python sqlite3不會執行連接，但sqlite3會單獨執行

Question

使用python 2.6.4中的sqlite3標准庫，以下查詢在sqlite3命令行上正常工作：

select segmentid, node_t, start, number,title  from 
    ((segments inner join position using (segmentid)) 
    left outer join titles using (legid, segmentid)) 
    left outer join numbers using (start, legid, version);

但是如果我通過python中的sqlite3庫執行它，我會收到一個錯誤：

>>> conn=sqlite3.connect('data/test.db')
>>> conn.execute('''select segmentid, node_t, start, number,title  from 
((segments inner join position using (segmentid)) left outer join titles using 
(legid, segmentid)) left outer join numbers using (start, legid, version)''')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
sqlite3.OperationalError: cannot join using column start - column not present 
in both tables

連接左側的（計算）表似乎具有相關列，因為如果我自己檢查它，我得到：

>>> conn.execute('''select *  from ((segments inner join position using 
(segmentid)) left outer join titles using 
(legid, segmentid)) limit 20''').description
(('segmentid', None, None, None, None, None, None), ('html', None, None, None, 
None, None, None), ('node_t', None, None, None, None, None, None), ('legid', 
None, None, None, None, None, None), ('version', None, None, None, None, None, 
None), ('start', None, None, None, None, None, None), ('title', None, None, 
None, None, None, None))

我的架構是：

CREATE TABLE leg (legid integer primary key,  t char(16), year char(16), 
    no char(16));
CREATE TABLE numbers (
    number char(16), legid integer, version integer, start integer, 
    end integer, prev integer, prev_number char(16), next integer, 
    next_number char(16), primary key (number, legid, version));
CREATE TABLE position (
    segmentid integer, legid integer, version integer, start integer, 
    primary key (segmentid, legid, version));
CREATE TABLE 'segments' 
    (segmentid integer primary key,  html text, node_t integer);
CREATE TABLE titles (legid integer, segmentid integer, title text, 
    primary key (legid, segmentid));
CREATE TABLE versions 
    (legid integer, version integer, primary key (legid, version));
CREATE INDEX idx_numbers_start on numbers (legid, version, start);

我感到困惑的是我做錯了什么。 我試過退出/重啟python和sqlite命令行，但看不出我做錯了什么。 這可能是完全明顯的。

Answer 1

一個解決方案（對我使用python庫的問題）似乎是引入一個完全虛假的表名：

SELECT legid, version, segmentid, html, node_t, start, number, title 
    from ((segments inner join position using (segmentid))  
    left outer join titles using (legid, segmentid)) as LT 
    left outer join numbers using (start, legid, version);

我認為這樣做是強制sqlite收集最外層連接左側的名稱，其中一個是“開始”，然后為最外層的外連接提供操作。 這對我有用 - 升級可能會引入更多問題而不是刪除它們，但是當它到達時我會越過那座橋。

Answer 2

你有一個有趣名字的表：

CREATE TABLE 'segments' 

但我認為這不是問題所在。 根據請求，這是我執行您的查詢，並將'segments'表重新創建為segments ：

$ sqlite3 junk.sqlite
SQLite version 3.6.22
Enter ".help" for instructions
Enter SQL statements terminated with a ";"
sqlite> .schema
CREATE TABLE leg (legid integer primary key,  t char(16), year char(16), 
    no char(16));
CREATE TABLE numbers (
    number char(16), legid integer, version integer, start integer, 
    end integer, prev integer, prev_number char(16), next integer, 
    next_number char(16), primary key (number, legid, version));
CREATE TABLE position (
    segmentid integer, legid integer, version integer, start integer, 
    primary key (segmentid, legid, version));
CREATE TABLE segments (segmentid integer primary key,  html text, node_t integer);
CREATE TABLE titles (legid integer, segmentid integer, title text, 
    primary key (legid, segmentid));
CREATE TABLE versions 
    (legid integer, version integer, primary key (legid, version));
CREATE INDEX idx_numbers_start on numbers (legid, version, start);
sqlite> select segmentid, node_t, start, number,title  from 
   ...>     ((segments inner join position using (segmentid)) 
   ...>     left outer join titles using (legid, segmentid)) 
   ...>     left outer join numbers using (start, legid, version);
Error: ambiguous column name: segmentid

Answer 3

SQLite版本3.6.22 - 看起來您需要限定“模糊列名稱”...

sqlite> select segmentid, node_t, start, number,title  from
   ...> ((segments inner join position using (segmentid))
   ...> left outer join titles using (legid, segmentid))
   ...> left outer join numbers using (start, legid, version);
Error: ambiguous column name: segmentid

sqlite> select segments.segmentid, node_t, start, number,title  from
   ...> ((segments inner join position using (segmentid))
   ...> left outer join titles using (legid, segmentid))
   ...> left outer join numbers using (start, legid, version);
Error: ambiguous column name: start

sqlite> select segments.segmentid, node_t, numbers.start, number,title  from
   ...> ((segments inner join position using (segmentid))
   ...> left outer join titles using (legid, segmentid))
   ...> left outer join numbers using (start, legid, version);
sqlite> 

SQLite版本3.6.23.1的行為相同