Question

代码如下。如何按变量值?替换[table, url]？

预期的SQL命令为select * from OTHER_URL where url="http://a.com/a.jpg"

此SQL命令在sqlite3命令行界面上不会发生错误。

import sqlite3
from contextlib import closing

dbname = "ng.db"

with closing(sqlite3.connect(dbname)) as conn:
    c = conn.cursor()
    c.execute("CREATE TABLE IF NOT EXISTS OTHER_URL (url TEXT)")
    conn.commit()

table = "OTHER_URL"
url = "http://a.com/a.jpg"

with closing(sqlite3.connect(dbname)) as conn:
    c = conn.cursor()
    c.execute('select * from ? where url="?"', [table, url])
    print c.fetchone()

Answer 1

这里有两个错误。首先，您不能对表名（或列名）使用参数替换，仅用于值。您需要将字符串插值用于其他任何内容。

其次，你不需要在value参数周围引用;替代将照顾到这一点。

所以：

from datetime import datetime

with open("file.txt", mode="r+") as file1:
    opened = datetime.now()
    # do stuff with file

duration = datetime.now() - opened

print('File opened for {}'.format(duration.total_seconds()))

Python2.7 - SQLite3库输出错误消息＆＃34; sqlite3.OperationalError：near＆＃34;？＆＃34;：语法错误＆＃34;

1 个答案: