为什么不是我的sqlite3外键工作？ [英] Why aren't my sqlite3 foreign keys working?

问题描述

我从python解释器运行下面的代码，并期望insert语句失败并抛出某种异常。但是并没有发生：

$ py $ 2.6 $（r265：79096，Mar 19 2010，21:48:26）[MSC win32
上的v.1500 32位（Intel）]键入help，copyright，credits或license以获取更多信息。
>>> import sqlite3
>>> conn = sqlite3.connect（test.db）
>>> conn.executescript（
... pragma foreign_keys = on;
... begin transaction;
... create table t1（i integer key，a）;
... create table t2（i，a，foreign key（i）references t1（i））;
... commit;
...）
< ;在0x0229DAA0的sqlite3.Cursor对象>
>>> c = conn.cursor（）
>>> c.execute（insert into t2 values（6，8））
< sqlite3.Cursor object at 0x0229DAD0>
>>> ＃???
...
>>> conn.commit（）
>>> ＃???????????
...
>>> c.execute（select * from t2）
< sqlite3.Cursor object at 0x0229DAD0>
>>> c.fetchall（）
[（6，8）]
>>> ＃但为什么！？
...
>>>

有谁知道为什么这不想工作？我的理解是插入应该失败，因为我为 t2（i）赋予的值不是 t1 ，但它很高兴地做到了这一点...？

解决方案

在SQLite中使用外键支持是非常新的仅在10月14日的3.6.19发布。你确定你正在使用SQLite 3.6.19或更高版本吗？

检查sqlite3模块中的sqlite_version常量。例如。在Mac OS X 10.6系统上安装默认的python / sqlite：

 >>> import sqlite3 
>>> sqlite3.sqlite_version 
'3.6.12'
>>> 
  

I run the following code from a python interpreter, and expect the insert statement to fail and throw some kind of exception. But it's not happening:

Python 2.6.5 (r265:79096, Mar 19 2010, 21:48:26) [MSC v.1500 32 bit (Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import sqlite3
>>> conn = sqlite3.connect("test.db")
>>> conn.executescript("""
... pragma foreign_keys=on;
... begin transaction;
... create table t1 (i integer primary key, a);
... create table t2 (i, a, foreign key (i) references t1(i));
... commit;
... """)
<sqlite3.Cursor object at 0x0229DAA0>
>>> c = conn.cursor()
>>> c.execute("insert into t2 values (6, 8)")
<sqlite3.Cursor object at 0x0229DAD0>
>>> #???
...
>>> conn.commit()
>>> #???????????
...
>>> c.execute("select * from t2")
<sqlite3.Cursor object at 0x0229DAD0>
>>> c.fetchall()
[(6, 8)]
>>> #but why!?
...
>>>

Does anyone know why this doesn't want to work? My understanding is that the insert should fail since the value I gave for t2(i) isn't a primary key in t1, but it happily does it anyway...?

解决方案

Working foreign key support in SQLite is very new -- it was only released in 3.6.19 on October 14th. Are you sure you're using SQLite 3.6.19 or later?

Check the sqlite_version constant in the sqlite3 module. E.g. on a Mac OS X 10.6 system with the default python/sqlite install:

>>> import sqlite3
>>> sqlite3.sqlite_version
'3.6.12'
>>> 

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