使用Python sqlite3模块时的OperationalError

Question

def StatusUpdate(self, table):
    inventoryCurs.execute('SELECT * from Table')
    for i in inventoryCurs:
        html = urlopen(i[5]).read()
        Soup = BeautifulSoup(html)

        if table.StockStatus(Soup) == 'Out of Stock':        
            inventoryCurs.execute('''UPDATE table SET status = 'Out of Stock' WHERE id = %s)''', i[0])

inventoryCurs.execute('''UPDATE table SET status = 'Out of Stock' WHERE id = %s)''', i[0]) OperationalError: near "%": syntax error

Answer 1

Without seeing more of the code, it's difficult to fix the problem completely, but looking at your code, I think the problem might be the %s in this line:

inventoryCurs.execute('''UPDATE table SET status = 'Out of Stock' WHERE id = %s)''', i[0])

According to the documentation for the SQLite module in both Python 2 and Python 3 , the sqlite3 module requires a ? as a placeholder, not %s or some other format string.

According to the Python 2 documentation , a %s placeholder could be used like this:

import sqlite3
conn = sqlite3.connect('example.db')
c = conn.cursor()
# Never do this -- insecure!
symbol = 'IBM'
c.execute("select * from stocks where symbol = '%s'" % symbol)

but that's a simple format string, not actually the database's placeholder. Also, as the comment shows, you should never build queries that way because it makes them vulnerable to SQL injection. Rather, you should build them like this, using a ? instead:

import sqlite3
conn = sqlite3.connect('example.db')
c = conn.cursor()
# Do this instead
t = (symbol,)
c.execute('SELECT * FROM stocks WHERE symbol=?', t)

The documentation has more details, but I believe that is the solution to the error you posted.