Jump to content

Query Problem, Where Clause


son
 Share

Recommended Posts

The following query works well in phpMyAdmin:

SELECT ageing_id, ageing_name FROM ageing WHERE ageing_id IN (SELECT age_id FROM productAge WHERE product_id = 30)

Running same query does not display the select that receives data from this query. The code for this is:

if (mysqli_num_rows($ageing_result) > 0)		{		echo "<select name=\"ageing\" id=\"ageing\">\n";		echo "<option value=\"0\">Please select...</option>";			while ($row = mysqli_fetch_array ($ageing_result, MYSQLI_ASSOC))			{			echo "<option value=\"{$row['ageing_id']}\"";					echo ">{$row['ageing_name']}</option>\n";			}		echo "</select>";

Taking out 'WHERE ageing_id IN (SELECT age_id FROM productAge WHERE product_id = 30)' from query makes it work, but I only want to have options of ages that are for selected product_id appliable (stored in productAge). As it is working without fault in phpMyAdmin (and shows right set of data) am a bit confused now as why it won't work. Any ideas?Son

Link to comment
Share on other sites

You're probably not running the same query in PHP. If you were running the same query it would return the same results, right? It's not going to return different results just because it was run in PHP (and, also, phpMyAdmin is also written in PHP).

Link to comment
Share on other sites

You're probably not running the same query in PHP. If you were running the same query it would return the same results, right? It's not going to return different results just because it was run in PHP (and, also, phpMyAdmin is also written in PHP).
The query actually uses a variable instead of '30' at the end. Still, when I echo the variable is correctly shows 30 (in line above the display of query result). The query is:
SELECT ageing_id, ageing_name FROM ageing WHERE ageing_id IN (SELECT age_id FROM productAge WHERE product_id = $pid

Son

Link to comment
Share on other sites

That query has an error in it, are you checking for errors when you run the query?
Sorry, forgot to copy the last closing bracket... Just now have it working. Replaced single quotation marks of query with double marks and works. What do you mean by checking for errors when you run the query? I run it in phpMyAdmin without problem before. Is this what you mean?Son
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share

×
×
  • Create New...