My guess is that there is a trailing space in the record. Try the following: sqlite> select save_id ||'<' from ae_objects where save_id like 165; 165<
And see where the "sean" save_id field ends. Regards, Noah -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Sean Riley Sent: Friday, July 11, 2008 5:04 PM To: General Discussion of SQLite Database Subject: Re: [sqlite] problem with simple select Thanks for the quick response. My application is using 3.4.1, but I grabbed the 3.5.9 executable and got the same thing. SQLite version 3.5.9 Enter ".help" for instructions sqlite> select * from ae_objects; 20086|sean|1|0|5.43301269412041|4.43301269412041|0.0|0.0|1|0|165|2 sqlite> select * from ae_objects where save_id=165; sqlite> select * from ae_objects where save_id like 165; 20086|sean|1|0|5.43301269412041|4.43301269412041|0.0|0.0|1|0|165|2 So I tried what your code from below and it worked for me in a new database. Strange thing though, if I do the insert from your code into my existing database, then that new record shows up when I do: sqlite> select * from ae_objects where save_id=165; But the existing record (the "sean" one) does not! CONFIDENTIALITY NOTICE: This message may contain confidential and/or privileged information. If you are not the addressee or authorized to receive this for the addressee, you must not use, copy, disclose, or take any action based on this message or any information herein. If you have received this message in error, please advise the sender immediately by reply e-mail and delete this message. Thank you for your cooperation. _______________________________________________ sqlite-users mailing list sqlite-users@sqlite.org http://sqlite.org:8080/cgi-bin/mailman/listinfo/sqlite-users
