如何在 MySQL 语句中使用 .format() 字符串？

Question

我正在尝试使用今天以及 7 天前的数据来格式化 MySQL 语句。我很确定日期的格式正确，所以我认为这不是问题所在。

报错为：

 Warning: (1292, "Incorrect datetime value: '{} 16:00:00' for column 'run_start_date' at row 1")
      result = self._query(query)
    Traceback (most recent call last):
         AttributeError: 'int' object has no attribute 'format'

例如

today = DT.date.today()
week_ago = today - DT.timedelta(days=7)
print(today.strftime('%Y-%m-%d'))
print(week_ago.strftime('%Y-%m-%d'))

cursor.execute(SELECT * FROM db WHERE run_start_date BETWEEN '{} 16:00:00' AND '{} 16:00:00'format(week_ago, today)

Answer 1

使用准备好的语句（安全方式）：

qry = """
    SELECT *
    FROM db
    WHERE run_start_date BETWEEN '%s 16:00:00' AND '%s 16:00:00'
""" # 

today = DT.date.today()
week_ago = today - DT.timedelta(days=7)
today = str(today.strftime('%Y-%m-%d'))  # Convert to string
week_ago = str(week_ago.strftime('%Y-%m-%d'))  # Convert to string

cursor.execute(qry, [today, week_ago])

使用 .format() 会使您面临 sql 注入的风险（例如，如果您将用户输入传递给 .format()）

qry = """
    SELECT *
    FROM db
    WHERE run_start_date BETWEEN '{today} 16:00:00' AND '{week_ago} 16:00:00'
""" # Use named placeholders, nicer to read, prevents you having to repeat variables multiple time when calling .format()

today = DT.date.today()
week_ago = today - DT.timedelta(days=7)
today = str(today.strftime('%Y-%m-%d'))  # Convert to string
week_ago = str(week_ago.strftime('%Y-%m-%d'))  # Convert to string

qry = qry.format(today=today, week_ago=week_ago)
cursor.execute(qry)

如何在 MySQL 语句中使用 .format() 字符串？

How to use .format() string in a MySQL statement?

python

mysql

string.format

使用准备好的语句（安全方式）：

使用 .format() 会使您面临 sql 注入的风险（例如，如果您将用户输入传递给 .format()）