添加链接
link之家
链接快照平台
  • 输入网页链接,自动生成快照
  • 标签化管理网页链接
Collectives™ on Stack Overflow

Find centralized, trusted content and collaborate around the technologies you use most.

Learn more about Collectives

Teams

Q&A for work

Connect and share knowledge within a single location that is structured and easy to search.

Learn more about Teams

How do I know if there's an error if I did $db = new SQLite3("somedb.db"); in PHP? Right now the $db doesn't really give me any sort of error?

I can check for file existance, but I'm unsure if there could be any other errors when I open a connection.

You should enable exceptions and instantiate in a try-catch block.

It is not obvious from the documentation but if you use the constructor to open the database it will throw an exception on error.

Further if you set the flag SQLITE3_OPEN_READWRITE in the second argument then it will also throw an exception when the database does not exist (rather than creating it).

class Database extends SQLite3
    function __construct($dbName)
        $this->enableExceptions(true);
            parent::__construct($dbName, SQLITE3_OPEN_READWRITE );
        catch(Exception $ex) { die( $ex->getMessage() ); }
        

Thanks for contributing an answer to Stack Overflow!

  • Please be sure to answer the question. Provide details and share your research!

But avoid

  • Asking for help, clarification, or responding to other answers.
  • Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.