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| 08.25.2008 at 08:52AM PDT, ID: 23675601 |
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Attachment Details
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The Solution Rating System
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CREATE table tblDiacritics
(
Country varchar(50),
Diacritic varchar(50),
DiacriticReplacment varchar(50)
)
CREATE table tblAddresses
(
Country varchar(50),
Address_1 varchar(50)
)
Note: The below values may not be the actual suggested replacement for diacritics per each country
INSERT INTO tblDiacritics
VALUES ('Germany', 'é', 'e')
INSERT INTO tblDiacritics
VALUES ('Germany', 'ü', ue')
INSERT INTO tblDiacritics
VALUES ('Germany', 'ö', 'oe')
INSERT INTO tblDiacritics
VALUES ('Denmark', 'ö', 'o')
INSERT INTO tblAddresses
VALUES ('10','Germany', 'Testéöüßé')
INSERT INTO tblAddresses
VALUES ('1','Germany', 'Testé')
INSERT INTO tblAddresses
VALUES ('2','Germany', 'Testö')
INSERT INTO tblAddresses
VALUES ('3','Denmark', 'Testö')
INSERT INTO tblAddresses
VALUES ('4','Test', 'Testéd')
INSERT INTO tblAddresses
VALUES ('5','Test', 'Tüstéd')
INSERT INTO tblAddresses
VALUES ('6','Germany', 'Tüstéd')
INSERT INTO tblAddresses
VALUES ('7','Germany', 'Test')
INSERT INTO tblAddresses
VALUES ('8','Germany', 'Teßed')
INSERT INTO tblAddresses
VALUES ('9','Germany', 'Tessed')
Here is my test query:
SELECT tblAddresses.RecordID, tblAddresses.Address_1, Replace(tblAddresses.Address_1,tblDiacritics.diacritic,tblDiacritics.DiacriticReplacment) AS Expr1, tblAddresses.Country
FROM tblAddresses INNER JOIN tblDiacritics ON tblAddresses.Country = tblDiacritics.Country
WHERE (((tblAddresses.Address_1) Like "*" & [tblDiacritics].[diacritic] & "*"));
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