Question

问题：UNION ALL有100多个表。名称中唯一的区别是日期。与其编写100多个代码块并仅更改日期，不如编写一个python脚本输出sql代码更有意义：

我们的SQL查询，查询一张表（即一个财务季度）：

SELECT 
  (SELECT
      PCR.repdte
      FROM 
      All_Reports_19921231_Performance_and_Condition_Ratios as 
      PCR) AS Quarter,
  (SELECT 
      Round(AVG(PCR.lnlsdepr))
      FROM 
      All_Reports_19921231_Performance_and_Condition_Ratios as 
      PCR) AS NetLoansAndLeasesToDeposits,
  (SELECT sum(CAST(LD.IDdepsam as int))
      FROM 
     'All_Reports_19921231_Deposits_Based_on_the_Dollars250,000
     _Reporting_Threshold' AS LD) AS 
     DepositAccountsWith$LessThan$250k

每个文件名的命名约定包括日期（财务季度）

All_Reports_19921231_Performance_and_Condition_Ratios
All_Reports_19921231_Performance_and_Condition_Ratios
All_Reports_19921231_Deposits_Based_on_the_Dollars250,000
_Reporting_Threshold

我们想查询包括19921231在内的所有财务季度

脚本将：

Step one:  find the yyyy/mm/dd in the file name (e.g. 19921231)

Step two:  copy the query

Step three: change the yyyy/mm/dd in the copied file name(s)
IF 1231 change to “+1”0331   (e.g. 19921231 to 19930331)
IF 0331 change to 0630       (e.g. 19930331 to 19930630)
IF 0630 change to 0930       (e.g. 19930630 to 19930930)
IF 0930 change to 1231       (e.g. 19930930 to 19931231)
IF 1231 change to +1 0331    (e.g. 19931231 to 19940331)
…..
…..
…..
IF 91231 change to 00331  (e.g. 19991231 to 20000331)
….
IF 91231 change to 0031 (e.g. 20091231 to 20100331) 

Step four: print new code block after UNION ALL 
Step five: repeat step three
Step six: repeat step four

输入将是单个财务季度（请参见上面的代码块），输出是该代码块重复100多次以上，每个文件名中仅更改yyyy / mm / dd。每个代码块都将与UNION ALL连接起来：

SELECT 
  (SELECT
      PCR.repdte
      FROM 
      All_Reports_19921231_Performance_and_Condition_Ratios as 
      PCR) AS Quarter,
  (SELECT 
      Round(AVG(PCR.lnlsdepr))
      FROM 
      All_Reports_19921231_Performance_and_Condition_Ratios as 
      PCR) AS NetLoansAndLeasesToDeposits,
 (SELECT sum(CAST(LD.IDdepsam as int))
      FROM 
     'All_Reports_19921231_Deposits_Based_on_the_Dollars250,000
      _Reporting_Threshold' AS LD) AS 
      DepositAccountsWith$LessThan$250k
UNION ALL
SELECT 
  (SELECT
      PCR.repdte
      FROM 
      All_Reports_19930330_Performance_and_Condition_Ratios as 
      PCR) AS Quarter,
   (SELECT 
     Round(AVG(PCR.lnlsdepr))
     FROM All_Reports_19930330_Performance_and_Condition_Ratios 
     as PCR) AS NetLoansAndLeasesToDeposits,
   (SELECT sum(CAST(LD.IDdepsam as int))
     FROM 
    'All_Reports_19930330_Deposits_Based_on_the_Dollars250,000
     _Reporting_Threshold' AS LD) AS 
     DepositAccountsWith$LessThan$250k

Answer 1

第一步是编写一个生成器，该生成器以给定日期字符串到现在的三个月为增量生成日期字符串。我们可以将初始日期存储在_, contours, _ = cv2.findContours(closing, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE) bgr_img = cv2.merge([img, img, img]) contour_img = cv2.drawContours(bgr_img, contours, -1, (255, 0, 255))对象中，使用该对象在每个步骤中生成新的日期字符串，并使日期不超过当前日期。然后，我们可以使用calendar.monthrange查找给定月份的最后几天。

datetime

然后我们可以使用字符串格式将该值注入模板字符串

from datetime import datetime
from calendar import monthrange

def dates_from(start):
    date  = datetime.strptime(start, "%Y%m%d")
    today = datetime.today()
    while date < today:
        yield date.strftime("%Y%m%d")
        month = date.month + 3
        year = date.year
        if month > 12:
            year += 1
            month -= 12
        _, day = monthrange(year, month)
        date = datetime(year, month, day)

在文件名（yyyy / mm / dd）中查找日期/财务季度，然后将文件名中的日期更改为yyyy / mm / dd + 3个月

1 个答案: