Re: [PHP-DB] Select query with Forein key Relation

[email protected] (Evert Lammerts)
Newsgroups php.db
Message-ID <[email protected]>
It'd be consistent to use uniform naming conventions for your columns. 
E.g., an ID column is called 'id' in every table, and if you use two 
words in a column name, to separate them either by an underscore or by a 
capital letter for the second word.

In PHP a dot is an append operator - to reference a member of a certain 
namespace you use '->'. In this case you don't need it though. To get a 
column 'vname' from the table 'venue' you use:

while ($row = mysql_fetch_assoc($result)) {
    $sub = $row["venue.vname"];
}

Remember that in this case, $sub will always only hold the value of the 
last result - or of the only result if there is only one result row.

Nasreen Laghari wrote:
> Hi Evert,
>  
> What to you mean by this "If I were you I'd go for one of them and 
> apply this to all."
> also to get the value of columns do i need to do following in php
>  
> /while ($row = mysql_fetch_array($result))
>   {/
> /        $sub= $row[venue].[vname];/
> /}/
>  
> Regards
>  
>
>  
> ----- Original Message ----
> From: Evert Lammerts <[email protected]>
> To: Nasreen Laghari <[email protected]>
> Cc: [email protected]
> Sent: Wednesday, April 23, 2008 11:48:39 AM
> Subject: Re: [PHP-DB] Select query with Forein key Relation
>
> SELECT *
> FROM gig
> LEFT JOIN genre ON gig.genreId = genre.genreId
> LEFT JOIN venue ON gig.venueID = venue.vid
> WHERE gig.gigid = $gigdetail
>
> I'd replace the dash with [table].[columnames]. Also, you're using four
> different naming conventions in your columns - gigid, genreId, venueID
> and vid. If I were you I'd go for one of them and apply this to all.
>
> Evert
>
> Nasreen Laghari wrote:
> > Hi,
> > 
> > I have a table which contains 2 foreign key relation columns. I'm 
> trying to get all columns from main table as well as all column from 
> those 2 foreign key relation tables.
> > 
> > The query i'm using is :
> > 
> > select * from gig where gig.gigid = $gigDetail LEFT JOIN genre ON 
> gig.genreId=genre.genreId LEFT JOIN venue ON gig.venueID = venue.vid 
> ORDER BY gig.gigid";
> > 
> > is this query OK?
> > 
> > I know  how to get value from gig table colums but how could i get 
> value of columns from venue table?
> > 
> > Regards
> > 
> > Nasreen
> > 
> > 
> > 
> >
> >
> >
> > Be a better friend, newshound, and know-it-all with Yahoo! Mobile. 
> Try it now.
> >
> >
> >      
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