Python，数据框sql连接

Question

我有两个直接查询数据库的 python 函数。 有没有办法在 python 中加入这两个函数？

我想做几个连接，但不确定如何在 python 中完成。

查询 1：

def query1(businessDate):
    con = pyodbc.connect(r'DSN='+'Stack',autocommit=True)  
    print('working')
    #businessDate = r"'2019-03-13'"
    #remember business date should be entered like "'2019-03-13'"

    sql = f"""

     SELECT 
       iddate, 
       businessdate, 
       stack, identifier
     FROM stackoverflow
     where stack is not null 
     and businessdate = {businessDate}

    """

    df_stack = pd.read_sql(sql,con)

    con.close()

    return(df_stack)

查询 2：

    def superuser(businessDate):  
    con = pyodbc.connect(r'DSN='+'super',autocommit=True)  
    print('working')
    #remember business date should be entered like "'2019-03-13'"

    sql = f"""
    SELECT 
      iddate,
      businessdate,
      stack, identifier
    FROM superuser
    WHERE stack is not null 
    and businessdate = {businessDate}


    """

    df_super = pd.read_sql(sql,con)

    con.close()

    return(df_super)

我想在identifier、stack、iddate 和 businessdate 上对表 2 进行左外联接

尝试：

def testjoin():
    con = pyodbc.connect(r'DSN='+'Stack',autocommit=True)  
    print('working') 

    pd.merge(df_stack,df_super, on = ['identifier','stack','iddate'])

    df_test = pd.read_sql(sql,con)

    con.close()

    return(df_test)

尝试2：

def testjoin():
    con = pyodbc.connect(r'DSN='+'Stack',autocommit=True)  
    print('working') 

    df_stack= query1("'2019-03-13'")
    df_super= superuser("'2019-03-13'")

    pd.merge(df_stack,df_super, on = ['identifier','stack','iddate'])

    df_test = pd.read_sql(sql,con)

    con.close()

    return(df_test)

收到错误name 'sql' is not defined'

Answer 1

左外连接

SELECT *
FROM df_stack
LEFT OUTER JOIN df_super
    ON df_stack.stack = df_super.stack
    ON df_stack.identifier= df_super.identifier 
    ON df_stack.iddate = df_super.iddate
    ON df_stack.businessdate = df_super.businessdate;

pd.merge(df_stack,df_super,
         on=['iddate','businessdate', 'stack', 'identifier'], 
         how='left')

Answer 2

好的，我将把它作为答案而不是 cmets 发布，因为有几种方法可以满足您的要求。 sql 没有像我在评论中指出的那样定义，因为它超出了函数 testjoin() 的范围。

一种方法是将 SQL 字符串视为全局变量，然后在函数内部访问它。

sql = '''
  SELECT 
    iddate, 
    businessdate, 
    stack, identifier
  FROM stackoverflow
  where stack is not null 
  and businessdate = {businessDate}
'''

def testjoin():
    con = pyodbc.connect(r'DSN='+'Stack',autocommit=True)  
    print('working') 

    df_stack= query1("'2019-03-13'")
    df_super= superuser("'2019-03-13'")

    pd.merge(df_stack,df_super, on = ['identifier','stack','iddate'])

    df_test = pd.read_sql(sql.format(businessDate="'2019-03-14'"),con)

    con.close()

    return(df_test)

我使用.format() 而不是f 字符串的原因是f 字符串需要在创建f 字符串时声明变量。如果您没有 businessdate 作为变量，那将是一个错误。 .format() 允许您将变量放在字符串中，然后随时更改它的值。如果您的查询的主要部分不会发生太大变化并且您只需要按日期过滤，我会这样做。

另一种方法是在函数外部构建字符串，然后将其作为参数传入

businessdate = "'2019-03-13'"
sql = f'''
  SELECT 
    iddate, 
    businessdate, 
    stack, identifier
  FROM stackoverflow
  where stack is not null 
  and businessdate = {businessDate}
'''

def testjoin(sql_string):
    con = pyodbc.connect(r'DSN='+'Stack',autocommit=True)  
    print('working') 

    df_stack= query1("'2019-03-13'")
    df_super= superuser("'2019-03-13'")

    pd.merge(df_stack,df_super, on = ['identifier','stack','iddate'])

    df_test = pd.read_sql(sql_string,con)

    con.close()

    return(df_test)

test_df = testjoin(sql)

您也可以继续在每个函数中构建字符串，但考虑到编码中的Don't Repeat Yourself paradigm，并且您已经在其他两个函数中构建它，最好避免这种情况。