MySQL：如何编写查询以选择 MAX 和今天的日期以从给定列表中获取数据并存储在数据库中

Question

我在 mysql 数据库中存储数据时遇到问题。具有列表df = [['Euro', 0.877641, '2018-10-24'], ['British Pound', 0.7752180000000001, '2018-10-24'], ['Indian Rupee', 73.21869000000001, '2018-10-24'], ['Australian Dollar', 1.4135360000000001, '2018-10-24'], ['Canadian Dollar', 1.299316, '2018-10-24']] 列表的变量希望将其存储在最后更新日期的货币名称、日期、值列中。那么我如何在代码中编写该逻辑（我是使用 sql 的 max()、min() 方法。）

试过了：

    import pandas as pd
     import pymysql
    import MySQLdb
        sqlconn = pymysql.connect(host='ip', port=port, user='username',
                                  passwd='pass', db='dbname', autocommit=True)
        # Create a cursor object
        if (sqlconn == None):
            print("conn not found")
        else:
            print("conn is found")


        cursor.execute("INSERT INTO TestCurrencyHistory (currency, date, value) VALUES (%s, %s, %s)") % (
                                              elements
                                            )
        element = cursor.fetchall()

        cursor.close()
        sqlconn.close()
Offcource this is not working. So please help ..

提前谢谢..（如果有任何 python/pandas 逻辑来选择最大和最小日期，请不要告诉我）

Answer 1

使用 mysql-connector，您可以使用 executemany()，但我不确定 pandas 是否支持。这篇文章可能会有所帮助：how to transform pandas dataframe for insertion via executemany() statement?

df = [['Euro', 0.877641, '2018-10-24'], ['British Pound', 0.7752180000000001, '2018-10-24'], ['Indian Rupee', 73.21869000000001, '2018-10-24'], ['Australian Dollar', 1.4135360000000001, '2018-10-24'], ['Canadian Dollar', 1.299316, '2018-10-24']]

获取df的最新日期：

date_list = []
for x in df:
    date_list.append(x[-1])
latest = max(date_list)

# turn into datetime format, not str
latest = datetime.datetime.strptime(latest, '%Y-%m-%d').date()

# increase by one day
next_day = latest + datetime.timedelta(days=1)

我不知道您的 SQL 查询是什么样的，但您可能会执行以下操作以获取第二天的数据：

latest_date_query_results = cursor.execute(
"""
SELECT currency, value, added_date
FROM your_table
WHERE added_date > {}
""".format(next_day)
)

结果应该变成一个元组来处理executemany()

elements = [] 
for x in df:
    elements.append(tuple((x)))
elements = tuple((elements))
print(elements)
# (('Euro', 0.877641, '2018-10-24'), ('British Pound', 0.7752180000000001, '2018-10-24'), ('Indian Rupee', 73.21869000000001, '2018-10-24'), ('Australian Dollar', 1.4135360000000001, '2018-10-24'), ('Canadian Dollar', 1.299316, '2018-10-24'))

那么，你可以这样做：

sql = "INSERT INTO TestCurrencyHistory (currency, date, value) VALUES (%s, %s, %s)"
cursor.executemany(sql, elements)