Python sqlite3 : UPDATE failure claims "no such column"

Question

以下脚本在 UPDATE 命令中引发以下异常。我认为这个简单的 UPDATE 命令应该发生的是 db_2.foo.bar 的值应该从 1 加倍到 2。我的猜测是我在 UPDATE 语句中有一些微妙的语法错误(或者是 Python 中的错误)或sqlite3级别)；然而，我仔细研究了 sqlite3 文档——尤其是“更新”和“表达式”页面——并没有发现我做错了什么。

db_1.foo_bar= (1,)
Traceback (most recent call last):
  File "try2.py", line 29, in <module>
    db_2.execute( 'UPDATE foo SET bar = bar + db_1.foo.bar WHERE rowid = db_1.foo.rowid' ) 
sqlite3.OperationalError: no such column: db_1.foo.bar

有什么建议或解决方法吗？

import sqlite3

# Create db_1, populate it and close it:
open( 'db_1.sqlite', 'w+' )
db_1 =  sqlite3.connect( 'db_1.sqlite' )
db_1.execute( 'CREATE TABLE foo(bar INTEGER)' )
db_1.execute( 'INSERT INTO foo (bar) VALUES (1)' )
db_1.commit()
db_1.close()

# Create db_2:
open( 'db_2.sqlite', 'w+' )
db_2 =  sqlite3.connect( 'db_2.sqlite' )
db_2.execute( 'CREATE TABLE foo(bar INTEGER)' )

# Attach db_1 to db_2 connection:
db_2.execute( 'ATTACH "db_1.sqlite" AS db_1' )

# Populate db_2 from db_1:
db_2.execute( 'INSERT INTO foo SELECT ALL * FROM db_1.foo' )

# Show that db_1.foo.bar exists:
cur_2 = db_2.cursor()
cur_2.execute( 'SELECT bar from db_1.foo' )
for result in cur_2.fetchall():
    print 'db_1.foo_bar=', result

# However, the following claims that db_1.foo.bar does not exist:
db_2.execute( 'UPDATE foo SET bar = bar + db_1.foo.bar WHERE rowid = db_1.foo.rowid' ) 

db_2.execute( 'DETACH db_1')
db_2.commit()
db_2.close()

Answer 1

要使用不同表中的值更新 foo，您可以使用嵌套 SELECT 表达式。注意，foo.rowid指的是外表的rowid，而t.rowid指的是内表的rowid:

cur_2.execute( '''\
    UPDATE foo SET bar = bar +
        IFNULL( (SELECT t.bar 
                 FROM db_1.foo AS t
                 WHERE foo.rowid = t.rowid), 0)''' )  

为了测试正确的 rowids 确实匹配在一起，我稍微修改了您的代码，以便 db_1.foo 的 rowids 执行此操作与 db_2.foo 的 rowids 不匹配:

import sqlite3

# Create db_1, populate it and close it:
open( 'db_1.sqlite', 'w+' )
db_1 =  sqlite3.connect( 'db_1.sqlite' )
db_1.execute( 'CREATE TABLE foo(bar INTEGER)' )
db_1.execute( 'INSERT INTO foo (rowid,bar) VALUES (2,1)' )
db_1.execute( 'INSERT INTO foo (rowid,bar) VALUES (3,2)' )
db_1.commit()
db_1.close()

# Create db_2:
open( 'db_2.sqlite', 'w+' )
db_2 =  sqlite3.connect( 'db_2.sqlite' )
cur_2 = db_2.cursor()
cur_2.execute( 'CREATE TABLE foo(bar INTEGER)' )

# Attach db_1 to db_2 connection:
cur_2.execute( 'ATTACH "db_1.sqlite" AS db_1' )

# Populate db_2 from db_1:
cur_2.execute( 'INSERT INTO foo SELECT * FROM db_1.foo' )

注意 foo 的 rowids 是 1 和 2:

cur_2.execute( 'SELECT rowid,bar from foo' )
for result in cur_2.fetchall():
    print('foo: {0}'.format(result))
    # foo: (1, 1)
    # foo: (2, 2)

请注意 db_1.foo 的 rowids 是 2 和 3:

# Show that db_1.foo.bar exists:
cur_2.execute( 'SELECT rowid,bar from db_1.foo' )
for result in cur_2.fetchall():
    print('db_1.foo: {0}'.format(result))
    # db_1.foo: (2, 1)
    # db_1.foo: (3, 2)

cur_2.execute( '''\
    UPDATE foo SET bar = bar +
        IFNULL( (SELECT t.bar 
                 FROM db_1.foo AS t
                 WHERE foo.rowid = t.rowid), 0)''' )  

UPDATE后，rowid = 1的行没有改变， 而 rowid = 2 的行已更新。

cur_2.execute( 'SELECT rowid,bar from foo' )
for result in cur_2.fetchall():
    print('foo after update: {0} '.format(result))
    # foo after update: (1, 1) 
    # foo after update: (2, 3) 

cur_2.execute('DETACH db_1')
db_2.commit()
db_2.close()

我发现这些页面有助于构建这个答案:here和 here ，尽管任何错误当然都是我自己的。