I have a database structure like so:
SELECT * FROM Culture;
------------------------
Id ShortName FullName Supported
22 en-US English (United States) 1
23 fr-FR French (France) 1
24 hi-IN Hindi (India) 0
SELECT * FROM ResourceKey;
----------------------------
Id Name
20572 HowAreYou
20571 Hello
20573 ThisKeyHasUSEnglishValueOnly
SELECT * FROM Strings;
-----------------------
Id CultureId ResourceKeyId ResourceValue
41133 22 20571 Hello
41134 22 20572 How are you?
41135 23 20571 Bonjour
41136 23 20572 Comment allez-vous?
41137 22 20573 This key has US English value only.
SELECT * FROM Category;
------------------------
Id Name
1 JavaScript
SELECT * FROM StringCategory;
------------------------------
Id ResourceKeyId CategoryId
1 20571 1
2 20572 1
3 20573 1
I want to display all resource key names and resource values, i.e. string values against each key, for, say, the French (France) culture, i.e. the culture with the ShortName
fr-FR but even if a key does not have a value in the culture, it must display the key name but NULL for the value. Like so:
Name ResourceValue
-------------------------------------------------------
Hello Bonjour
HowAreYou Comment allez-vous?
ThisKeyHasUSEnglishValueOnly NULL
It seems like a simple LEFT OUTER JOIN application to me, but my code isn't working. Could someone please help correct my code?
My query is:
SELECT ResourceKey.Name AS Name, ResourceValue
FROM
ResourceKey LEFT OUTER JOIN Strings
ON
Strings.ResourceKeyId = ResourceKey.Id
INNER JOIN StringCategory
ON
StringCategory.ResourceKeyId = Strings.ResourceKeyId
INNER JOIN Category
ON
StringCategory.CategoryId = Category.Id
LEFT OUTER JOIN Culture
ON
Strings.CultureId = Culture.Id AND Culture.Id = (SELECT Id FROM Culture WHERE ShortName = 'fr-FR')
AND
Category.Name = 'JavaScript';
Somehow, the last join in the above-query turns out to become an inner join, eliminating those rows where there is no value in the said culture.
SELECT a.name, b.ResourceValue
FROM ResourceKey a
LEFT JOIN
(
SELECT b.ResourceKeyID, b.ResourceValue
FROM Strings b
INNER JOIN Culture c
ON b.CultureID = c.ID
WHERE c.shortname = 'fr-FR'
) b ON a.ID = b.ResourceKeyId
UPDATED
USE SSTOBMAY;
SELECT a.name, b.ResourceValue
FROM ResourceKey a
LEFT JOIN
(
SELECT b.ResourceKeyID, b.ResourceValue
FROM Strings b
INNER JOIN Culture c
ON b.CultureID = c.ID
WHERE c.shortname = 'fr-FR'
) b ON a.ID = b.ResourceKeyId
INNER JOIN
StringCategory sc ON
sc.ResourceKeyId = a.Id
INNER JOIN Category c ON c.Id = sc.CategoryId
WHERE c.Name = 'JavaScript';
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