[RESOLVED] How can I create this sql query?

thara

hi.. everyone..

This is a sql query problem that I have uncounted and have been in the same script for hours. But Still I couldn't get it to work.

I need to get a image from database and its type should be one of a value from my $designation array. The query should check $designation array's value with database and if its match with one of a value in database then query can retrieve one row.

This is my designation array. its value I use for image type in my db.

$designation = array ( 'Managing Director', 'Manager', 'Director', 'The Principal', 'Deputy Principal', 'Proprietor', 'Assistant Manager', 'Head Master');

My problem is how can I check these value exit in my table and how can I make a query for this. In query where condition is confusing to me...

any comments are greatly appreciated.
Thank you..

Weedpacket

...WHERE what_ever_the_column_is_called IN (list of possible values).

bradgrafelman

Which DBMS are you using? I'll assume MySQL for now...

Sounds like the IN() function is what you need, giving you a query that contains something like:

WHERE designation_column IN ('Value1', 'Value2', ..., 'ValueN')

To create that comma-separated list of values, you could use [man]implode/man on your array above, e.g.:

"('" . implode("', '", $desgination) . "')"

thara

thanks for replies..

Then Can I use it like this?

		// Make the designaton array for check Image Type:
		$desig = array ( 'Managing Director', 'Manager', 'Director', 'The Principal', 'Deputy Principal', 'Proprietor', 'Assistant Manager', 'Head Master');	 

	$q = "SELECT image_id FROM institute_images 
		 WHERE institute_id = $instituteId AND image_type IN ('" . implode("', '", $desig ) . "')"; 

	$r = mysqli_query ( $dbc, $q );			
	$numRows = mysqli_num_rows($r);

Weedpacket

Have you tried to?

thara

Yes.. Now my problem is solved.... Thanks for all answers.