Query addition

Hi,

I use the following query to count unique pageviews, grouped by date.

$sqlStr = '
SELECT
	date(viewdate) AS myDate,
	COUNT(DISTINCT userip) AS myCount
FROM
	archive_views
GROUP BY
	myDate
ORDER BY
	myDate DESC;
';

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I need to add this part to it, but not sure how.

SUM(earnings) AS mySum WHERE approved='true' DISTINCT userip

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The idea is to count all approved earnings from a specific date, but only from unique visitors (my affiliates only get paid per 24 hour unique visitor).
kgp43Asked:
Who is Participating?
 
Cornelia YoderConnect With a Mentor ArtistCommented:
Can you just use

SELECT
      date(viewdate) AS myDate,
      COUNT(DISTINCT userip) AS myCount
FROM
      archive_views
GROUP BY
      myDate, userip
ORDER BY
      myDate DESC;
';
0
 
kgp43Author Commented:
Update: let me test a bit and get back to you.

 - I do not need the approved='true'
0
 
kgp43Author Commented:
Does not work.

And "SUM(earnings) AS mySum," does not count earnings per unique 24 hour IP either... I think. Hmm, hard one.

<?php
$sqlStr = '
SELECT
	date(viewdate) AS myDate,
	SUM(earnings) AS mySum,
	COUNT(DISTINCT userip) AS myCount
FROM
	archive_views
GROUP BY
	myDate, userip
ORDER BY
	myDate DESC;
';

# Query
$view_query = mysql_query($sqlStr) or die(mysql_error());
while($view_fetch = mysql_fetch_array($view_query)) {	
	?>

	<tr>
		<td>Date: <?php echo date("F jS, Y", strtotime($view_fetch['myDate'])); ?></td>
		<td>Views: <?php echo $view_fetch['myCount'] ?></td>
		<td>Amount: <?php echo "$".round($view_fetch['mySum'], 2); ?></td>
	</tr>
	
<?php
}
?>

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0
 
kgp43Author Commented:
I changed my script a bit, so I can just use this instead.

$sqlStr = '
SELECT
	date(viewdate) AS myDate,
	SUM(earnings) AS mySum,
	COUNT(DISTINCT userip) AS myCount
FROM
	archive_views
GROUP BY
	myDate
ORDER BY
	myDate DESC;
';

Open in new window


Going to close the question soon, unless something else comes up regarding this query.
0
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