我正在尝试创建具有2个现有表的联接表。如下所示:
这是第一个表查询,看起来像这样
alphabets = {"A":1,"B":2,"C":3, "D":4,"E":5,"F":6,"G":7,"H":8,"I":9,"J":1,"K":2,"L":3,"M":4,"N":5,"O":6,"P":7,"Q":8,"R":9,"S":1,
"T":2,"U":3,"V":4,"W":5, "X":6,"Y":7,"Z":8}
def digit_sum(num):
return sum( [ int(char) for char in str(num) ] )
def numerology(word):
total = 0
for letter in word:
total += alphabets[letter]
total = digit_sum(total)
return total
fan = 'PRINCELY'
def vowels(fan):
vowels=[]
if 'I' in fan:
vowels.append(9)
fan1=fan[:fan.index('I')]+fan[fan.index('I')+1:]
consonant = fan1
if 'E' in fan:
vowels.append(5)
fan2=fan1[:fan1.index('E')]+fan1[fan1.index('E')+1:]
consonant = fan2
if 'A' in fan:
vowels.append(1)
fan3=fan2[:fan2.index('A')]+fan2[fan2.index('A')+1:]
consonant = fan3
if 'O' in fan:
vowels.append(6)
fan4=fan3[:fan3.index('O')]+fan3[fan3.index('O')+1:]
consonant = fan4
if 'U' in fan:
vowels.append(3)
fan5=fan4[:fan4.index('U')]+fan4[fan4.index('U')+1:]
consonant = fan5
print(vowels)
print(consonant)
print(digit_sum(sum(vowels)))
cons = numerology(consonant)
print(cons)
vowels(fan)
#outputs
#[9, 5]
#PRNCLY
#5
#7
这是第二个表查询,看起来像这样
SELECT
DATEPART( week, dbo.Income.IncomeDate ) AS [Week Income],
DATEPART( YEAR, dbo.Income.IncomeDate ) AS [Year],
SUM ( dbo.Income.CardAmount ) AS [Total Card],
SUM ( dbo.Income.CashAmount ) AS [Total Cash],
SUM ( dbo.Income.TipsAmount ) AS [Total Tip],
SUM ( dbo.Income.SalaryAmount ) AS [Total Salary],
SUM ( dbo.Income.Adjustment ) AS [Total Adjustment]
FROM
dbo.Income
GROUP BY
DATEPART( week, dbo.Income.IncomeDate ),
DATEPART( YEAR, dbo.Income.IncomeDate )
ORDER BY
DATEPART(YEAR, dbo.Income.IncomeDate )
我所期望的就是这样。表1和表2相加,总费用相加。
答案 0 :(得分:0)
正如SO人士所说,我们无法访问描述您预期结果的图像。但是,我们了解到您正在寻找JOIN
您所显示的两个查询的结果。
操作方法如下:将每个查询转换为子查询,然后将其JOIN
一起放在其公共键字段上;在您的用例中,必须为Year
和Week
。 ORDER BY
子句需要移到外部查询。在SELECT
,WHERE
和ORDER BY
子句中,您可以使用定义的别名(此处为A
和{{1},从子查询中自由访问所有字段。 }。
示例代码(您将不得不修改B
,我无法告诉您您希望它的外观如何):
SELECT