SQLite递归CTE更新适用于3.18.0，SQL_约束适用于3.19.0

我们有一个SQLite CTE更新，它在3.18.0之前的版本中工作，但在3.19.0中由于外键约束而失败（错误19）

下面是一个示例玩具数据库，显示了该行为

sqlite> .version
SQLite 3.18.0 2017-03-28 18:48:43 424a0d380332858ee55bdebc4af3789f74e70a2b3ba1cf29d84b9b4bcf3e2e37
sqlite> .dump t2
PRAGMA foreign_keys=OFF;
BEGIN TRANSACTION;
CREATE TABLE T2 (Id INTEGER PRIMARY KEY AUTOINCREMENT , ParentId INTEGER NOT NULL , Name TEXT NOT NULL , FavoriteState INT NOT NULL DEFAULT 0, FOREIGN KEY (ParentId) REFERENCES T2(Id) );
INSERT INTO T2 VALUES(1,0,'/',2);
CREATE INDEX idx_pid_t2 ON T2 (ParentId);
CREATE INDEX idx_files_favstate_t2 on T2 (FavoriteState);
CREATE UNIQUE INDEX idx_pnc_t2 ON T2 (ParentId, Name);
COMMIT;
sqlite> PRAGMA foreign_keys=ON;
sqlite> WITH RECURSIVE under_favorite_path(parent) AS ( VALUES(1) UNION ALL SELECT T2.Id FROM T2 JOIN under_favorite_path ON T2.ParentId = under_favorite_path.parent ) UPDATE T2 SET FavoriteState = 1
WHERE Id IN under_favorite_path;
sqlite> select * from t2;
1|0|/|1
sqlite> WITH RECURSIVE under_favorite_path(parent) AS ( VALUES(1) UNION ALL SELECT T2.Id FROM T2 JOIN under_favorite_path ON T2.ParentId = under_favorite_path.parent ) UPDATE T2 SET FavoriteState = 2
WHERE Id IN under_favorite_path;
sqlite> select * from t2;
1|0|/|2

在3.19.0中禁用外键可以使更新成功

sqlite> .version
SQLite 3.19.0 2017-05-22 13:58:13 28a94eb282822cad1d1420f2dad6bf65e4b8b9062eda4a0b9ee8270b2c608e40
sqlite> .dump t2
PRAGMA foreign_keys=OFF;
BEGIN TRANSACTION;
CREATE TABLE T2 (Id INTEGER PRIMARY KEY AUTOINCREMENT , ParentId INTEGER NOT NULL , Name TEXT NOT NULL , FavoriteState INT NOT NULL DEFAULT 0, FOREIGN KEY (ParentId) REFERENCES T2(Id) );
INSERT INTO T2 VALUES(1,0,'/',2);
CREATE INDEX idx_pid_t2 ON T2 (ParentId);
CREATE INDEX idx_files_favstate_t2 on T2 (FavoriteState);
CREATE UNIQUE INDEX idx_pnc_t2 ON T2 (ParentId, Name);
COMMIT;
sqlite> PRAGMA foreign_keys=ON;
sqlite> WITH RECURSIVE under_favorite_path(parent) AS ( VALUES(1) UNION ALL SELECT T2.Id FROM T2 JOIN under_favorite_path ON T2.ParentId = under_favorite_path.parent ) UPDATE T2 SET FavoriteState = 1
WHERE Id IN under_favorite_path;
Error: FOREIGN KEY constraint failed
sqlite> PRAGMA foreign_keys=OFF;
sqlite> WITH RECURSIVE under_favorite_path(parent) AS ( VALUES(1) UNION ALL SELECT T2.Id FROM T2 JOIN under_favorite_path ON T2.ParentId = under_favorite_path.parent ) UPDATE T2 SET FavoriteState = 1
WHERE Id IN under_favorite_path;
sqlite> select * from t2;
1|0|/|1

---

这是我们的CTE以前应该标记的问题，还是在以后的SQLite版本中可能会回归的问题？

问题不在于CTE，而是表包含无效数据（

0

不是有效ID）：

---

显然，旧版本在FK列未被修改时没有检查约束。

是的，现在看起来肯定是这样。我想知道这一更改是有意的还是无意的，基于3.19.0发行说明中的这一行，因为我读到的内容与正在发生的事情相反：在不涉及受外键约束的列的UPDATE语句中避免不必要的外键处理。实际上是“不必要的外键处理”。似乎还有其他一些变化… sqlite> pragma foreign_key_check; table rowid parent fkid ---------- ---------- ---------- ---------- T2 1 T2 0

sqlite> update t2 set name = name;
Error: FOREIGN KEY constraint failed