mysql的简单数据库查询

Question

i am having a database in that 'users' table contain 4 fields id , name , gender , language
ex. (database entries)

1   jhon     male     en-sp-gr  
2   mira    female    sp-  
3   mike     male     en-  
4   shel    female    sp-gr  

etc...

here en,sp,gr are languages English , Spanish , German what i am trying to fetch the count of users who speaks only en,sp (english and spanish languages) the database contains more than 300 users and i am more more confused how to get the count of users who speaks only en and sp the language field of the table contains string

 en-sp-gr 

how can i omit -gr and hyphen ( - ) form the string and this only for one row i have to perform this for each row and then count of those users only who speaks en and sp. i am not expert in sql :(

Answer 1

You could do something like this:

SELECT * 
FROM users 
WHERE language IN ('en-sp', 'sp-en')

However, you really should look into normalizing your database:

Users
ID|Name|Gender

Languages
ID|Code

User_Language_Map
UserID|LanguageID

Then you could do something like this:

SELECT *
FROM Users
WHERE ID IN
(
    SELECT UserID 
    FROM User_Language_Map
    WHERE LanguageID IN (IDFORSP, IDFORGR)
    GROUP BY UserID
    HAVING COUNT(DISTINCT LanguageID) = 2
)

If you cannot do that, and the language filter is going to be dynamic, then you really would need to create a function that will create the appropriate string combinations (or you could build a regex on the fly) and pass that into your IN . Otherwise, I don't know that you have many options as you are parsing strings at that point.

Answer 2

This is why you should not store multiple values in one field. They should be stored in a normalized table instead.

But you can accomplish this by replacing the - with , and using FIND_IN_SET() . This is necessary because FIND_IN_SET() expects a comma-separated list of values. This can only be reasonably managed with a handful of languages though, since you need to code all the permutations into the query. For this reason (among others like indexing) it is very strongly recommended to refactor this into a related table linking users to their spoken languages.

SELECT COUNT(*) FROM users 
WHERE 
  FIND_IN_SET('en', REPLACE(language,'-',',')) > 0
  AND FIND_IN_SET('sp', REPLACE(language,'-',',')) > 0
  AND FIND_IN_SET('gr', REPLACE(language,'-',',')) = 0

Answer 3

SELECT 
      count(case when language = 'en' then language end) as English, 
      count(case when language = 'sp' then language end) as Spanish 
FROM users

In above query can count the English Language Users and Spanish Laguage User.