Tough One?

[Anon]

Ok, look at this one:

I am programming a download-archive-script. Theres is one option where a user can select a rubric (like overclocking-tools, benchmark tools, etc.). When he selects that rubrik, a variable ($auswahl) containg the name of that rubric is passed to a new php page.

I made this work perfectly. Now the tough one (or maybe not):

The variable (named "$auswahl) is passed to the new page. In that page the database shall look for downloads (programs) in my table called "downloadarchiv" but only those that are within the rubric that the user selected before. All these specific downloads have to be display afterwards.

I know how to look through a database by using:

$result = mysql_query("SELECT * FROM downloadarchiv",$db);

How do I have to change the upper code to make it work the way I explained above?

I tried the following:

$result= mysql_query(SELECT FROM downloadarchiv WHERE rubrik=$auswahl,$db);

Info: "rubrik" is a variable of my table where the rubrics (benchmarks, overclocking-tools) are saved in.

Any hints? I am a bloody beginner so help is appreciated. Thank you!

[Anon]

So what was wrong with what you did? It should work fine?

scott d~

[Anon]

Scott: So you are saying that this should work uh ? Then tell me what he wants to select ? =)

$result= mysql_query(SELECT FROM downloadarchiv WHERE rubrik=$auswahl,$db);

Jens: Swede huh? In that query you are not selecting anything at all..should be something like:

$result= mysql_query(SELECT dowloads FROM downloadarchiv WHERE rubrik=$auswahl,$db);

where 'downlaods' are the column you are looking in or the whole row by using * if thats what you want..

Hope this helps...

[Anon]

I made it after a little time:

$result = mysql_query("SELECT * FROM downloadarchiv WHERE rubrik='$auswahl'",$db);

I forgot to use the '' in my query. Now it works.

Next time I will try a little harder before I start posting in this forum...sorry folks.