[PHP] newbY prob

[Phillip Blancher]

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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RE: [PHP] newbY prob

[Jennifer Goodie]

 ! 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.


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Re: [PHP] newbY prob

[R'twick Niceorgaw]

On 23 Jul 2003 at 15:38, Phillip Blancher wrote:

 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);

shouldn't the second argument be $dbquerymeal ?

R'twick


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Re: [PHP] newbY prob

[CPT John W. Holmes]

From: Jennifer Goodie [EMAIL PROTECTED]
  ! 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...


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RE: [PHP] newbY prob

[Carl Furst]

$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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Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.501 / Virus Database: 299 - Release Date: 7/14/2003


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