根據其他列包含多個LEFT JOIN Mysql總結不同

Question

我有5個表，我想LEFT JOIN在一起。表格爲： 訪問者，優惠，合同1，合同2和合同3。

QUERY：這裏

SELECT 
     count(DISTINCT visitors.ID) as visitors, 
     sum(
     CASE 
     WHEN offers.ACTIVE = 1 THEN 1 
     ELSE 0 
     END) as offers, 
     count(contracts1.ID) as contracts1, sum(contracts1.PRICE) as sum_contracts1, 
     count(contracts2.ID) contracts2, 
     sum(
     CASE 
     WHEN contracts2.PAYMENT = 'YEARLY' THEN contracts2.PRICE 
     WHEN contracts2.PAYMENT = 'TWICE' THEN contracts2.PRICE*2 
     ELSE contracts2.PRICE*4 
     END) as sum_contracts2, 
     count(contracts3.ID) as contracts3, sum(contracts3.PRICE) as sum_contracts3 
     FROM visitors 
     LEFT JOIN offersON offers.VISITOR_ID = visitors.ID AND (offers.IP > 100 OR offers.IP < 0) 
     LEFT JOIN contracts1 ON 
     (offers.ID = contracts1.ID_OFFER) 
     LEFT JOIN contracts2 ON 
     (offers.ID = contracts2.ID_OFFER) 
     LEFT JOIN contracts3 ON 
     (offers.ID = contracts3.ID_OFFER) 
     WHERE visitors.TIME >= '2017-01-01 00:00:00' AND visitors.TIME <= '2017-05-25 23:59:59' 

問題是，contracts1，contracts2和contracts3沒有共同的列，以便被接合在一起。所以不是20行的合約1，30的是合同2和50的合約3，我得到了所有這些組合。因爲他們根據訪問者加入並提供表格。簡單的GROUP BY在查詢結束時通常會解決問題，但是如果我在END中爲其中一個表（或全部）使用GROUP BY，它將創建多行而不是1。而且它也會清除我通過ID計數訪問者的部分​​的所有其他結果以及ID提供的部分...我可以在SELECT的count（）部分上使用DISTINCT，但不能使用sum（），因爲合同的PRICE可能是相同的，即使身份證不是（你知道，例如2個巧克力是2行不同的ID，但價格相同，每個10美元）。

所以我的問題是：

有什麼辦法來總結只有contracts1，contracts2和contracts3的那些價格，即有不同的ID，雖然擺脫加起來的複製品？這有可能沒有創建VIEW？

我也嘗試了GROUP BY裏面的LEFT JOIN，但是再次當我左側連接所有3個合同表時，即使我在他們之前對它們進行了重新分組。預期的結果的

實施例：

在我指出上述我期望的時間範圍：具有35個報價和5 contracts1與千歐元總和，12 contracts2與686歐元總和 80訪問者3份合同3，總計12歐元。它是8行數據的一行。

取而代之的是我得到的結果： 80個訪客，35個offer，180個contract1（總數也不好），180個contract2（總數也不好），180個contract3（總數也不好）。

Answer 1

由於熱膨脹係數（Supported by MariaDB 10.2.1）我會寫這樣的事：

WITH v AS (
    SELECT ID as VISITOR_ID 
    FROM visitors 
    WHERE visitors.TIME >= '2017-01-01 00:00:00' 
     AND visitors.TIME <= '2017-05-25 23:59:59' 
), o AS (
    SELECT offers.ID as ID_OFFER 
    FROM v 
    JOIN offers USING(VISITOR_ID) 
    WHERE offers.ACTIVE = 1 
     AND (offers.IP > 100 OR offers.IP < 0) 
), c1 AS (
    SELECT count(*) as contracts1, sum(contracts1.PRICE) as sum_contracts1 
    FROM o JOIN contracts1 USING(ID_OFFER) 
), c2 AS (
    SELECT 
     count(*) contracts2, 
     sum(CASE contracts2.PAYMENT 
      WHEN 'YEARLY' THEN contracts2.PRICE 
      WHEN 'TWICE' THEN contracts2.PRICE*2 
      ELSE contracts2.PRICE*4 
     END) as sum_contracts2 
    FROM o JOIN contracts2 USING(ID_OFFER) 
), c3 AS (
    SELECT count(*) as contracts3, sum(contracts3.PRICE) as sum_contracts3 
    FROM o JOIN contracts3 USING(ID_OFFER) 
) 
    SELECT c1.*, c2.*, c3.*, 
     (SELECT count(*) FROM v) as visitors, 
     (SELECT count(*) FROM o) as offers, 
    FROM c1, c2, c3; 

