From: Gavin Towey (gtoweyffn.com)
Date: Mon Jun 01 2009 - 20:06:53 CDT

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A LEFT OUTER JOIN in that query in not necessary. An inner join should be used.

-----Original Message-----
From: Shiv [mailto:shivamygmail.com]
Sent: Saturday, May 30, 2009 10:18 PM
To: bharani kumar
Cc: mysql
Subject: Re: Left join query

Hi,
   Along with tables, you should also provide details on how they are
related. Assuming "Code" is unique in both tables and left joined on
Airport table, you can do something like this

SELECT A.Code, C.Code, A.SlNo, C.SlNo, A.Name, C.Location, A.status,
C.status, C.type
FROM Airport A
LEFT OUTER JOIN Common C
ON A.Code = C.Code
WHERE A.status = 1 and C.status = 1 and C.type = 'airport'
Cheers,
Shiva

On Fri, May 29, 2009 at 8:52 PM, bharani kumar <
bharanikumariyerphpgmail.com> wrote:

> Airport table
>
> SlNo Name Code Auth Location status
> 1 Chennai CHN Yes India 1
> 2. Hydarabed HYD Yes India 0
> 3 walker WAK Yes uk 1
>
>
> common table
>
> SlNo Name Code Type Location
> status
> 1 Chennai CHN Airport India
> 1
> 2. guest Road GR Seaport India 1
> 3. Hyderabad HYD Airport
> India 0
> 4. John garden JG Seaport India 0
> 5 walker WAK airport uk 1
>
>
> Hi All
>
> Can u please tell the query for the above table ,
>
> Display the record which satisfy below condtions ,
>
> 1.Need to display airport name , Where airport status = 1 and common
> table status = 1 and common table type = airport
>
> Output somthing like below
>
> AirportCode CommonCode AirportSlNo CommonSlNo AirportName
> CommonLocation Status Type
>
> CHN CHN 1 1
> Chennai india 1 airport
>
> WAK WAK 3 5
> walker uk 1 airport
>
>
> Any idea ,,,,,,,,,
>
>
>
>
> Thnaks
>
>
>
>
>
>
>
> --
> Regards
> B.S.Bharanikumar
> http://php-mysql-jquery.blogspot.com/
>
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>
>

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