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Quick PHP Query Project

$30-100 USD

In Progress
Posted about 18 years ago

$30-100 USD

Paid on delivery
I am building a topsites list, and in the process of ranking sites have stumbled on a problem. I list sites in order of the sum of their views in the last 7 days, from topsite_views which contains the number of views for any given day there is more than 1 view. I know how to do this. My trouble is how to assign an actual rank number to the sites. I need to find how many sites are above them and add 1 (I know this much), but I don't know how to compare them when the actual calculation takes place inside the query. Here is an example: First, I find out the sum of views in last 7 days for the site in question ($valuex), then I want to see how many sites are better: $qry = "select sum(views) as viewsum from topsite_views where viewsum >= '$valuex'"; Obviously, the query fails because it doesnt know what viewsum is until the query is finished... yada yada.. To clarify: The query needs to count the number of sites in the toplist_views table that have a sum of views column [where date = DATE_SUB(curdate(),INTERVAL 7 day)] which is greater than a specific value. I cannot use subselects or subqueries because of my mysql version. For example, if I was just counting the sum of views for one site it would be: $prerankqry = "select sum(views) as viewsum from toplist_views where siteid='$siteid' and date = DATE_SUB(curdate(),INTERVAL 7 day) group by siteid"; Query needs to be in format that is compatible with using TemplatePower. THIS IS ONE LINE OF CODE NEEDED TODAY - THAT IS THE WHOLE PROJECT AVAILABLE ON MSN FOR DISCUSSION IF NECCESSARY
Project ID: 46994

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1 proposal
Remote project
Active 18 yrs ago

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Hello. Please see PMB for more details. Regards Zend Certified PHP Developer Vladimir Kuznetsov.
$30 USD in 0 day
5.0 (49 reviews)
4.9
4.9

About the client

Flag of UNITED KINGDOM
Crieff, United Kingdom
5.0
47
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Member since May 19, 2005

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