variable interpolation in mysql queries

[Ryan Snow]

Hi, Im kinda new to this list. Can anyone tell me what is the proper way
to interpolate my php variables into my mysql queries?

I've been trying $query = "SELECT password FROM users WHERE login='$login'";

then:
mysql_query($result);
$pd = mysql_result($result);

but I get a message that says: "Supplied Argument is not a valid
MySQL-Link resource"

any ideas, anyone?

[Bogdan Stancescu]

PS. You should consider using mysql_fetch_row() or at least
mysql_fetch_array() for perforance reasons.

B

Bogdan Stancescu wrote:

> $query="query";
> $result=mysql_query($query);
> $pd=mysql_result($result);
>
> Notice the zig-zag - $query is first on the left side, then on the
> right side, then $result is first on the left side, then on the right
> side.
>
> Bogdan

[Bogdan Stancescu]

$query="query";
$result=mysql_query($query);
$pd=mysql_result($result);

Notice the zig-zag - $query is first on the left side, then on the right
side, then $result is first on the left side, then on the right side.

Bogdan

[William Fong]

Would it be faster if all you wanted was one variable, or your SQL statement
only returned on cell?

--
William Fong - wf...@mookandblanchard.com
Phone: 626.968.6424 x210 | Fax: 626.968.6877
Wireless #: 805.490.7732 | Wireless E-mail: 80549...@mobile.att.net

----- Original Message -----
From: "Bogdan Stancescu" <m...@fx.ro>
To: <bog...@lanifex.com>
Cc: "Ryan Snow" <sn...@cosmic.utah.edu>; <php...@lists.php.net>
Sent: Tuesday, February 26, 2002 2:32 PM
Subject: Re: [PHP-DB] variable interpolation in mysql queries

: PS. You should consider using mysql_fetch_row() or at least

: >>
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: >
: >
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: --
: PHP Database Mailing List (http://www.php.net/)
: To unsubscribe, visit: http://www.php.net/unsub.php
:

[Ryan Snow]

Cool, I didn't know you could embed mysql_query(...) inside
mysql_fetch_row(...). That's kinda nice--I hate having a line each for
1-connecting, 2-selecting, 3-querying, and 4-resulting. I'll hafta try
that. Thanks.

Ryan

On Tue, 26 Feb 2002, Stewart Gateley wrote:

> First of all that should throw an error, correct syntax is mysql_result
> ($query, 0) meaning to grab the 0 index returned.
>
> I am not sure about performance wise, however I dislike mysql_result
> since if nothing is returned then you get a runtime error. Instead I
> like to use
>
> list($pd) = mysql_fetch_row (
> mysql_query (
> "select password
> from users
> where login = '$login'"
> )
> );
>
> Then you can test $pd without getting sql errors.
>
> -- Stewart

>
> --- Ryan Snow <sn...@cosmic.utah.edu> wrote:
> > Hi, Im kinda new to this list. Can anyone tell me what is the proper
> > way
> > to interpolate my php variables into my mysql queries?
> >
> > I've been trying $query = "SELECT password FROM users WHERE
> > login='$login'";
> >
> >
> > then:
> > mysql_query($result);
> > $pd = mysql_result($result);
> >
> > but I get a message that says: "Supplied Argument is not a valid
> > MySQL-Link resource"
> >
> > any ideas, anyone?
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
> > To unsubscribe, visit: http://www.php.net/unsub.php
> >
>
>

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[Stewart Gateley]

First of all that should throw an error, correct syntax is mysql_result
($query, 0) meaning to grab the 0 index returned.

I am not sure about performance wise, however I dislike mysql_result
since if nothing is returned then you get a runtime error. Instead I
like to use

list($pd) = mysql_fetch_row (
mysql_query (
"select password
from users
where login = '$login'"
)
);

Then you can test $pd without getting sql errors.

-- Stewart

--- Ryan Snow <sn...@cosmic.utah.edu> wrote: