从多个sql中获得结果作为列明智

-4

我有不同的SQL，它返回不同的结果。我需要获得所有列作为明智的。但我不知道该怎么做。请任何人都可以帮我解决这个问题吗？这里是我的SQL低于>>>从多个sql中获得结果作为列明智

SELECT "one" AS value_1, 
    "two" AS value_2 

SELECT "three" AS value_3, 
    "four" AS value_4 

SELECT "five" AS value_5, 
    "six" AS value_6

主查询1 >>>

SELECT -- lat.`group_info_id`, gi.`group_name`, 
    SUM(lat.`principal_outstanding`) AS principalOutstanding -- , SUM(lat.`accrued_interest`) AS interestReceivable, SUM(lat.`realizable_interest`) AS interestRealizable 
    -- , SUM(lat.`principal_outstanding`)+SUM(lat.`accrued_interest`)+SUM(lat.`realizable_interest`) AS loanDue, SUM(lat.`outstanding_balance`) AS totalLoanDue 
FROM loan_account_transaction AS lat 
INNER JOIN loan_account AS la ON (la.id = lat.`loan_account_id`) 
LEFT JOIN group_info AS gi ON (gi.id = lat.`group_info_id`) 
INNER JOIN 
(
SELECT MAX(ilat.id) AS max_lat_id, ilat.loan_account_id 
FROM loan_account_transaction AS ilat 
WHERE ilat.`domain_status_id` = 1 AND ilat.`office_info_id` = (SELECT id FROM office_info WHERE office_code = 'UG500002') 
AND ilat.`project_info_id` IN ('BI0000000000000000000004') AND ilat.transaction_date <= '2013-04-30' 
GROUP BY ilat.loan_account_id 
) AS tlat ON (tlat.loan_account_id = lat.`loan_account_id` AND tlat.max_lat_id = lat.id) 
WHERE lat.`office_info_id` = (SELECT id FROM office_info WHERE office_code = 'UG500002') 
AND lat.`project_info_id` IN ('BI0000000000000000000004') 
AND lat.`loan_status_id` != 2 AND lat.`loan_status_id` != 7 AND la.`domain_status_id` = 1 AND lat.`domain_status_id` = 1

主查询2 >>>

SELECT  SUM(CASE WHEN la.loan_duration = 12 THEN 1 ELSE 0 END) AS total_count_40, 
    SUM(CASE WHEN la.loan_duration = 12 THEN IFNULL(la.disbursed_amount,0) ELSE 0 END) AS disbursed_amount_40, 
    SUM(CASE WHEN la.loan_duration = 6 THEN 1 ELSE 0 END) AS total_count_20, 
    SUM(CASE WHEN la.loan_duration = 6 THEN IFNULL(la.disbursed_amount,0) ELSE 0 END) AS disbursed_amount_20 
FROM loan_account AS la 
INNER JOIN member_info AS mi ON (la.disbursement_date <='2013-04-30' 
    AND la.project_info_id IN ('BI0000000000000000000004') 
    AND la.domain_status_id = 1 AND la.member_id=mi.id) 
INNER JOIN office_info AS oi 
ON(oi.id = 'BI0000000000000000000363' AND mi.branch_info_id = oi.id)

**新>>>

SELECT COUNT(1), SUM(principal_outstanding) 
FROM 
(

/* Current Loan */ 
     SELECT tempData.country_id, 
     tempData.project_info_id, 
     tempData.prodId, 
     tempData.loan_name, 
     tempData.office_info_id, 
     office_info.office_code, 
     office_info.office_name, 
     tempData.group_info_id, 
     group_info.group_code, 
     group_info.group_name, 
     missed_days, 
     SUM(tempData.principal_outstanding) AS principal_outstanding, 
     COUNT(DISTINCT member_id) AS borrower_count 

     FROM (
       SELECT lat.country_id,lat.office_info_id,lat.project_info_id, lp.id AS prodId, lp.`loan_name`, 
       lat.group_info_id, 
       lat.principal_outstanding,la.member_id, 

       (CEIL(lat.overdue_amount/la.installment_amount) * 
         CASE WHEN lpp.policy_repayments_plan_id = 1 THEN 
          1 
         WHEN lpp.policy_repayments_plan_id = 2 THEN 
          7 
         WHEN lpp.policy_repayments_plan_id = 3 THEN 
          30 
         WHEN lpp.policy_repayments_plan_id = 4 THEN 
          365 
         END) AS missed_days 
       FROM loan_account_transaction AS lat 
       INNER JOIN loan_account AS la ON (la.id = lat.loan_account_id) 
       INNER JOIN loan_product AS lp ON (la.`product_info_id` = lp.`id`) 
       INNER JOIN loan_product_policy AS lpp ON(lpp.id = la.policy_info_id) 
       INNER JOIN (SELECT MAX(lat2.id) AS max_lat2_id,lat2.loan_account_id FROM loan_account_transaction AS lat2 
       WHERE lat2.transaction_date <= '2013-03-31' AND 
       lat2.domain_status_id = 1 AND 
       lat2.country_id IN (1) AND 
       lat2.office_info_id IN ('BI0000000000000000000363') AND 
       lat2.project_info_id IN ('BI0000000000000000000004') 
       GROUP BY lat2.loan_account_id 
      ) AS temp ON lat.id = temp.max_lat2_id 
      WHERE (lat.loan_status_id= 1 OR lat.loan_status_id IS NULL) 
     ) AS tempData 
     INNER JOIN office_info ON (office_info.id = tempData.office_info_id) 
     LEFT JOIN group_info ON (group_info.id = tempData.group_info_id) 
     WHERE missed_days > 30 
     GROUP BY 
     tempData.country_id, 
     tempData.office_info_id, 
     tempData.project_info_id, 
     tempData.prodId, 
     tempData.group_info_id, 
     missed_days 

UNION 

/* Late and NIBL Loan */ 
     SELECT tempData.country_id,tempData.project_info_id, 
       tempData.prodId, 
     tempData.loan_name, 
       tempData.office_info_id,office_info.office_code,office_info.office_name, 
       tempData.group_info_id,group_info.group_code,group_info.group_name, 
       tempData.missed_days,SUM(tempData.principal_outstanding) AS principal_outstanding, 
       COUNT(DISTINCT tempData.member_id) AS borrower_count 
     FROM 
     (SELECT lat.country_id,lat.office_info_id,lat.project_info_id, lp.id AS prodId, lp.`loan_name`, 
       lat.group_info_id, 
        (CASE WHEN lat.loan_status_id = 3 THEN 
         (la.loan_duration*30) -- +(la.loan_duration*30/2) 
        WHEN lat.loan_status_id = 4 THEN 
         (la.loan_duration*30) -- *2 
        WHEN lat.loan_status_id = 5 THEN 
         (la.loan_duration*30) -- *2+7 
        WHEN lat.loan_status_id = 6 THEN 
         (la.loan_duration*30) -- *2+7 
        END) AS missed_days, 
       lat.principal_outstanding,la.member_id 
       FROM loan_account_transaction AS lat 
       INNER JOIN loan_account AS la ON (la.id = lat.loan_account_id) 
       INNER JOIN loan_product AS lp ON (la.`product_info_id` = lp.`id`) 
       INNER JOIN (SELECT MAX(lat2.id) AS max_lat2_id,lat2.loan_account_id FROM 
       loan_account_transaction AS lat2 
       WHERE lat2.transaction_date <= '2013-03-31' AND 
       lat2.domain_status_id = 1 AND 
       lat2.country_id IN (1) AND 
       lat2.office_info_id IN ('BI0000000000000000000363') AND 
       lat2.project_info_id IN ('BI0000000000000000000004') 
       GROUP BY lat2.loan_account_id 
      ) AS temp ON lat.id = temp.max_lat2_id 
       WHERE (lat.loan_status_id != 2 AND lat.loan_status_id != 7 AND lat.loan_status_id = 1) 
     ) AS tempData 
     INNER JOIN office_info ON (office_info.id = tempData.office_info_id) 
     LEFT JOIN group_info ON (group_info.id = tempData.group_info_id) 
     GROUP BY tempData.country_id, 
     tempData.office_info_id, 
     tempData.project_info_id, 
     tempData.prodId, 
     tempData.group_info_id, 
     missed_days 

     ) AS t 
     GROUP BY t.prodId

来源

2013-05-27 Sumon Bappi

MySQL或SQL服务器还是两者兼而有之？另外，如果3个查询给出不同数量的结果，您能否添加所需的结果？ –

什么是列明智？ – Alex

让他们顺行，并使用编程语言来做矩阵转置。（懒惰的解决方案） – diegoperini

郁可使用下面的查询，

select * from 
    (
     (SELECT 'one' AS value_1, 'two' AS value_2) A 

    JOIN (SELECT 'three' AS value_3, 'four' AS value_4) B ON 1=1 

    JOIN (SELECT 'five' AS value_5, 'six' AS value_6) C ON 1=1 
    )

SQL FIDDLE DEMO

答案的新问题：

我不知道F该是去工作。但试试吧。

SELECT * from 
(SELECT COUNT(1) COUNT1, SUM(principal_outstanding) SUM1 
FROM 
(

/* Current Loan */ 
     SELECT tempData.country_id, 
     tempData.project_info_id, 
     tempData.prodId, 
     tempData.loan_name, 
     tempData.office_info_id, 
     office_info.office_code, 
     office_info.office_name, 
     tempData.group_info_id, 
     group_info.group_code, 
     group_info.group_name, 
     missed_days, 
     SUM(tempData.principal_outstanding) AS principal_outstanding, 
     COUNT(DISTINCT member_id) AS borrower_count 

     FROM (
       SELECT lat.country_id,lat.office_info_id,lat.project_info_id, lp.id AS prodId, lp.`loan_name`, 
       lat.group_info_id, 
       lat.principal_outstanding,la.member_id, 

       (CEIL(lat.overdue_amount/la.installment_amount) * 
         CASE WHEN lpp.policy_repayments_plan_id = 1 THEN 
          1 
         WHEN lpp.policy_repayments_plan_id = 2 THEN 
          7 
         WHEN lpp.policy_repayments_plan_id = 3 THEN 
          30 
         WHEN lpp.policy_repayments_plan_id = 4 THEN 
          365 
         END) AS missed_days 
       FROM loan_account_transaction AS lat 
       INNER JOIN loan_account AS la ON (la.id = lat.loan_account_id) 
       INNER JOIN loan_product AS lp ON (la.`product_info_id` = lp.`id`) 
       INNER JOIN loan_product_policy AS lpp ON(lpp.id = la.policy_info_id) 
       INNER JOIN (SELECT MAX(lat2.id) AS max_lat2_id,lat2.loan_account_id FROM loan_account_transaction AS lat2 
       WHERE lat2.transaction_date <= '2013-03-31' AND 
       lat2.domain_status_id = 1 AND 
       lat2.country_id IN (1) AND 
       lat2.office_info_id IN ('BI0000000000000000000363') AND 
       lat2.project_info_id IN ('BI0000000000000000000004') 
       GROUP BY lat2.loan_account_id 
      ) AS temp ON lat.id = temp.max_lat2_id 
      WHERE (lat.loan_status_id= 1 OR lat.loan_status_id IS NULL) 
     ) AS tempData 
     INNER JOIN office_info ON (office_info.id = tempData.office_info_id) 
     LEFT JOIN group_info ON (group_info.id = tempData.group_info_id) 
     WHERE missed_days > 30 
     GROUP BY 
     tempData.country_id, 
     tempData.office_info_id, 
     tempData.project_info_id, 
     tempData.prodId, 
     tempData.group_info_id, 
     missed_days) g1 
     GROUP BY g1.prodId)t1 



