select count(*) questions

Anon

Hi friends;

I am using php3 and mysql 3.22, in linux redhat. I have a problem with the command select count (*). When I am executing this command inside mysql I receive the correct answer about it , like:

SELECT COUNT(*) FROM pc;

+----------+
| COUNT(*) |
+----------+
| 3 |
+----------+

but, when I use the comand in php
mysql_query("select count(*) from pc"), I receive the answer :

Anyone can tell me what is the problem. There are another structure to count values in PHP.

Regards

Anon

I assume you are printint what was returned from mysql_query. That is not correct. mysql_query returns an index that points to the result set. You have to use one of the various fetch or result functions to get the actual result...

$result=mysql_query("select count(*) from pc");
list($count=mysql_fetch_row($result);
print $count;

Anon

Rod's should work as this should as well. I think the following could give you some flexibility.

<?
$db = "database";
$host = "localhost";
$user = "mysqluser";
$pass = "mysqlpass";

mysql_pconnect($host,$user,$pass);

$query = "SELECT COUNT(*) AS num from pc";

$result = mysql_db_query($db,$query);

$mycount = mysql_result($result,0,"num");

echo $mycount;

Anon

Hi friends;

Thank you for your help. This notes are very good for my implementation.

Regards

Anon

This is very helpful information...
It helps me too!

saheed

Hi Anon,
wonderful notes. can you pls post the oracle version of the above question.

Regards

Weedpacket

After eight years you'd be lucky to find that Anon is still around 🙂.
But all the information already posted is available in the manual. For MySQL it's in the [man]mysql[/man] section, and for Oracle it's in the [man]oracle[/man] section (e.g., http://www.php.net/manual/en/oci8.examples.php).