counting all values in column without creating a new column

Smudly

Hi,

I have downloads that are available for users. They can click the links to download the file. Once they do this, I have a query that updates the number of downloads that file has received. I need to create a query that adds up all the rows Total Downloads.

So for example: i have 3 rows:
Name - Total Downloads
First - 2
Second - 23
Third - 7

Using the query for these rows would give me an amount of 32.

Any idea how to do this?

I don't want to have to create a new column to find the total number of downloads. Any other ways?

laserlight

Use the SUM aggregate function in SQL, perhaps? 🙂

Smudly

Alright, I've been trying that and not getting the right results.

Here is the code I'm using.

$downloadscount = mysql_query("SELECT SUM(timesdownloaded) FROM sheets WHERE timesdownloaded>='0'");
$totaldownloads = mysql_num_rows($downloadscount);

My sql table looks like:

id int(11) No
artist varchar(50) No
title varchar(50) No
url varchar(2083) No
timesdownloaded int(11) No
lastdownloaded date No
date date No

Right now I have 20 rows in there. Most are set to 0, but five have a value of at least 1

thanks

laserlight

Well, the result of that query is a single row with a single column containing the sum of fields under the timesdownloaded column. Clearly, finding out the number of rows does not give you $totaldownloads. You need to retrieve the value of that field in the result set.

Smudly

Is this correct? (bare with me lol)

$downloadscount = mysql_query("SELECT SUM(timesdownloaded) FROM sheets WHERE timesdownloaded>='0'");
$totaldownloadsresult = mysql_fetch_assoc($downloadscount);
$totaldownloads = $totaldownloadsresult['timesdownloaded'];

it isn't echoing out, so i obviously did something wrong. 🙁

laserlight

For example:

$query = mysql_query("SELECT SUM(timesdownloaded) FROM sheets WHERE timesdownloaded>='0'");
$row = mysql_fetch_assoc($query);
$totaldownloads = $row['SUM(timesdownloaded)'];

Smudly

Yeah! working. Thanks for your patience.