Question

我在此代码中收到错误：

try
        {
            $db = parent::getConnection();
            if($this->id == 0 )
            {
                $query = 'insert into articles (modified, username, url, title, description, points )';
                $query .= " values ('$this->getModified()', '$this->username', '$this->url', '$this->title', '$this->description', '$this->points' )";

            }
            else if($this->id != 0)
            {

                $query = "update articles set modified = CURRENT_TIMESTAMP, username = '$this->username', url = '$this->url', title = '$this->title', description = '$this->description', points = '$this->points', ranking = '$this->ranking' where id = '$this->id' ";
            }

            $lastid = parent::execSql2($query);

            if($this->id == 0 )
                $this->id = $lastid;

        }
        catch(Exception $e){
            error_log($e);
        }

我需要添加什么以便获得一些有意义的SQL错误消息？

（某些查询似乎没有得到用户名）

编辑：我收到此错误日志：

[18-Mar-2011 05:19:13] exception 'Exception' in /home1/mexautos/public_html/kiubbo/data/model.php:90
Stack trace:
#0 /home1/mexautos/public_html/kiubbo/data/article.php(276): Model::execSQl2('update articles...')
#1 /home1/mexautos/public_html/kiubbo/data/article.php(111): Article->save()
#2 /home1/mexautos/public_html/kiubbo/pages/frontpage.php(21): Article->calculateRanking()
#3 /home1/mexautos/public_html/kiubbo/pages/frontpage.php(27): FrontPage->updateRanking()
#4 /home1/mexautos/public_html/kiubbo/index.php(15): FrontPage->showTopArticles('426')
#5 {main}

谢谢，

此致

卡洛斯

Answer 1

处理此问题的最佳方法是使用数据库处理程序抛出的自定义异常。

class DatabaseErrorException{
    public function __construct( $errorMesssage, $query ){
        throw new Exception( $errorMessage . " for query: " . $query );
    }
}

所以你可以检测数据库中的错误并从那里抛出，或者在try语句中你可以：

if( $db->someError )
    throw new DatabaseErrorException( $db->someError, $query );

并且您的catch语句将变为

catch( DatabaseErrorException $e ){
    error_log( $e->getMessage( ) );
    //Or whatever handling you wish to do with it.
}

Answer 2

error_log('Failed to set record in articles table: '.
          $e->getMessage().
          "\n".$query
         );

使用SQL错误消息引发异常

2 个答案: