RE: [PHP] URGENT - Help in mySql query
| From: | Jason Murray | Date: | Thu, 18 Oct 2001 02:34:21 +0000 |
| Subject: | RE: [PHP] URGENT - Help in mySql query | ||
| Groups: | php.general | ||
| Request: | Send a blank email to php-general+get-71435@lists.php.net to get a copy of this message | ||
Pretty simple really:
SELECT *
FROM newsletter, newsletterArticle
WHERE newsletter.ID = newsletterArticle.refID;
... will only pull out data when the condition is satisfied.
Jason
--
Jason Murray
jasonm@melbourneit.com.au
Web Developer, Melbourne IT
"Work now, freak later!"
> -----Original Message-----
> From: Manisha [mailto:manisha@aurica.com]
> Sent: Thursday, October 18, 2001 12:29 PM
> To: php-general@lists.php.net
> Subject: [PHP] URGENT - Help in mySql query
>
>
> hi,
>
> I want to write one query. I have 2 tables one as a master
> and another as
> detail.
>
> 1) newsletter - master table - ID and title is stored
> 2) newsletterArticle - details table - artID, refID (ref of
> master table)
> and body text is stored.
>
> I want to display the newsletter only if I found
> corresponding record in
> newsletterArticle.
>
> i.e if in newsletter there is an entry for ID 1, but no article in
> newsletterArticle, then I do not want to display that record.
>
> This I want to achieve in single query. Something like below
>
> select ID from newsletter where (select count(*) from
> newsletterArticle
> where refID = the ID of newsletter ???? > 0)
>
> Platform - PHP / mySql /BSDi
>
>
> Any help ? It is very urgent
>
> manisha
>
>
>
>
>
>
>
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