Hi,
when I use the following code to count number of articles, I get "Resource id #2" as output. How come???
----
$num = mysql_query("SELECT COUNT(*) FROM article WHERE auid=5");
----
I don't get it. What's wrong?
Thanks
</jppr>
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Hi,
when I use the following code to count number of articles, I get "Resource id #2" as output. How come???
----
$num = mysql_query("SELECT COUNT(*) FROM article WHERE auid=5");
----
I don't get it. What's wrong?
Thanks
</jppr>
I believe the correct syntax for the mysql count is
$result=mysql_query("select * from article where auid=5");
$num=mysql_num_rows($result);
Actually the correct syntax would be
$num = mysql_query("SELECT COUNT(*) as totalnum FROM article");
$totalnum = mysql_result($num, 0);
Okay, that is probably a better way of doing it, but the way I wrote seems to work okay.
I have never seen it written like that before.. but you know what people say, "you learn something new everday" :)
It's just a matter of using less resource intensive functions whenever possible
Why does count(*) needs an alias like "as blabla" ??
</jppr>
because its a function and the result of that function needs to be assigned to something how else would you go about getting the results
I see.
I thought that when I did something like
$var = SELECT COUNT(*) blabla
that var would automatically get the result of count.
Thanks freddy
Or when getting the result, you could use:
$row = mysql_fetch_array($result)
then $counter = $row["COUNT(*)"]; // because COUNT(*) will be the field