基於文本字段加入2個mysql select

Question

我在Mysql中有以下2個選擇：

第一選：

(SELECT DISTINCT `Online_playerdatabase_v2`.`Player`, 
Online_playerdatabase_v2.First_Deposit_Date As FirstDep, 
TRUNCATE(Online_playerdatabase_v2.Balance,2) as Balance

FROM Online_playerdatabase_v2 
WHERE `Online_playerdatabase_v2`.`Player`<>'Player'
ORDER BY `Online_playerdatabase_v2`.`Balance` DESC;

2d選擇：

SELECT DISTINCT(Online_customer_activity_v2.Customers) as Player,
max(Online_customer_activity_v2.Date) as LastAction
FROM Online_customer_activity_v2
WHERE `Online_customer_activity_v2`.`Total_Bets`>0
Group by Online_customer_activity_v2.Customers

輸出選擇1

Player      FirstDep    Balance
Ray         2014-10-19  9100.00
Ramzi       2014-11-02  9.61
tareq       2014-11-06  805.00
STAN        2014-10-17  7.50
Bill        2014-03-25  68.40
karam       2014-11-16  676.50
Abdul       2014-11-13  650.00
Renaud      2014-03-12  507.00
John        2014-11-22  500.00

輸出選擇2

Player  LastAction
John    2015-11-13
Bill    2014-12-14
Renaud  2015-03-14
Abdul   2015-11-16
Ray     2015-11-22
STAN    2015-10-29
Ramzi   2015-11-10
Tarek   2015-05-10
karam   2014-12-10
Abdul   2015-02-10

期望的輸出，兩者上的連接選擇添加以下計算： active days (FirstDep-LastAction)和active days (FirstDep-LastAction) Days_last_Visit (CurrentDate - Last Action)

總結如下表：

Player  FirstDep    Balance LastAction  Active_days Days_last_Visit
Ray     2014-10-19  9100.00 2015-11-22  399          1
Ramzi   2014-11-02  9.61    2015-11-10  373          13
tareq   2014-11-06  805.00  2015-05-10  185          197
STAN    2014-10-17  7.50    2015-10-29  377          25
Bill    2014-03-25  68.40   2014-12-14  264          344
karam   2014-11-16  676.50  2014-12-10  24           348
Abdul   2014-11-13  650.00  2015-02-10  89           286
Renaud  2014-03-12  507.00  2015-03-14  367          254
John    2014-11-22  500.00  2015-11-13  356          10

非常感謝您的幫助！ 謝謝

Answer 1

以下查詢應該給出您想要的結果。 我將添加我使用Player字段從上面的中間查詢加入兩個表。 這不是一種非常強大的加入方式，因為該名稱在表格中的所有玩家中可能不是唯一的。 更好的加入方式是使用某種唯一標識符。

SELECT t1.Player, t1.FirstDep, t1.Balance, t2.LastAction,
    DATEDIFF(t2.LastAction, t1.FirstDep) AS Active_days,
    DATEDIFF(NOW(), t2.LastAction) AS Days_last_Visit
FROM
(
    SELECT DISTINCT `Online_playerdatabase_v2`.`Player`, 
        Online_playerdatabase_v2.First_Deposit_Date AS FirstDep, 
        TRUNCATE(Online_playerdatabase_v2.Balance,2) AS Balance
    FROM Online_playerdatabase_v2 
    WHERE `Online_playerdatabase_v2`.`Player` <> 'Player'
) t1
INNER JOIN
(
    SELECT DISTINCT(Online_customer_activity_v2.Customers) AS Player,
        MAX(Online_customer_activity_v2.Date) AS LastAction
    FROM Online_customer_activity_v2
    WHERE `Online_customer_activity_v2`.`Total_Bets` > 0
    GROUP BY Online_customer_activity_v2.Customers
) t2
ON t1.`Player` = t2.`Player`

Answer 2

您需要使用播放器字段將2個選擇作為子查詢加入第3個選擇中。 可以使用DateDiff（）函數計算Active_days和Days_last_Visit字段。

SELECT *
,DateDiff(t2.LastAction,t1.FirstDep) AS Active_days
,DateDiff(CURDATE(), t2.LastAction) AS  Days_last_Visit
FROM
(SELECT DISTINCT `Online_playerdatabase_v2`.`Player`, 
Online_playerdatabase_v2.First_Deposit_Date As FirstDep, 
TRUNCATE(Online_playerdatabase_v2.Balance,2) as Balance   
FROM Online_playerdatabase_v2 
WHERE `Online_playerdatabase_v2`.`Player`<>'Player'
ORDER BY `Online_playerdatabase_v2`.`Balance` DESC) t1
LEFT JOIN
(SELECT DISTINCT(Online_customer_activity_v2.Customers) as Player,
max(Online_customer_activity_v2.Date) as LastAction
FROM Online_customer_activity_v2
WHERE `Online_customer_activity_v2`.`Total_Bets`>0
Group by Online_customer_activity_v2.Customers) t2
ON t1.Player=t2.Player

但是，您必須考慮如何加入2個數據集。 我使用左連接，因為玩家表可能會占用所有玩家，但您可能希望根據您的要求和數據進行內部聯接或模擬完整的外部聯接。