$dbquerymeal = "select COUNT(*) from mealtype";
$resultmeal = mysql_db_query($dbname,$dbqueryshipping1);
if(mysql_error()!=""){echo mysql_error();}
$mealcount = mysql_fetch_row($resultmeal);
echo $mealcount;
YOUR query is not stored in ,$dbqueryshipping1 but in $dbquerymeal I
believe... switch the vars and you should solve the problem.. at least
that's a perfunctory glance..
You are actually better off not using that function, anyway
You are probably better off doing something like:
$mealcountrst = mysql_query($query);
if ($mealcountrst != FALSE) {
$mealcount = mysql_fetch_row($resultmeal);
}
else {
print "yo, you messed up! " . mysql_error();
} // there are "nicer" ways to do this of course..
one is:
$mealcountrst = mysql_query($query);
$mealcountrst ? $mealcount = mysql_fetch_row($resultmeal) : print
mysql_error();
// ps... none of this code has been tested use at your own risk.
Carl Furst
Chief Technical Officer
Vote.com
50 Water St.
South Norwalk, CT. 06854
203-854-9912 x.231
-----Original Message-----
From: Phillip Blancher [mailto:[EMAIL PROTECTED]
Sent: Wednesday, July 23, 2003 3:38 PM
To: PHP List
Subject: [PHP] newbY prob
Problem with Count.
! 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?
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