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命令应该发生的事情是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Python上的一个错误）或sqlite3级别）;但是，我仔细阅读了sqlite3文档 - 尤其是“UPDATE”和“expression”页面 - 并且没有看到我做错了什么。

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

有任何建议或解决方法吗？

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

要使用其他表中的值更新foo，可以使用嵌套的SELECT表达式。请注意，foo.rowid引用外部表的rowid，而t.rowid引用内部表的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，因此rowids的{{1}}与db_1.foo rowids不匹配1}}：

db_2.foo

请注意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' )的{{1}}为1和2：

rowids

请注意foo的{{1}}为2和3：

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)

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

rowids

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

Answer 2

嗯...问题很可能是我正在查看3.7.3文档，而我的安装版本是3.6.16。调查。

Python sqlite3：UPDATE失败声明“没有这样的列”

2 个答案: