PHP SQLite3 error?

Question

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.

rkulla · Accepted Answer · 2011-06-25 21:57:33Z

7

Try:

echo $db->lastErrorMsg();

answered Jun 25, 2011 at 21:57

rkulla

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Gavin · Accepted Answer · 2012-10-16 17:48:34Z

7

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);

        try
        {
            parent::__construct($dbName, SQLITE3_OPEN_READWRITE );
        }
        catch(Exception $ex) { die( $ex->getMessage() ); }
    }

answered Oct 16, 2012 at 17:48

Gavin

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