[support] SQL Query - "LIKE" usage

[Liviu Nicolicioiu]

Try this:

db_query('SET @select_ip="%s"', db_escape_string($ip));
db_query('SELECT name, id, lft, rgt FROM resubmt_skill_ids AS node
WHERE ( node.name LIKE CONCAT_WS( "", "%%", @select_ip, "%%" )") ORDER
BY node.lft')

Liviu.

On Fri, May 20, 2011 at 4:46 AM, Ms. Nancy Wichmann
<nan_wich at bellsouth.net> wrote:
> The problem is not the "s."  You wanted a "%" to be in the query. The
> corrects way to do that, as the API doc states, is with "%%". That's why it
> worked when you changed it.
>
>
> Nancy
>
>
>
> Injustice anywhere is a threat to justice everywhere. -- Dr. Martin L. King,
> Jr.
>
> ________________________________
> From: Austin Einter
>
> str = SELECT name, id, lft, rgt FROM resubmt_skill_ids AS node WHERE (
> node.name LIKE "%sip%") ORDER BY node.lft
>
> On debugging, looks the query went as
> SELECT name, id, lft, rgt FROM resubmt_skill_ids AS node WHERE ( node.name
> LIKE "ip%") ORDER BY node.lft
>
> Then I changed the query to
>
> SELECT name, id, lft, rgt FROM resubmt_skill_ids AS node WHERE ( node.name
> LIKE "%%sip%") ORDER BY node.lft
> Above query returns expected records.
>
> My question is , when a token starts with "s", we will be facing above
> issue.
> What is the best way to resolve it.
>
> --
> [ Drupal support list | http://lists.drupal.org/ ]
>



-- 
regards, mit freundlichen Grüßen, cu stima,
Liviu Nicolicioiu

Chief Technichal Officer
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