Question

我的数据采用这种格式

CALENDAR    CLIENTID    TOTAL
20170801    1700          2
20170801    1800    
20170801    1900          2
20170801    1990    
20170801    2000          0
20170801    2090          0
20170802    2090          0
20170803    2090          0

当一个特定的客户连续3个总数为零时，我试图获得最小的calendar day。因此，在上面的示例中，我的输出为20170801，因为客户端2090在total，20170801和20170802天的20170803为零。到目前为止，我有以下查询，但是a和b的计数不准确。

WITH cte AS (
    SELECT *,COUNT(1) OVER(PARTITION BY clientid) b  FROM
    (
        SELECT tt.*
           ,(SELECT COUNT(CALENDAR) FROM STATS 
                WHERE total = 0 
            ) AS a
        FROM STATS tt
        WHERE total = 0
    ) t1
)
SELECT * FROM cte WHERE b >= 3

Answer 1

使用lead()。以下查询获取所有此类日期，按日期排序：

select s.*
from (select s.*,
             lead(total) over (partition by clientid order by calendar) as total_1,
             lead(total, 2) over (partition by clientid order by calendar) as total_2
      from stats s
     ) s
where total = 0 and total_1 = 0 and total_2 = 0
order by date;

您可以为数据库添加fetch first 1 row only或等同名称，以仅获得一行。

Answer 2

方法是

from PyQt4 import QtGui, QtCore

class Window(QtGui.QWidget):
    def __init__(self):
        QtGui.QWidget.__init__(self)
        layout = QtGui.QVBoxLayout(self)
        self.combo = QtGui.QComboBox()
        self.combo.setEditable(True)
        self.combo.lineEdit().setAlignment(QtCore.Qt.AlignCenter)
        self.combo.addItems('One Two Three Four Five'.split())
        layout.addWidget(self.combo)


if __name__ == '__main__':

    import sys
    app = QtGui.QApplication(sys.argv)
    window = Window()
    window.show()
    sys.exit(app.exec_())

结果

WITH cte AS (
  Select * From dbo.Consequtive Where TOTAL = 0
  )

SELECT r1.*, r2.*, r3.*
  FROM cte as r1
Left Join cte as r2
  On r1.CLIENTID = r2.CLIENTID and DATEADD(day, 1, r1.CALENDAR)  = r2.CALENDAR
Left Join cte as r3
  On r1.CLIENTID = r3.CLIENTID and DATEADD(day, 2, r1.CALENDAR)= r3.CALENDAR

Where r1.TOTAL = 0 and r2.TOTAL = 0 and r3.TOTAL = 0

查找具有相同值的连续天

2 个答案: