NYCPHP Meetup

[Mark Armendariz]

So you have an array of zips, like array(11201, 11206, 11301), right?  
and you need to find all the jobs within that array of zips?

If so your query would look something like this:

mysql_query('SELECT * FROM Providers INNER JOIN Employers ON Jobs.empid = 
Employers.EmpID WHERE Jobs.zip IN (' . implode(',', $aZips) . ')');

Which becomes 
SELECT * FROM Providers INNER JOIN Employers ON Jobs.empid = 
Employers.EmpID WHERE Jobs.zip IN (11201,11206,11301)

Not sure if that's what you're looking for.

Mark


Kristina Anderson wrote:
> OK, this really has me flummoxed.  I'm not sure if I'm going about this 
> the right way.
>
> I have three tables, an Employer table, a Provider table and a Jobs 
> table.  Each Provider (worker) has up to nine zip codes and a possible 
> metro area which contains all the zip codes in their city which is 
> their "service area".
>
> When they log on, I need to display only the Jobs in any of their zip 
> codes. 
>
> In the Jobs table, there is a remote zip field (remzip) and if that has 
> a value, that is the zip where the job is, otherwise we use the zip 
> field in the Employers table.
>
> I've built two arrays one by 
> "SELECT * FROM Providers INNER JOIN Employers ON Jobs.empid = 
> Employers.EmpID ";
>
> which gives me both the remzip and zip fields and the jobid field.
>
> And one by querying the database for all the zips in the service area 
> of that particular Provider and dumping them all into an array.
>
> So now I have these two arrays I can compare...is there an easier way 
> to do this or ... ?
>
> I need to compare these two arrays and if any value from array 2 is in 
> array 1, pull out the entire row by jobid from the database and display 
> it.
>
> Thanks in advance,
> ---- Kristina 
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