Forum Moderators: coopster
Supplied argument is not a valid MySQL result resource
in my PHP pages..the thing is I am following a tutorial line by line. ¦ did a search and found other web pages with this very same thing check out. > [dearones.com...]
Now surely all these people would realise they have a problem? Is it because there is something missing from the PHP installation?
Its really starting to do my head in.
You have a fault in your SQL-query.
Post the query and table structure and we will see if we can help you.
/Grodan
An SQL query looks like so
SELECT * FROM table_name WHERE table_id = 68;
And table structure means what was used to make the table you are calling in your query. For example:
CREATE TABLE pet (
name VARCHAR(20),
owner VARCHAR(20),
species VARCHAR(20),
sex CHAR(1),
birth DATE,
death DATE
);
That was taken from the mysql website manual [mysql.com]
Hope this helps some.
[added]I could be misinterpreting what GrodanBoll meant by saying table structure. [/added]