Jump to content

Checking Existence


WarKirby

Recommended Posts

I'm looking for a simple way to check if something exists in a table.

 

For example, my products table has articleNumber as the primary key. I want to check if a given product exists, so I'm trying to check if a supplied articlenumber is found in the database

 

What I'm thinking of doing:

 

$var = mysql_query("SELECT FROM tblProducts WHERE articleNumber=".$TestProduct);

if (isset($var))

...

 

Am I correct in thinking that isset($var) will return false if no matching record was found in the table? or will I need to do this some other way ?

Link to comment
Share on other sites

For this situation isset does not work, cause either way $var is going to be set to a resource or a boolean. mysql_query will tell you what it will return. Returns false on error and Resource on a successful query.  Where isset is valid is usually in POST / GET data to test if you received a certain data item and allows you to assign it without causing a notice error.

Link to comment
Share on other sites

Afraid I'm having  an issue with this solution. unsolving for now :(

 

This is the function I wrote, seems simple enough:

function isValidProduct($input)
{
$test = mysql_query("SELECT FROM ".$dbProductsTable." WHERE articleNumber=".QuoteSmart(trim($input)));
if (mysql_num_rows($test) > 0)
{
	return TRUE;
}
else
{
	return FALSE;
}
}

 

it's returning an error when it gets called though:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /usr/www/users/warkirby/testing/db.inc on line 41

 

any idea what might be wrong here ?

Link to comment
Share on other sites

Change the query line to this:

 

$test = mysql_query("SELECT FROM ".$dbProductsTable." WHERE articleNumber=".QuoteSmart(trim($input))) or trigger_error("Query Failed: " . mysql_error());

 

The issue is going to be with $dbProductsTable, as it has not been defined within the scope of the function. Check out Variables Scope for more indepth information.

Link to comment
Share on other sites

Change the query line to this:

 

$test = mysql_query("SELECT FROM ".$dbProductsTable." WHERE articleNumber=".QuoteSmart(trim($input))) or trigger_error("Query Failed: " . mysql_error());

 

The issue is going to be with $dbProductsTable, as it has not been defined within the scope of the function. Check out Variables Scope for more indepth information.

 

I'm not stupid  :P

Of course it's defined. This is just a function out of the whole script. That variable is defined at the top of the script,  not shown in this snippet.

 

See anything else that might be causing it ?

 

Edit: Oops, I missed the other part syou said. trying it now

Link to comment
Share on other sites

I'm not stupid  :P

Of course it's defined. This is just a function out of the whole script. That variable is defined at the top of the script,  not shown in this snippet.

 

I never said you were.

 

Yes, it is a function, and I do not doubt you have it defined. But it is not defined in the scope of the function, read my whole line for more information on Variables Scope (note that was a link above).

Link to comment
Share on other sites

Ok, gave that a try. the resulting extra error:

 

Notice: Query Failed: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM WHERE articleNumber='\'4v3vc3y2c\''' at line 1 in /usr/www/users/warkirby/testing/db.inc on line 40

 

But I swear it's defined. I have this at the top of the file

 

$dbProductsTable	= 'tblProducts';

 

Possibly relevant, this function is in a file which I'm referencing from my main script, via require_once. As far as I understand from how it works, that shouldn't affect anything though.

Link to comment
Share on other sites

Ok, I had a look, you were right.

 

I amended my query to this, but it still doesn't seem to work:

$test = mysql_query("SELECT FROM ".$GLOBALS['$dbProductsTable']." WHERE articleNumber=".QuoteSmart(trim($input))) or trigger_error("Query Failed: " . mysql_error());

 

Link to comment
Share on other sites

Ok, read what I write please and look at the link.

 

Variables Scope means that you can only use variables that are:

A. Global

B. Super Global

C. Defined in the Function

D. Passed in as a parameter.

 

Given this, you do not have $dbProductsTable defined, passed in or set as a global. Thus it is not set in the function. To fix this error, you either have to define $dbProductsTable as a global or pass it in as a parameter. Here is an example of each:

 

function isValidProduct($input, $dbProductsTable) {
   $test = mysql_query("SELECT FROM ".$dbProductsTable." WHERE articleNumber=".QuoteSmart(trim($input))) or trigger_error("Query Failed: " . mysql_error());
   if (mysql_num_rows($test) > 0)  {
      return TRUE;
   }

   return FALSE;
}

 

 

// where the $dbProductsTable variable is defined:
global  $dbProductsTable;
$dbProductsTable = 'products';

function isValidProduct($input) {
   global  $dbProductsTable;
   $test = mysql_query("SELECT FROM ".$dbProductsTable." WHERE articleNumber=".QuoteSmart(trim($input))) or trigger_error("Query Failed: " . mysql_error());
   if (mysql_num_rows($test) > 0)  {
      return TRUE;
   }

   return FALSE;
}

 

Or a 3rd option, making it a CONSTANT with define

 

// where the $dbProductsTable variable is defined change to:
define('DB_PRODUCTS_TABLE', 'products');

function isValidProduct($input) {
   $test = mysql_query("SELECT FROM ". DB_PRODUCTS_TABLE ." WHERE articleNumber=".QuoteSmart(trim($input))) or trigger_error("Query Failed: " . mysql_error());
   if (mysql_num_rows($test) > 0)  {
      return TRUE;
   }

   return FALSE;
}

 

Please read this thoroughly: http://php.net/manual/en/language.variables.scope.php

Link to comment
Share on other sites

Does this apply to functions too?

 

Because I seem to have gotten that part working but I'm still getting an error from the same line, just apparently at a different part of it now.

 

I used the function exactly as you put it in the second example, as well as the global definition at the top.

 

global $dbProductsTable;
$dbProductsTable	= 'tblProducts';

//stuff in between

function isValidProduct($input) {
   global  $dbProductsTable;
   $test = mysql_query("SELECT FROM ".$dbProductsTable." WHERE articleNumber=".QuoteSmart(trim($input))) or trigger_error("Query Failed: " . mysql_error());
   if (mysql_num_rows($test) > 0)  {
      return TRUE;
   }

   return FALSE;
}

 

The errors I'm now getting:

 

Notice: Query Failed: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM tblProducts WHERE articleNumber='\'aaaaaaaaa\''' at line 1 in /usr/www/users/warkirby/testing/db.inc on line 43

 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /usr/www/users/warkirby/testing/db.inc on line 44

 

The string there is just 9 a's in a row, for testing.

I can see the table name is going in properly, do that's now working, but I'm not sure what else isn't. QuoteSmart is a function in the same file to protect against injection attacks, which reads:

 

// Quote variable to make safe
function QuoteSmart($value)
{
// Stripslashes
if (get_magic_quotes_gpc())
     {
     $value = stripslashes($value);
     }
// Quote if not integer
if (!is_numeric($value) || $value[0] == '0')
     {
     $value = "'" . mysql_real_escape_string($value) . "'";
     }
return $value;
}

 

any idea what's wrong now? Did I misunderstand something again ?

Link to comment
Share on other sites

This thread is more than a year old. Please don't revive it unless you have something important to add.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.