SQLite FTS表可以支持一個視圖嗎？

Question

我創建通過使用「內容」的說法一VIEW支持的FTS表，但它不生產，我會期望的結果：

CREATE TABLE tracks(docid INTEGER PRIMARY KEY, track_name varchar, ref_aid integer, ref_rid integer); 
CREATE TABLE artists(artist_name varchar, aid integer); 
CREATE TABLE releases(release_name varchar, rid integer); 

CREATE VIEW v AS 
SELECT docid, track_name, artist_name, release_name 
FROM tracks, artists, releases 
WHERE ref_aid = aid AND ref_rid = rid; 

CREATE VIRTUAL TABLE t USING fts4(content="v", track_name, artist_name, release_name); 

INSERT INTO tracks VALUES(0, 'xxx', 1, 1); 
INSERT INTO tracks VALUES(1, 'yyy', 1, 1); 
INSERT INTO artists VALUES('aaa', 1); 
INSERT INTO releases VALUES('rrr', 1); 

INSERT INTO t (docid, track_name, artist_name, release_name) 
SELECT docid, track_name, artist_name, release_name 
FROM v; 

-- this prints 0 as expected: 
SELECT docid FROM t WHERE t MATCH 'xxx'; 

-- this prints ||, indicating that it failed to look up the content in the view: 
SELECT * FROM t WHERE t MATCH 'xxx'; 

我希望第二個SELECT語句返回xxx|aaa|rrr ，但它不會返回任何東西。

Answer 1

外部內容表需要有一個名爲rowid的列，以便可以檢索數據。這工作：

CREATE TABLE tracks(rowid INTEGER PRIMARY KEY, track_name varchar, ref_aid integer, ref_rid integer); 
CREATE TABLE artists(artist_name varchar, aid integer); 
CREATE TABLE releases(release_name varchar, rid integer); 

CREATE VIEW v AS 
SELECT rowid, track_name, artist_name, release_name 
FROM tracks, artists, releases 
WHERE ref_aid = aid AND ref_rid = rid; 

CREATE VIRTUAL TABLE t USING fts4(content="v", track_name, artist_name, release_name); 

INSERT INTO tracks VALUES(0, 'xxx', 1, 1); 
INSERT INTO tracks VALUES(1, 'yyy', 1, 1); 
INSERT INTO artists VALUES('aaa', 1); 
INSERT INTO releases VALUES('rrr', 1); 

INSERT INTO t (rowid, track_name, artist_name, release_name) 
SELECT rowid, track_name, artist_name, release_name 
FROM v; 

-- this prints 0 as expected: 
SELECT docid FROM t WHERE t MATCH 'xxx'; 

-- this prints xxx|aaa|rrr as expected: 
SELECT * FROM t WHERE t MATCH 'xxx';