尝试使用每个现有字段值加上传递的参数的总和来更新所有记录。如果记录 A 中的字段 = 4 且记录 B = 5,我想更新所有记录,因此 A = 4 + X 和 B = 5 + X。这样,我可以快速将所有记录值增加给定数字 (X)。到目前为止,我还不知道如何让更新语句扩展传递的参数。
我可以用这行代码添加硬编码值:c.execute("UPDATE level SET sell = sell + 1;")
下面的代码按原样工作。但是当我尝试用这行代码传递一个值update_sell(1)
并def update_sell(margin):
像这样引用传递的值时,c.execute("UPDATE level SET sell = sell + margin;")
它失败了。我在这里遗漏了什么?
这里完全是 Python 菜鸟。
要运行的脚本:
#MainScript.py
import sqlite3
from LevelClass import level
conn = sqlite3.connect(':memory:')
c = conn.cursor()
c.execute("""CREATE TABLE level (
buy integer,
sell integer,
quan integer
)""")
def insert_level(level):
with conn:
c.execute("INSERT INTO level VALUES (:buy, :sell, :quan)", {'buy': level.buy, 'sell': level.sell, 'quan': level.quan})
def get_level_by_sell(sell):
c.execute("SELECT * FROM level WHERE sell=:sell", {'sell': sell})
return c.fetchall()
def update_sell(margin):
with conn:
# below works with "sell + 1", but fails with "sell + margin"
c.execute("UPDATE level SET sell = sell + 1;")
trans_1 = level(1, 5, 50)
trans_2 = level(2, 10, 60)
insert_level(trans_1)
insert_level(trans_2)
# value to pass to update_sell as var "margin"
update_sell(1)
find = get_level_by_sell(6)
print(find)
find = get_level_by_sell(11)
print(find)
conn.close()
班级:
# saved as LevelClass.py
class level:
def __init__(self, buy, sell, quan):
self.buy = buy
self.sell = sell
self.quan = quan
def __repr__(self):
return "Level('{}', '{}', {})".format(self.buy, self.sell, self.quan)
答案1
我认为您忘记了:
更新函数。这对我来说有用。
def update_sell(margin):
with conn:
# below works with "sell + 1", but fails with "sell + margin"
#c.execute("UPDATE level SET sell = sell + 1")
c.execute("UPDATE level SET sell = sell + :margin", {"margin":margin})