Solved

mysql - like statement

Posted on 2012-04-13
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Last Modified: 2012-06-27
Hi,

Trying to list from table courier all couriers whose courier IDs are in table1. However, I am using the like statement because it will not be an exact match up. E.g. in table courier I may have UK00062884, whereas in table1 the entry may just be 62884. Each number is unique though, for instance in this instance, 62884 is definitely courier UK00062884. This code works for just a single hardcoded one

SELECT * from courier where `courier`.`CourierID` LIKE CONCAT('%',"62884",'%');

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but neither of these work

SELECT * from courier where `courier`.`CourierID` LIKE CONCAT('%','table1.courier2','%');

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or

SELECT * from courier where `courier`.`CourierID` LIKE CONCAT('%',table1.courier2,'%');

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somebody help me out?

thanks.
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Question by:RupertA
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Expert Comment

by:stavros41
ID: 37842782
Try this:
SELECT * from courier where `courier`.`CourierID` LIKE '%62884%';

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Author Comment

by:RupertA
ID: 37842916
stavros,

ahh no, I may not have been clear here. What I need to do is to list all the couriers from table courier that are in table1. u cud use this.

SELECT * from courier where `courier`.`CourierID` IN( select courier2 from table1);

However the problem is that they are not exact, table1 may have 1234, whereas courier may have IR0001234. They are def same person though, hence why I was trying to use "like".

thanks.
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Accepted Solution

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stavros41 earned 500 total points
ID: 37843018
Ok, how about this:

SELECT * FROM courier JOIN table1 ON `CourierID` like CONCAT('%',courier2,'%')

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Author Comment

by:RupertA
ID: 37843075
yeaaaaah, that works great :)

Thanks!!
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LVL 6

Expert Comment

by:stavros41
ID: 37843240
No problem, please accept the answer to allocate points.
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