Re: query help
От | brian |
---|---|
Тема | Re: query help |
Дата | |
Msg-id | 46E95C3C.9060307@zijn-digital.com обсуждение исходный текст |
Ответ на | Re: query help (volunteer@spatiallink.org) |
Список | pgsql-general |
volunteer@spatiallink.org wrote: > hello > i add more column not row for new user. i want all "last like 'J%'". I get the feeling that the result as you've laid it out is not what we all think it is. For example: >>table is >>+-------+-------+------+-------+ >>| id | one | two | three | >>+-------+-------+------+-------+ >>| first | Jack | Jill | Mary | >>| last | Ja | Ji | Ma | >>+-------+-------+------+-------+ I took that to meant that you have columns 'id', 'one', two', three', and that 'first' & 'last' are field values. However, it now seems that 'first' & 'last' are column names. If so, this makes no sense. I think what you wanted to give us was: +-------+-------+------+ | id | first | last | +-------+-------+------+ | one | Jack | Ja | | two | Jill | Ji | | three | Mary | Ma | result: +-------+-------+------+ | id | first | last | +-------+-------+------+ | one | Jack | Ja | | two | Jill | Ji | So, the query you want is, in fact: SELECT * FROM your_table WHERE last LIKE ('J%'); If that's not working for you, it's perhaps because you have rows for columns and columns for rows. > http://www.nabble.com/an-other-provokative-question---tf4394285.html > sincerely > siva > What the heck does this have to do with anything? Please don't top-post. brian
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