/* Late and NIBL Loan */ 
     JOIN (SELECT COUNT(1) COUNT2, SUM(principal_outstanding) SUM2 
FROM(SELECT tempData.country_id,tempData.project_info_id, 
       tempData.prodId, 
     tempData.loan_name, 
       tempData.office_info_id,office_info.office_code,office_info.office_name, 
       tempData.group_info_id,group_info.group_code,group_info.group_name, 
       tempData.missed_days,SUM(tempData.principal_outstanding) AS principal_outstanding, 
       COUNT(DISTINCT tempData.member_id) AS borrower_count 
     FROM 
     (SELECT lat.country_id,lat.office_info_id,lat.project_info_id, lp.id AS prodId, lp.`loan_name`, 
       lat.group_info_id, 
        (CASE WHEN lat.loan_status_id = 3 THEN 
         (la.loan_duration*30) -- +(la.loan_duration*30/2) 
        WHEN lat.loan_status_id = 4 THEN 
         (la.loan_duration*30) -- *2 
        WHEN lat.loan_status_id = 5 THEN 
         (la.loan_duration*30) -- *2+7 
        WHEN lat.loan_status_id = 6 THEN 
         (la.loan_duration*30) -- *2+7 
        END) AS missed_days, 
       lat.principal_outstanding,la.member_id 
       FROM loan_account_transaction AS lat 
       INNER JOIN loan_account AS la ON (la.id = lat.loan_account_id) 
       INNER JOIN loan_product AS lp ON (la.`product_info_id` = lp.`id`) 
       INNER JOIN (SELECT MAX(lat2.id) AS max_lat2_id,lat2.loan_account_id FROM 
       loan_account_transaction AS lat2 
       WHERE lat2.transaction_date <= '2013-03-31' AND 
       lat2.domain_status_id = 1 AND 
       lat2.country_id IN (1) AND 
       lat2.office_info_id IN ('BI0000000000000000000363') AND 
       lat2.project_info_id IN ('BI0000000000000000000004') 
       GROUP BY lat2.loan_account_id 
      ) AS temp ON lat.id = temp.max_lat2_id 
       WHERE (lat.loan_status_id != 2 AND lat.loan_status_id != 7 AND lat.loan_status_id = 1) 
     ) AS tempData 
     INNER JOIN office_info ON (office_info.id = tempData.office_info_id) 
     LEFT JOIN group_info ON (group_info.id = tempData.group_info_id) 
     GROUP BY tempData.country_id, 
     tempData.office_info_id, 
     tempData.project_info_id, 
     tempData.prodId, 
     tempData.group_info_id, 
     missed_days 

     ) AS g2 
     GROUP BY g2.prodId) t2 ON 1=1

来源

2013-05-27 06:53:03 Maximus

它的工作原理。谢谢。请你再帮我多一点。我有一个SQL返回两行和两个字段的结果。但我希望结果在一行和四个字段中。我正在将查询上传为新的查询。你能帮我解决这个问题吗？ –

请更新。如果可以的话，我会帮你的。 – Maximus

sql post is updated –

这可能是你的意思吗？

SELECT id, column1 AS data 
FROM table1 
UNION 
SELECT id, column2 AS data 
FROM table2 
UNION 
SELECT id, column3 AS data 
FROM table3;

列column1，column2，column3都应该有相同的数据类型，并走样他们具有相同的名称是什么让UNION的工作。

来源

2013-05-27 04:46:33

感谢您的回复。但是如果我使用union，我不会在结果中获得适当的列名称。我正在给我原来的查询。它很大。这就是为什么我这样发布。我需要的是我想要每个排队的结果作为一个列。就这样。 –

  
WITH cte (value1, value2) 
         AS (SELECT 'ONE' AS value_1 , 
            'TWO' AS value_2 
         ), 
        cte2 
         AS (SELECT 'THREE' AS value_3 , 
            'FOUR' AS value_4 
         ), 
        cte3 
         AS (SELECT 'FIVE' AS value_5 , 
            'SIX' AS value_6 
         ) 
       SELECT * 
       FROM cte , 
         cte2 , 
         cte3

来源

2013-05-27 05:38:26 Ahmed

我认为你正在寻找这个。 – Ahmed

亲爱的朋友，没有运气 –

从多个sql中获得结果作为列明智

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