MySQL Selection problem

Question

I have two tables on database user and ads when user visit the ads page I want to show him only the ads related to his country in the ads table , there is field called contry where I insert related countries like that country="Country1;Coutry2;Country3;..." and every user have a field with his country name How to select ads from ads table where the user's coutry is in the country field I guess something like that

SELECT * FROM ads WHERE $usercoutry IN ads.country

Thanks for All , I found the solution

SELECT * FROM ads WHERE coutry LIKE '%{$usercountry}%'

That Code solved the problem :)

Thanks for All , I found the solution SELECT * FROM ads WHERE coutry LIKE '%{$usercountry}%' That Code solved the problem :) — webneat
– webneat, Commented Aug 24, 2011 at 23:08
Can you mark the correct solution by checking the checkbox outline next to the answer? — CrackerJack9
– CrackerJack9, Commented Aug 25, 2011 at 5:22

Andrey · Accepted Answer · 2011-08-24 17:13:59Z

1

SELECT * FROM ads WHERE country like '%$usercoutry%'

answered Aug 24, 2011 at 17:13

Andrey

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John M · Accepted Answer · 2011-08-24 16:54:28Z

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You could try:

SELECT * FROM ads WHERE ads.country = $usercountry

answered Aug 24, 2011 at 16:54

John M

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Vincent · Accepted Answer · 2011-08-24 17:08:04Z

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Are you trying to do something like the following?

SELECT * FROM ads WHERE ads.country like '%' || $usercountry || '%';

Nota: you should use a parameter (eg a prepared statement with '?' in the query), and set the concatenated string as its value, in order to avoid SQL injection - if it's a concern.

answered Aug 24, 2011 at 17:08

Vincent

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CrackerJack9 · Accepted Answer · 2011-08-25 05:25:58Z

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Need to use LIKE with the variable encased in braces:

SELECT * FROM ads WHERE country LIKE '%{$usercountry}%';

answered Aug 25, 2011 at 5:25

CrackerJack9

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