

You need to "catch" this fatal exception via a try/catch statement. This allows you to handle the fatal error and then instead of your script crashing due to it, you can output an error message and exit instead :D

PHP Exceptions: http://php.net/manual/en/language.exceptions.php

I hope this helps.

try
{
    $appDB = new mysqli($data[0]['MYSQL']['HOST'], $data[0]['MYSQL']['BENUTZER'], $data[0]['MYSQL']['PW'], $data[0]['MYSQL']['TABELLE']);
      if ($appDB->connect_error) 
      { $res['code']=FALSE; } 
      else 
      { 
        $appDB -> close();
        $res['code']=TRUE; 
      }
}
catch (Exception $e)
{
    echo 'Caught exception: ',  $e->getMessage(), "
";
    exit;
}

This will change the ERROR to a WARNING, which will not break your script. From there, if you want to hide everything EXCEPT fatal errors, you can use this line of PHP:

error_reporting(E_ERROR);