RE: [PHP] newbY prob

> ! am trying to count the number of items in this table. The table
> has one field in it.
>
> The code I am using is:
>
> $dbquerymeal = "select COUNT(*) from mealtype";
> $resultmeal = mysql_db_query($dbname,$dbqueryshipping1);
> if(mysql_error()!=""){echo mysql_error();}
> $mealcount = mysql_fetch_row($resultmeal);
> echo $mealcount;
>
> The result I am getting is:
>
> Query was empty
> Warning: mysql_fetch_row(): supplied argument is not a valid
> MySQL result resource in search.php
>
> Any suggestions?

Where is $dbqueryshipping1 set? I see $bdquerymeal, but not
$dbqueryshipping1.

From: "Jennifer Goodie" <goodie@apollointeractive.com>
> > ! am trying to count the number of items in this table. The table
> > has one field in it.
> >
> > The code I am using is:
> >
> > $dbquerymeal = "select COUNT(*) from mealtype";
> > $resultmeal = mysql_db_query($dbname,$dbqueryshipping1);
> > if(mysql_error()!=""){echo mysql_error();}
> > $mealcount = mysql_fetch_row($resultmeal);
> > echo $mealcount;
> >
> > The result I am getting is:
> >
> > Query was empty
> > Warning: mysql_fetch_row(): supplied argument is not a valid
> > MySQL result resource in search.php
> >
> > Any suggestions?
>
> Where is $dbqueryshipping1 set? I see $bdquerymeal, but not
> $dbqueryshipping1.

Also, once you fix that, $mealcount is going to be an array. You need to
display $mealcount[0] to get the value.

---John Holmes...