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 <[EMAIL PROTECTED]> 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?
>
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> PHP Database Mailing List (http://www.php.net/)
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>
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