How to eliminate duplicates in drop-down?

zzz

I am trying to build a dynamic selection tool for publications that will have 3 steps:

Step 1 - where user first has to select publication

Step 2 - select volume (there could be more then one volumes in a publication)

Step 3 - select number (multiple numbers per volume)

I try to put results into a drop-down menu, but due to the fact that I use one table where I store entries for each issue, when I get results into a drop-down I do it via foreach loop and results get all row entries and sometimes repeat.

Here are the fields I use:

id
publication parent
volume_number
issue_number

So when I get volume_number of the rows that have the same publication parent I get results as:

<option>1
<option>2
<option>2
<option>3
<option>2
<option>3

Question: How can i show each volume_number just once, even if it appears in the table multiple times?

pohopo

I believe all you need to do is use DISTINCT in your select statement

SELECT DISTINCT pub_parent
FROM your_table

bradgrafelman

If you use a foreach loop to go through and store all the output into a variable before you actually echo() or print() it, you could do a simple eregi() to see if you've already added a volume to the output variable...

foreach (...) {
  if(eregi("<option>$volume_number", $output)) continue; else $output .= "<option>$volume_number\n";
}

etc. (I used '...' where you should already have your foreach statement set up, just replace with what you have)

I believe that answers your question on duplicate entries... if not, post some code for me/the community to work with, and we can go from there.

zzz

Wow, two great replies in such a short time. I've tried DISTINCT and it works great!

bradgrafelman

nods and gives props to pohopo