[英]CRUD with Python … IndexError: tuple index out of range
我必須使用Python做幾個選擇和更新。 這是一種全新的語言,在執行以下(簡單)查詢時,我在語法上遇到了一些麻煩:
SELECT A.CUSTOMER_NAME,
A.CUSTOMER_CITY,
B.POPULATION
FROM
CONTRACTS AS A
JOIN CITIES AS B ON
A.CUSTOMER_CITY = B.IDENT
WHERE B.POPULATION <=500000
SELECT A.IDENT,
A.MAKE,
A.MODEL,
A.LUXURY,
B.CAR_IDENT
FROM CARS AS A
JOIN CONTRACTS AS B ON
A.IDENT = B.CAR_IDENT
WHERE LUXURY = 'Y'
UPDATE CONTRACTS
SET BASE_PRICE=1000
WHERE CONTRACT_CLASS >=10
我從更新開始...我認為它比代碼更短...。當我執行更新語句時,出現以下錯誤:
>'update {} set BASE_PRICE = ? where {}'.format(self._table), (row['BASE_PRICE'], >row['where']))
>IndexError: tuple index out of range
這是我的方法
def retrieve(self, key):
cursor = self._db.execute('select from {}'.format(self._table))
return dict(cursor.fetchone())
def update(self, row):
self._db.execute(
'update {} set BASE_PRICE = ? where {}'.format(self._table),
(row['BASE_PRICE'], row['where']))
self._db.commit()
def disp_rows(self):
cursor = self._db.execute('select IDENT, CONTRACT_CLASS, BASE_PRICE from {} order
by BASE_PRICE'.format(self._table))
for row in cursor:
print(' {}: {}: {}'.format(row['IDENT'], row['CONTRACT_CLASS'],
row['BASE_PRICE']))
def __iter__(self):
cursor = self._db.execute('select * from {} order by
BASE_PRICE'.format(self._table))
for row in cursor:
yield dict(row)
def main():
db_cities = database(filename = 'insurance.sqlite', table = 'CITIES')
db= database(filename = 'insurance.sqlite', table = 'CONTRACTS')
db_cars = database(filename = 'insurance.sqlite', table = 'CARS')
print('Retrieve rows')
for row in db: print(row)
print('Update rows')
db.update({'where': 'CONTRACT_CLASS' >= 10', 'BASE_PRICE'= 1000})
for row in db: print(row)
print('Retrieve rows after update')
for row in db: print(row)
Thanks in advance for your help!
您在self._table
之后放了一個括號。
'UPDATE {table_name} SET BASE_PRICE = {base_price} WHERE {condition}'.format(
table_name=self._table,
base_price=row['BASE_PRICE'],
condition=row['where']
)
我還做了一些代碼清理。
另一個問題:為什么不使用ORM?
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.