php - 如何捕获 pg_connect() 函数错误？

Question

pg_connect() 以表格格式显示错误。不是以表格格式显示错误消息，而是需要错误消息警报。

Error Message
Warning: pg_connect() [function.pg-connect]: Unable to connect to PostgreSQL server: FATAL: password authentication failed for user "test" in /home/test/public_html/QueueManager/Modules/Database.php on line 41

如果以表格格式显示错误之后。
执行 pg_connect() 后抛出异常。
但是它不起作用。

代码

function connect()
{
  $HOST = $GLOBALS[Database_Conn][Db_Host];     # Host name 
  $USER = $GLOBALS[Database_Conn][Db_User];     # database user name 
  $DBNAME = $GLOBALS[Database_Conn][Db_Name];   # name of the database
  $PASSWORD = $GLOBALS[Database_Conn][Db_Pass]; # password the database user.

  try 
  {
    $conn = pg_connect("host=$HOST dbname=$DBNAME user=$USER ".
                       "password=$PASSWORD sslmode=disable");
    if(!$conn)
    {
      throw new Exception("Database Connection Error");
    }
    return $conn;
  }
  catch (Exception $e) 
  {
    print <<<_HTML_
    <script> alert('Caught exception'); 
    </script> _HTML_;
    die();
  }
}

请给我解决方案

Answer 1

pg_connect 不会抛出异常，因此您必须像下面这样转换为异常。

function exception_error_handler($errno, $errstr, $errfile, $errline ) {
    throw new ErrorException($errstr, $errno, 0, $errfile, $errline);
}
set_error_handler("exception_error_handler");

try {
    $conn=@pg_connect("host=dbhost user=dbuser dbname=db password=dbpass");
} Catch (Exception $e) {
    Echo $e->getMessage();
}

请引用这个更多的细节

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