[Resolved] Outputting the value from a Count query

ExpendableDecoy

Hi there. 🙂

I know this is reletively simply but I cannot seem to get it right.

I need to find out how many rows in the column 'clan_name' equal 'Expendable Decoys'.

						$sql = " SELECT COUNT(*) FROM guild_db
							WHERE clan_name = 'Expendable Decoys' ";
						if( !($result = $db->sql_query($sql)) )
						{
							message_die(GENERAL_ERROR, 'Could not query count for info page', '', __LINE__, __FILE__, $sql);
						}

If that is correct I then need to figure out how to output that value into a seperate file which holds the HTML part of the code.

'U_CLAN_TOTAL_MEMBERS' => [what goes here?],

Any help would be appriciated. 😉

Thanks!. 🙂

TheDefender

$sql = mysql_query("SELECT * FROM guild_db WHERE clan_name = 'Expendable Decoys'");
$countemup = mysql_num_rows($sql);
echo $countemup;

You can then take that $countemup variable and hide it in a form you are posting to the next page, or include it in a link as a get function... That's up to you.

ExpendableDecoy

Thanks. 🙂

That works great if the name clan name is hard coded, but when I try to replace the name 'Expendable Decoys' with a variable:

$sql = mysql_query("SELECT * FROM guild_db WHERE clan_name = '$clan_name' ");
$countemup = mysql_num_rows($sql);

...it doesn't work. I have the variable set earlier in the script and others are working ok off it.

Should it be looking something like this instead?.

' . $clan_name . '

dnast

It looks okay. You should try echoing the query and using [man]mysql_error[/man]

i.e.

$query = "SELECT * FROM guild_db WHERE clan_name = '$clan_name'";
echo $query;
$sql = mysql_query($query)
  or die(mysql_error());
$countemup = mysql_num_rows($sql);

It'll make it easier to debug.

TheDefender

You should be able to do it like dnast wrote:

$query = "SELECT * FROM guild_db WHERE clan_name = '$clan_name'";

or you could do it like so:

$query = "SELECT * FROM guild_db WHERE clan_name = '".$clan_name."'";

Either way, you should debug it like dnast suggested by echoing out the variable to make sure it has a value, and using the mysql_error().

ExpendableDecoy

Ok, thanks guys, all working good now. 🙂

TheDefender

Good deal, be sure to mark this thread resolved using the "Mark Thread Resolved" link at the bottom of the page.