Hey all, Here's the table definition:
CREATE TABLE bsp_options ( optionsID INTEGER NOT NULL PRIMARY KEY, modelID INT(11) NOT NULL, startyear INT(11) NOT NULL, endyear INT(11) NOT NULL, options TEXT NOT NULL, productcodesize VARCHAR(10), productdesc VARCHAR(200), pattern VARCHAR(10) ); The insert statement that creates the records looks like this: $sql="INSERT INTO bsp_options (modelID,startyear,endyear,options, productcodesize,productdesc) VALUES ($modelID, ".$arr[3].", ".$arr[4].", '".sqlite_escape_string($arr[5])."', '".$arr[7]."', '".$arr[6]."')"; After this statement is used in a loop that inserts 14,000 records I can see in SQLiteAdmin there are dozens of records that fit the criteria. SELECT * FROM bsp_options WHERE startyear=1990 Produces no results. SELECT * FROM bsp_options WHERE startyear='1990' Produces no results. SELECT * FROM bsp_options WHERE startyear like '1990%' Produces results!!! Is it treating the data like a string? Why is the like qualifier working??? -- Skip Evans Big Sky Penguin, LLC 503 S Baldwin St, #1 Madison, WI 53703 608-250-2720 http://bigskypenguin.com =-=-=-=-=-=-=-=-=-= Check out PHPenguin, a lightweight and versatile PHP/MySQL, AJAX & DHTML development framework. http://phpenguin.bigskypenguin.com/ _______________________________________________ sqlite-users mailing list sqlite-users@sqlite.org http://sqlite.org:8080/cgi-bin/mailman/listinfo/sqlite-users