Что с кодом этим?

Question

[Марат Мхитарян]

I have a code:

import requests
import sqlite3
conn = sqlite3.connect('g.db')
c = conn.cursor()

for row in c.execute('SELECT * FROM urls ORDER BY utl'):
 r = requests.get(row[3])
 c.execute('UPDATE urls SET utl = ? WHERE ID = ?' , ([sqlite3.Binary(r.content)], row[0]))
conn.commit()

conn.close()

I have error:

Traceback (most recent call last):
  File "/Users/mkhitaryan/Documents/303.py", line 8, in <module>
    c.execute('UPDATE urls SET utl = ? WHERE ID = ?' , ([sqlite3.Binary(r.content)], row[0]))
sqlite3.InterfaceError: Error binding parameter 0 - probably unsupported type.
[Finished in 1.4s with exit code 1]
[shell_cmd: python -u "/Users/mkhitaryan/Documents/303.py"]
[dir: /Users/mkhitaryan/Documents]
[path: /usr/bin:/bin:/usr/sbin:/sbin]

Why!?

Comments

[Александр Ручкин]

Почему [sqlite3.Binary(r.content)], а не просто sqlite3.Binary(r.content)?

[Марат Мхитарян]

Александр Ручкин: без [] я бы не смог сохранить 107 000 байт

[Roman K]

Желание выёживаться в мелочах?

Answer 1

[Довольный Айтишникъ]

Error binding parameter 0 - probably unsupported type