On Sun, Jun 27, 2010 at 4:08 AM, Tanel Tammik keevit...@gmail.com wrote:
Hello,
how to select only if value is present?
$query = $db-query(select menus.id, menus.name,
case
when panels.id is not null then '1'
end as hiddenpanel
from . \DB_MENUS . as menus
On Monday 28 June 2010 09:49:55 Andrew Ballard wrote:
On Sun, Jun 27, 2010 at 4:08 AM, Tanel Tammik keevit...@gmail.com wrote:
Hello,
how to select only if value is present?
$query = $db-query(select menus.id, menus.name,
case
when panels.id is not null then '1'
On Sunday 27 June 2010 22:12:41 Brandon Rampersad wrote:
no
At least smack me and give us an explanation. :-)
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Blessings,
David M.
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On Mon, Jun 28, 2010 at 10:27 AM, David McGlone da...@dmcentral.net wrote:
Tanel, we both learned something. I didn't fully understand join myself yet,
but I think I do now.
but let me ask this if the join wasn't there would an if statement like I
mentioned have worked?
Blessings,
David M.
On Monday 28 June 2010 11:14:53 Andrew Ballard wrote:
On Mon, Jun 28, 2010 at 10:27 AM, David McGlone da...@dmcentral.net wrote:
Tanel, we both learned something. I didn't fully understand join myself
yet, but I think I do now.
but let me ask this if the join wasn't there would an if
On Sunday 27 June 2010 04:08:24 Tanel Tammik wrote:
Hello,
how to select only if value is present?
$query = $db-query(select menus.id, menus.name,
case
when panels.id is not null then '1'
end as hiddenpanel
from . \DB_MENUS . as menus
left join
no
On Sun, Jun 27, 2010 at 8:29 PM, David McGlone da...@dmcentral.net wrote:
On Sunday 27 June 2010 04:08:24 Tanel Tammik wrote:
Hello,
how to select only if value is present?
$query = $db-query(select menus.id, menus.name,
case
when panels.id is not null then
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