Re: [PHP] Warning Extract
I meant there is no row in table members where username='$logname',
whatever value $logname contains.
Don't forget to run session_start() at the begining of every script.
Andy wrote:
logname is not in my database.
It is a session variable used to hold the members login name.
In the form it is session_register('logname'); which is then set to a
variable with this code:
if ($num2 > 0) { // password is correct
$auth="yes";
$logname=$fusername;
$today = date("Y-m-d h:m:s");
$sql = "INSERT INTO login (username,loginTime)
VALUES ('$logname','$today')";
does that make any sense?
Andy
"Marek Kilimajer" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
Andy wrote: (and Marek fixed)
$connection = mysql_connect($host, $user,$password)
or die ("Couldn't connect to server.");
$db = mysql_select_db($database, $connection)
or die ("Couldn't select database.");
$sql = "SELECT name FROM members
WHERE username='$logname'";
$result = mysql_query($sql)
or die("Couldn't execute query 1.");
if($row = mysql_fetch_array($result,MYSQL_ASSOC)) extract($row);
if $logname is not in members table, then your query returns zero rows
and mysql_fetch_array returns false. First argument to extract (see
manual) must be an array, but is false in your case, and php warns you
about that.
echo "
New Member Welcome etc
Thank you all again
Andy
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Re: [PHP] Warning Extract
logname is not in my database.
It is a session variable used to hold the members login name.
In the form it is session_register('logname'); which is then set to a
variable with this code:
if ($num2 > 0) { // password is correct
$auth="yes";
$logname=$fusername;
$today = date("Y-m-d h:m:s");
$sql = "INSERT INTO login (username,loginTime)
VALUES ('$logname','$today')";
does that make any sense?
Andy
"Marek Kilimajer" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
>
>
> Andy wrote: (and Marek fixed)
>
> >$connection = mysql_connect($host, $user,$password)
> > or die ("Couldn't connect to server.");
> >$db = mysql_select_db($database, $connection)
> > or die ("Couldn't select database.");
> >$sql = "SELECT name FROM members
> > WHERE username='$logname'";
> >$result = mysql_query($sql)
> > or die("Couldn't execute query 1.");
> >if($row = mysql_fetch_array($result,MYSQL_ASSOC)) extract($row);
> >
> if $logname is not in members table, then your query returns zero rows
> and mysql_fetch_array returns false. First argument to extract (see
> manual) must be an array, but is false in your case, and php warns you
> about that.
>
> > echo "
> >New Member Welcome etc
> >
> >Thank you all again
> >
> >Andy
> >
> >
> >
> >
> >
>
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Re: [PHP] Warning Extract
Andy wrote: (and Marek fixed)
$connection = mysql_connect($host, $user,$password)
or die ("Couldn't connect to server.");
$db = mysql_select_db($database, $connection)
or die ("Couldn't select database.");
$sql = "SELECT name FROM members
WHERE username='$logname'";
$result = mysql_query($sql)
or die("Couldn't execute query 1.");
if($row = mysql_fetch_array($result,MYSQL_ASSOC)) extract($row);
if $logname is not in members table, then your query returns zero rows
and mysql_fetch_array returns false. First argument to extract (see
manual) must be an array, but is false in your case, and php warns you
about that.
echo "
New Member Welcome etc
Thank you all again
Andy
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[PHP] Warning Extract
Help AGAIN!!!
Thank you for all the help so far, i am close to getting this members
section working correctly now i think!
I think i would have given up by now if it were not for the help i get from
you all, thank you.
I have a warning that i don't understand:
Warning: extract() expects first argument to be an array in
/home/.sites//New_member.php on line 23
I also find that although i get through to the new members page OK when
filling in the form it does not write to the mysql database with the new
details, yet if i enter a username that already exists it tells me, so the
php and mysql must be interacting.
Here is the code for the warning message, hope it helps:
$connection = mysql_connect($host, $user,$password)
or die ("Couldn't connect to server.");
$db = mysql_select_db($database, $connection)
or die ("Couldn't select database.");
$sql = "SELECT name FROM members
WHERE username='$logname'";
$result = mysql_query($sql)
or die("Couldn't execute query 1.");
$row = mysql_fetch_array($result,MYSQL_ASSOC);
extract($row);
echo "
New Member Welcome etc
Thank you all again
Andy
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