"Sandeep Hundal" <[EMAIL PROTECTED]> wrote:
> > my query is "SELECT group, name FROM $tablename WHERE approved='yes'
> > GROUP BY folder ORDER BY folder;
>
> Sorry, substitute folder for Group. I just wrote that by mistake.
>
> Data is such:
> 1 | friends | name1 | yes
> 2 | misc| name2 | yes
-Original Message-
From: Steve Werby [mailto:[EMAIL PROTECTED]]
Sent: 10 May 2001 17:04
To: Sandeep Hundal; php
Subject: Re: [PHP] i need some help with extracting data from mysql
"Sandeep Hundal" <[EMAIL PROTECTED]> wrote:
> i've got a simple problem, yet it
"Sandeep Hundal" <[EMAIL PROTECTED]> wrote:
> i've got a simple problem, yet it doesn't work, so i hope someone can
> help me out here.
>
> i've got a table with id, group, and name. each group has many names
> ofcourse.
Are these *all* of the field names?
> now i need to output the data like :
age-
From: bill [mailto:[EMAIL PROTECTED]]
Sent: 10 May 2001 14:58
To: [EMAIL PROTECTED]
Subject: Re: [PHP] i need some help with extracting data from mysql
Sandeep,
Try:
$sql="SELECT group, name FROM $tablename WHERE approved='yes'
ORDER BY group, name";
Don't know wh
>name1
>name2
>name3
>group2
>name4
>name5
thanks for any help, i'm sure this is something really straightforward, but
i can't put my finger to it
tia!
/sunny
-Original Message-
From: bill [mailto:[EMAIL PROTECTED]]
Sent: 10 May 20
Sandeep,
Try:
$sql="SELECT group, name FROM $tablename WHERE approved='yes'
ORDER BY group, name";
Don't know what fieldname "folder" is in your original SQL. You said that
the table had only id, group, & name. But I note that you must also have a
fieldname called "approved".
In any case, as
hi all,
i've got a simple problem, yet it doesn't work, so i hope someone can
help me out here.
i've got a table with id, group, and name. each group has many names
ofcourse.
now i need to output the data like :
> group1
>name1
>name2
>name3
>group2
>name4
>name5
>group3
>n
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