QuePID

This is very helpful as I desire to produce code which works equally well across platforms.

Switching to / instead of \ as directory separators worked like a charm. Thanks guys!

What you suggested worked except for this line (#14): $speciesname='C:\herbariumdb.eku.edu\images\'.$family.'\'.$genus.'\'.$species; which gives error: PHP Parse error: syntax error, unexpected T_STRING in C:\herbariumdb.eku.edu\qpadmin\foldercreationtest.php on line 14

Here is the code: <?php //Define variables for a test drive of folder detection/creation. $family="Family"; $genus="Genus"; $species="Species"; //Create variable to hold family name along with path and make the folder should it not exist. $familyname="C:\herbariumdb.eku.edu\images\$family"; if(!is_dir($familyname)) mkdir ('$familyname',777); echo "$familyname<br>"; //Create variable to hold genus name along with path and make the folder should it not exist. $genusname='C:\herbariumdb.eku.edu\images\$family\$genus'; if(!is_dir($genusname)) mkdir ("$genusname",777); echo "$genusname<br>"; //Create variable to hold genus name along with path and make the folder should it not exist. $speciesname='C:\herbariumdb.eku.edu\images\$family\$genus\$species'; if(!is_dir($species)) mkdir ("$species",777); echo "$speciesname<br>"; ?> Here is the output: C:\herbariumdb.eku.edu\images$family C:\herbariumdb.eku.edu\images\$family\$genus C:\herbariumdb.eku.edu\images\$family\$genus\$species The $vars should be their values not their names, but something is askew.

Hi, I am wanting to check to see if a folder exists, if not create it. PHP on a windows machine so I would like to know how to handle the folder separators. I am wanting to use a path which includes the drive letter, such as... $foldername ="C:\folder1\images\$checkfolder" I was thinking about using something like this: if(!is_dir($foldername)) mkdir ("$foldername",777); What would be the best way of tackling this using windows paths?

Worked like a charm! Thanks.

It's not working for some reason here is the code: if ($date=NULL) $date="0000-00-00"; if ($col_datet=NULL) $col_datet="0000-00-00"; Here is the error given: Incorrect date value: '' for column 'Col_Datet' at row 1 What I am doing is retrieving a record from a mysql database in which the date and col_datet columns contain either a date or NULL. When I go to update the query with either of those two fields in which they contain a null the update query fails. So what I am wanting to do with PHP is to use a default date of 0000-00-00 instead of a NULL for the update query.

How can I check a variable to see if it is NULL and if so set to 0000-00-00? I've tried: if ($var IS NULL) THEN $var='0000-00-00'; to no avail.

I've tried it with a single SET and still gives the same error.

Hi, I am building an update query which will update a record in a table, but I am getting an error which doesn't make sense to me. Here is my code: $query = "UPDATE iknew SET ID='$id', SET Herbarium='$herbarium', SET Taxon='$taxon', SET Bar_Code='$bar_code', SET GUID='$guid', SET Private='$private', SET Image='$image', SET Kingdom='$kingdom', SET Genus='$genus', SET Species='$species', SET Variety_Name='$variety_name', SET Taxon_Author='$taxon_author', SET Accession='$accession', SET State='$state', SET County='$county', SET Habitat='$habitat', SET Locality='$locality', SET Latitude='$latitude', SET Longitude='$longitude', SET Quadrangle='$quadrangle', SET Collector='$collector', SET Collection_Number='$collection_number', SET Country='$country', SET Family='$family', SET Annotation='$annotation', SET Col_Date='$col_date', SET Comments='$comments', SET Natural_Area='$natural_area', SET Col_datet='$col_datet', SET Symbol='$symbol' WHERE ID='$id' LIMIT 1"; Here is a sample update query: UPDATE iknew SET ID='62547', SET Herbarium='EKY', SET Taxon='Salix exigua Nutt.', SET Bar_Code='3288', SET GUID='', SET Private='1', SET Image='', SET Kingdom='Plantae', SET Genus='Salix', SET Species='exigua', SET Variety_Name='', SET Taxon_Author='Nutt.', SET Accession='EKY3288', SET State='Kentucky', SET County='Madison', SET Habitat='stream bank', SET Locality='Hagan Mill Road at bridge over Silver Creek.', SET Latitude=' 0', SET Longitude=' 0', SET Quadrangle='', SET Collector='John Hornback', SET Collection_Number='38', SET Country='USA', SET Family='Salicaceae', SET Annotation='', SET Col_Date='1976-04-18', SET Comments='RRP', SET Natural_Area='', SET Col_datet='', SET Symbol='SAEX' WHERE ID='62547' LIMIT 1 And here is the error given: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'SET Herbarium='EKY', SET Taxon='Salix exigua Nutt.', SET Bar_Code='3288', SET' at line 3 Any help would be quite appreciated.

Worked like a charm, and was quite fast. How would I do similar for a count of all records within the table?

When I echo the results of your query string I get "Resource id #6", and when I use mysql_num_rows($results) I get a value of 1 which isn't correct (should be 224). What is the type of data returned in the results of the query?

Hi, I have a database with roughly 73k rows (records). I am using a query string such as: SELECT DISTINCT Column FROM databasename I am then using mysql_num_rows($results) to count the number of distinct rows. This method works, but it is slow. Is there a faster way of doing this perhaps internal to the mysql without having to send the entire results to the PHP when I only need a count?