沒有熱膨脹係數，你可以重寫它使用臨時表：

CREATE TEMPORARY TABLE v AS 
    SELECT ID as VISITOR_ID 
    FROM visitors 
    WHERE visitors.TIME >= '2017-01-01 00:00:00' 
     AND visitors.TIME <= '2017-05-25 23:59:59'; 

CREATE TEMPORARY TABLE o AS 
    SELECT offers.ID as ID_OFFER 
    FROM v 
    JOIN offers USING(VISITOR_ID) 
    WHERE offers.ACTIVE = 1 
     AND (offers.IP > 100 OR offers.IP < 0); 

CREATE TEMPORARY TABLE c1 AS 
    SELECT count(*) as contracts1, sum(contracts1.PRICE) as sum_contracts1 
    FROM o JOIN contracts1 USING(ID_OFFER); 

CREATE TEMPORARY TABLE c2 AS 
    SELECT 
     count(*) contracts2, 
     sum(CASE contracts2.PAYMENT 
      WHEN 'YEARLY' THEN contracts2.PRICE 
      WHEN 'TWICE' THEN contracts2.PRICE*2 
      ELSE contracts2.PRICE*4 
     END) as sum_contracts2 
    FROM o JOIN contracts2 USING(ID_OFFER); 

CREATE TEMPORARY TABLE c3 AS 
    SELECT count(*) as contracts3, sum(contracts3.PRICE) as sum_contracts3 
    FROM o JOIN contracts3 USING(ID_OFFER); 

SELECT c1.*, c2.*, c3.*, 
    (SELECT count(*) FROM v) as visitors, 
    (SELECT count(*) FROM o) as offers, 
FROM c1, c2, c3; 

Answer 2

只是一個概念證明，我沒有考慮時間和活動約束以及支付類型，但是不可能是這些方面的東西？

SELECT 
    VISITOR_ID, 
    SUM(CASE WHEN TYPE="contract1" THEN 1 else 0 END) as c1_count, 
    SUM(CASE WHEN TYPE="contract1" THEN PRICE else 0 END) as c1_total_price, 
    SUM(CASE WHEN TYPE="contract2" THEN 1 else 0 END) as c2_count, 
    SUM(CASE WHEN TYPE="contract2" THEN PRICE else 0 END) as c2_total_price, 
    SUM(CASE WHEN TYPE="contract3" THEN 1 else 0 END) as c3_count, 
    SUM(CASE WHEN TYPE="contract3" THEN PRICE else 0 END) as c3_total_price 
FROM (
    (SELECT "contract1" as TYPE, ID, PRICE, ID_OFFER, PAYMENT FROM contracts1) 
    UNION 
    (SELECT "contract2" as TYPE, ID, PRICE, ID_OFFER, PAYMENT FROM contracts2) 
    UNION 
    (SELECT "contract3" as TYPE, ID, PRICE, ID_OFFER, PAYMENT FROM contracts3) 
) as all_contracts 
JOIN offers on offers.id = all_contracts.ID_OFFER 
JOIN visitors on visitors.ID = offers.VISITOR_ID 
GROUP BY visitors.ID 

的想法是，首先你合併不同的合同到一個結果，你存儲的類型在一個名爲「類型」列（這是UNION查詢的目的），一旦你擁有這樣一個漂亮的表，其中每份合約只有一次，您可以非常直接地得到您想要的結果。我剛剛概述了你如何得到每種合同的總和和數量。當然，最終的查詢會更復雜一些，但核心思想應該是一樣的。

儘管您聲明您不想使用（臨時）視圖，但我會鼓勵您嘗試一下 - 我有一種感覺，將這些「all_contracts」與優惠和訪問者結合進臨時視圖可以改善性能，如果這是您的擔心，但不會使查詢太難看，主要是在您希望僅爲一位訪問者查看統計信息或進一步過濾它們（按時間，活動等）的情況下，因爲不必要的行贏得了沒有物化。但這只是一個印象，因爲我沒有在更大的數據集上嘗試查詢 - 您可以使用它。