i need a small help on a query

hello,

i am trying for ages and i do not have any luck. i am trying to make a simple query.

if only all item_done == "Yes" then do something. How can i know if all rows are "Yes". (all of them needs to be yes, i shouldn`t have any no)

$query="SELECT * FROM item WHERE orderr_reference = '".$ref."' ";

			$res=mysql_query($query);
			while($row = mysql_fetch_array($res)) {
			if ($row['item_done'] == "Yes") {
			// do something
			 }
			}

PHP:

 

Hi, try this:

$query="SELECT *,
IF(!((SELECT COUNT(*) FROM item)-(SELECT COUNT(*) FROM item WHERE item_done='Yes')),1,0) ALL_YES 
FROM item 
WHERE orderr_reference='{$ref}'";
$res=mysql_query($query) or die(mysql_error());
            while($row = mysql_fetch_array($res)) {
            if (intval($row['ALL_YES'])) {
            // do something
             }
            }

PHP:

 

And what doesn't work in this code?

 

@s_ruben he needs to know if there is any record with "No", it works, but doesn't returns desired result.

OK, to simplify my previous query:

$query="SELECT *,
EXISTS(SELECT 1 FROM item WHERE orderr_reference='{$ref}' AND item_done='No') ALL_YES
FROM item 
WHERE orderr_reference='{$ref}'";
$res=mysql_query($query) or die(mysql_error());
            while($row = mysql_fetch_array($res)) {
            if (intval($row['ALL_YES'])) {
            // do something
             }
            }

PHP:

Edit: Should be EXISTS instead of NOT EXISTS -> the above code edited...

 

And you can use this code too:

$query="SELECT * FROM item WHERE orderr_reference = '".$ref."' AND item_done='No' ";

$res=mysql_query($query);

if(mysql_num_rows($res)==0){
   // do something
}

Code (markup):

 

thanks for help. i got it now.