Re: [sqlite] DATETIME data type

2008-02-28 Thread Trey Mack
Store it in '-mm-dd' format, or use the julian date that's suggested at: http://www.sqlite.org/cvstrac/wiki?p=DateAndTimeFunctions Yong Zhao wrote: > It seems that sqlite3 does not support DATETIME data type. > > If I have the following data in table t1, how do I select people who is > older

Re: [sqlite] DATETIME data type

2008-02-28 Thread Igor Tandetnik
"Yong Zhao" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > It seems that sqlite3 does not support DATETIME data type. > > If I have the following data in table t1, how do I select people who > is older than certain date? > > create table t1(dob text, name text); > insert into

Re: [sqlite] DATETIME data type

2008-02-28 Thread Neville Franks
Friday, February 29, 2008, 8:29:16 AM, you wrote: YZ> It seems that sqlite3 does not support DATETIME data type. YZ> If I have the following data in table t1, how do I select people who is YZ> older than certain date? YZ> create table t1(dob text, name text); YZ> insert into t1('11/12/1930',

Re: [sqlite] DATETIME data type

2008-02-28 Thread John Stanton
Try using the Sqlite date functions. Yong Zhao wrote: > It seems that sqlite3 does not support DATETIME data type. > > If I have the following data in table t1, how do I select people who is > older than certain date? > > create table t1(dob text, name text); > insert into t1('11/12/1930',

Re: [sqlite] DATETIME data type

2008-02-28 Thread Steven Fisher
On 28-Feb-2008, at 1:29 PM, Yong Zhao wrote: > It seems that sqlite3 does not support DATETIME data type. > > If I have the following data in table t1, how do I select people who > is > older than certain date? Use -MM-DD instead of M/D/Y. Available formats described here under Time

[sqlite] DATETIME data type

2008-02-28 Thread Yong Zhao
It seems that sqlite3 does not support DATETIME data type. If I have the following data in table t1, how do I select people who is older than certain date? create table t1(dob text, name text); insert into t1('11/12/1930', 'Larry'); insert into t1('2/23/2003', 'Mary'); select * from t1 where