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
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