[SOLVED] Is it possible to...?

[dre]

Run a if statement on a query??

what i mean is i have the following piece of code:

          <?php 
$db = mysql_connect("localhost", "Username", "Password"); 
mysql_select_db("phone",$db); 
$result = mysql_query
("SELECT
pd.category,
SUM(phone_sales.quantity) * t.r AS '%'
FROM phone_sales, phone_details pd 
JOIN (SELECT 100/SUM(quantity) AS r FROM phone_details) AS t
WHERE pd.phone_id=phone_sales.phone_id
GROUP BY pd.category",$db)or die(mysql_error());

echo "<TABLE BORDER=3 BORDERCOLOR=black><TBODY BGCOLOR=white>"; 
echo"<TR><TD><B>Category</B></TD><TD><B>Category Sales Performance</B></TD></TR>";
while($myrow = mysql_fetch_array($result)) 
{ 
echo "<TR><TD>"; 
echo $myrow["category"];
echo "<TD>";
echo $myrow["%"];
echo "</TD>";
} 
echo "</TABLE>"; 
?>

this generates a table that places a list of mobile phones in one of three categories along with the sales performance (%), if a higher percentage of phones sold are category 1 then i would like to echo "recommendation" on screen.

Any help/ideas will be much appreciated!!

[JJohnsenDK]

I dont understand your request.. if a highere percentage of phones are category 1??? what do you mean? please explain your self in ohter way...

 

[dre]

Sorry,

say for example i get this result from my code:

 

category    sales(%)

    1                  30

    2                  21

    3                  12

 

i would then like a message printed on screen saying category 1 is most popular.

 

[roopurt18]

If you want to know which is most popular, slap:

 

ORDER BY '%' DESC LIMIT 1

 

onto the end of your query.

[dre]

lol knew it would be somthing simmple like that! cheers!