What does this error message means - Warning: Supplied argument is not a valid MySQL result resource and how do you solve it?

I think you got this message from PHP.

You get this message when you are trying to access $result variable, but there is nothing in this variable because of WRONG SQL command before.

Hjalp dette svar dig?

 Print denne artikel

Læs også

Does mysql support foreign keys?

The answer to this is two-fold. Yes. MySQL _DOES_ allow the creation of foreign key...

How can I select random rows from a table?

SELECT * FROM table_name ORDER BY RAND(); Found at...

I'm new to MySQL, where should I start?

The best source of MySQL information is the excellent on-line manual : ...

Do you provide phpMyAdmin web interface to manage the MySQL Database?

Yes, we provide that. You can login to DotNetPanel first Create MySQL Database, create MySQL...

Select statement to join tables?

For a simple join: SELECT a.*, b.*FROM tblAlpha a, tblBeta bWHERE (a.keyfield = b.foreignkey);...