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.

Esta resposta lhe foi útil?

 Imprimir este Artigo

Veja também

Select statement to join tables?

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

How can I select random rows from a table?

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

How do I delete a table from a database?

Use this SQL command : "DROP TABLE yourtablename"

Does mysql support foreign keys?

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

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

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