J,
Try this (it assumes that the softid you want to search for is in the
variable $softid):
$sql = "SELECT Expert.*
FROM Expert
LEFT JOIN Apliexpert ON Expert.id = Apliexpert.exptid
WHERE Apliexpert.softid = '$softid'
ORDER BY Expert.Name";
----------------------
Beverly Steiner
[EMAIL PROTECTED]
-----Original Message-----
From: Jason End [mailto:[EMAIL PROTECTED]
Sent: Thursday, March 06, 2003 8:52 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] mysql statement: two tables, two comparisons
I'm looking for a mysql select statement that does the
following:
- Check if the value of each expert.id on the table
experts matches a value expt.id in the table
apliexpert.
- For those values where this is true check whether
the softID value for that row matches the variable
$softId.
- return the names that are left after those 2 filters
So for tables:
Expert
id Name
1 Peter
2 Paul
3 Mary
4 Frank
Apliexpert
exptid softid
1 3
2 5
2 8
3 9
3 8
1. If the softID is 2, the select should return:
peter, paul, mary and frank (frank will always be
returned no matter what, because he isn't in
apliexpert)
2. If the softID is 3, the select should return: paul,
mary and frank
3. If the softID is 8, the select should return: peter
and frank
4. If the softID is 9, the select should return:
peter, paul and frank
J
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