[英]SQL GROUP_CONCAT split in different columns
我搜索了很多,但沒有找到解決問題的正確方法。
我想做什么?
我在MySQL中有兩個表: - 國家 - 貨幣(我通過CountryCurrency將它們連接起來 - >由於多對多關系)
請參閱此示例以獲取一個工作示例: http : //sqlfiddle.com/#!2/317d3/8/0
我想使用連接將兩個表鏈接在一起,但我希望每個國家只顯示一行(某些國家/地區有多種貨幣,因此這是第一個問題)。
我找到了group_concat函數:
SELECT country.Name, country.ISOCode_2, group_concat(currency.name) AS currency
FROM country
INNER JOIN countryCurrency ON country.country_id = countryCurrency.country_id
INNER JOIN currency ON currency.currency_id = countryCurrency.currency_id
GROUP BY country.name
這有以下結果:
NAME ISOCODE_2 CURRENCY
Afghanistan AF Afghani
Åland Islands AX Euro
Albania AL Lek
Algeria DZ Algerian Dinar
American Samoa AS US Dollar,Kwanza,East Caribbean Dollar
但我現在想要的是將貨幣分成不同的列(貨幣1,貨幣2,......)。 我已經嘗試過像MAKE_SET()這樣的函數,但這不起作用。
你可以用substring_index()
來做到這一點。 以下查詢將您的子查詢用作子查詢,然后應用此邏輯:
select Name, ISOCode_2,
substring_index(currencies, ',', 1) as Currency1,
(case when numc >= 2 then substring_index(substring_index(currencies, ',', 2), ',', -1) end) as Currency2,
(case when numc >= 3 then substring_index(substring_index(currencies, ',', 3), ',', -1) end) as Currency3,
(case when numc >= 4 then substring_index(substring_index(currencies, ',', 4), ',', -1) end) as Currency4,
(case when numc >= 5 then substring_index(substring_index(currencies, ',', 5), ',', -1) end) as Currency5,
(case when numc >= 6 then substring_index(substring_index(currencies, ',', 6), ',', -1) end) as Currency6,
(case when numc >= 7 then substring_index(substring_index(currencies, ',', 7), ',', -1) end) as Currency7,
(case when numc >= 8 then substring_index(substring_index(currencies, ',', 8), ',', -1) end) as Currency8
from (SELECT country.Name, country.ISOCode_2, group_concat(currency.name) AS currencies,
count(*) as numc
FROM country
INNER JOIN countryCurrency ON country.country_id = countryCurrency.country_id
INNER JOIN currency ON currency.currency_id = countryCurrency.currency_id
GROUP BY country.name
) t
表達式substring_index(currencies, ',' 2)
將貨幣列表中的貨幣列為第二個貨幣。 對於美國的Somoa來說,那將是'US Dollar,Kwanza'
。 以-1
作為參數的下一個調用采用列表的最后一個元素,即'Kwanza'
,這是currencies
的第二個元素。
另請注意,SQL查詢返回一組明確定義的列。 查詢不能具有可變數量的列(除非您通過prepare
語句使用動態SQL)。
使用此查詢計算出您需要的貨幣列數:
SELECT MAX(c) FROM
((SELECT count(currency.name) AS c
FROM country
INNER JOIN countryCurrency ON country.country_id = countryCurrency.country_id
INNER JOIN currency ON currency.currency_id = countryCurrency.currency_id
GROUP BY country.name) as t)
然后動態創建並執行預准備語句以生成結果,使用帶有上述查詢結果的Gordon Linoff解決方案在此線程中。
Ypu可以使用動態SQL,但您必須使用過程
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