Question

在Python 3.4.1中，为什么sqlite3不将自动递增的ID插入到下面程序的表中？根据{{3}}，一个整数主键列应该自动递增，我可以看到cur.lastrowid返回一个有效的整数，但相同的值没有插入表中（它变为NULL）。 / p>

import sqlite3


with sqlite3.connect(':memory:') as conn:
    cur = conn.cursor()
    # Note that column 'id' is an integer primary key
    cur.execute('create table test (id int primary key , name text)')
    cur.execute('insert into test (name) values (?)', ('Test',))
    last_id = cur.lastrowid
    assert last_id is not None

    id_, = cur.execute('select id from test').fetchone()
    assert id_ == last_id, '{} != {}'.format(id_, last_id)

Answer 1

显然，我错误地认为＆＃39; int＆＃39;是＆＃39;整数＆＃39;的同义词。在SQLite中。实际上，列are typeless in SQLite和integer primary key是此规则的一个例外，有效地声明了一个自动递增的列：

import sqlite3


with sqlite3.connect(':memory:') as conn:
    cur = conn.cursor()
    # Note that column 'id' is an integer primary key
    cur.execute('create table test (id integer primary key , name text)')
    cur.execute('insert into test (name) values (?)', ('Test',))
    last_id = cur.lastrowid
    assert last_id is not None

    id_, = cur.execute('select id from test').fetchone()
    assert id_ == last_id, '{} != {}'.format(id_, last_id)

为什么Python sqlite3没有插入自动递增的ID

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