Sqlite 虚拟表中记录的顺序不正确

在测试

SQLite虚拟表

机制的实现过程中，我们遇到了一个意外的行为。 对于以下

虚拟表

结构：

create table X(ID int, RL real)

此查询按RL的正确降序返回所有记录 场

查询1：

select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t1.RL desc;

explain query plan select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t1.RL desc;  

0|0|0| SCAN TABLE VTab AS t1 VIRTUAL TABLE INDEX 0:D1; (~0 rows)
0|1|1| SCAN TABLE VTab AS t2 VIRTUAL TABLE INDEX 4:C0=0; (~0 rows)

select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t2.RL desc

explain query plan select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t2.RL desc;  

0|0|0| SCAN TABLE VTab AS t1 VIRTUAL TABLE INDEX 0: (~0 rows)
0|11| SCAN TABLE VTab AS t2 VIRTUAL TABLE INDEX 4:C0=0; (~0 rows)

select * from Table1 t1 left outer join Table1 t2 on t1.ID = t2.ID order by t2.RL desc

explain query plan select * from Table1 t1 left outer join Table1t2 on t1.ID = t2.ID order by t2.RL desc;  

0|0|0| SCAN TABLE Table1 AS t1 (~1000000 rows)
0|1|0| SEARCH TABLE Table1 AS t2 USING AUTOMATIC COVERING INDEX (ID=?)
0|1|0| (~7 rows)
0|0|0| USE TEMP B-TREE FOR ORDER BY

执行计划1：

select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t1.RL desc;

explain query plan select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t1.RL desc;  

0|0|0| SCAN TABLE VTab AS t1 VIRTUAL TABLE INDEX 0:D1; (~0 rows)
0|1|1| SCAN TABLE VTab AS t2 VIRTUAL TABLE INDEX 4:C0=0; (~0 rows)

select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t2.RL desc

explain query plan select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t2.RL desc;  

0|0|0| SCAN TABLE VTab AS t1 VIRTUAL TABLE INDEX 0: (~0 rows)
0|11| SCAN TABLE VTab AS t2 VIRTUAL TABLE INDEX 4:C0=0; (~0 rows)

select * from Table1 t1 left outer join Table1 t2 on t1.ID = t2.ID order by t2.RL desc

explain query plan select * from Table1 t1 left outer join Table1t2 on t1.ID = t2.ID order by t2.RL desc;  

0|0|0| SCAN TABLE Table1 AS t1 (~1000000 rows)
0|1|0| SEARCH TABLE Table1 AS t2 USING AUTOMATIC COVERING INDEX (ID=?)
0|1|0| (~7 rows)
0|0|0| USE TEMP B-TREE FOR ORDER BY

D1

这里是由我们的

xBestIndex

方法实现生成的值，表示按字段#1=RL降序排序。

C0=0

这里表示字段#0=ID的操作相等。 这正如预期的那样有效

但是，下一个查询返回的行（RL字段的别名不同）没有任何排序

查询2：

select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t1.RL desc;

explain query plan select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t1.RL desc;  

0|0|0| SCAN TABLE VTab AS t1 VIRTUAL TABLE INDEX 0:D1; (~0 rows)
0|1|1| SCAN TABLE VTab AS t2 VIRTUAL TABLE INDEX 4:C0=0; (~0 rows)

select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t2.RL desc

explain query plan select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t2.RL desc;  

0|0|0| SCAN TABLE VTab AS t1 VIRTUAL TABLE INDEX 0: (~0 rows)
0|11| SCAN TABLE VTab AS t2 VIRTUAL TABLE INDEX 4:C0=0; (~0 rows)

select * from Table1 t1 left outer join Table1 t2 on t1.ID = t2.ID order by t2.RL desc

explain query plan select * from Table1 t1 left outer join Table1t2 on t1.ID = t2.ID order by t2.RL desc;  

0|0|0| SCAN TABLE Table1 AS t1 (~1000000 rows)
0|1|0| SEARCH TABLE Table1 AS t2 USING AUTOMATIC COVERING INDEX (ID=?)
0|1|0| (~7 rows)
0|0|0| USE TEMP B-TREE FOR ORDER BY

执行计划2：

select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t1.RL desc;

explain query plan select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t1.RL desc;  

0|0|0| SCAN TABLE VTab AS t1 VIRTUAL TABLE INDEX 0:D1; (~0 rows)
0|1|1| SCAN TABLE VTab AS t2 VIRTUAL TABLE INDEX 4:C0=0; (~0 rows)

select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t2.RL desc

explain query plan select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t2.RL desc;  

0|0|0| SCAN TABLE VTab AS t1 VIRTUAL TABLE INDEX 0: (~0 rows)
0|11| SCAN TABLE VTab AS t2 VIRTUAL TABLE INDEX 4:C0=0; (~0 rows)

select * from Table1 t1 left outer join Table1 t2 on t1.ID = t2.ID order by t2.RL desc

explain query plan select * from Table1 t1 left outer join Table1t2 on t1.ID = t2.ID order by t2.RL desc;  

0|0|0| SCAN TABLE Table1 AS t1 (~1000000 rows)
0|1|0| SEARCH TABLE Table1 AS t2 USING AUTOMATIC COVERING INDEX (ID=?)
0|1|0| (~7 rows)
0|0|0| USE TEMP B-TREE FOR ORDER BY

正如您所看到的，没有提到要排序的索引。 针对实际表（与我们的虚拟表具有完全相同的结构）执行的查询如下所示：

查询3：

select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t1.RL desc;

explain query plan select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t1.RL desc;  

0|0|0| SCAN TABLE VTab AS t1 VIRTUAL TABLE INDEX 0:D1; (~0 rows)
0|1|1| SCAN TABLE VTab AS t2 VIRTUAL TABLE INDEX 4:C0=0; (~0 rows)

select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t2.RL desc

explain query plan select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t2.RL desc;  

0|0|0| SCAN TABLE VTab AS t1 VIRTUAL TABLE INDEX 0: (~0 rows)
0|11| SCAN TABLE VTab AS t2 VIRTUAL TABLE INDEX 4:C0=0; (~0 rows)

select * from Table1 t1 left outer join Table1 t2 on t1.ID = t2.ID order by t2.RL desc

explain query plan select * from Table1 t1 left outer join Table1t2 on t1.ID = t2.ID order by t2.RL desc;  

0|0|0| SCAN TABLE Table1 AS t1 (~1000000 rows)
0|1|0| SEARCH TABLE Table1 AS t2 USING AUTOMATIC COVERING INDEX (ID=?)
0|1|0| (~7 rows)
0|0|0| USE TEMP B-TREE FOR ORDER BY

执行计划3：

select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t1.RL desc;

explain query plan select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t1.RL desc;  

0|0|0| SCAN TABLE VTab AS t1 VIRTUAL TABLE INDEX 0:D1; (~0 rows)
0|1|1| SCAN TABLE VTab AS t2 VIRTUAL TABLE INDEX 4:C0=0; (~0 rows)

select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t2.RL desc

explain query plan select * from VTab t1 left outer join VTab t2 on t1.ID = t2.ID order by t2.RL desc;  

0|0|0| SCAN TABLE VTab AS t1 VIRTUAL TABLE INDEX 0: (~0 rows)
0|11| SCAN TABLE VTab AS t2 VIRTUAL TABLE INDEX 4:C0=0; (~0 rows)

select * from Table1 t1 left outer join Table1 t2 on t1.ID = t2.ID order by t2.RL desc

explain query plan select * from Table1 t1 left outer join Table1t2 on t1.ID = t2.ID order by t2.RL desc;  

0|0|0| SCAN TABLE Table1 AS t1 (~1000000 rows)
0|1|0| SEARCH TABLE Table1 AS t2 USING AUTOMATIC COVERING INDEX (ID=?)
0|1|0| (~7 rows)
0|0|0| USE TEMP B-TREE FOR ORDER BY

正如您所看到的，这里有使用

B-tree的排序

我们检查了
sqlite3\u index\u orderby
结构（sqlite3\u index\u info

结构的一部分）中的内容，这些结构在

xBestIndex

中没有包含任何排序信息。 另外，

orderbyconsumered

out参数返回False（因为我们的输出没有排序，我们假设它本身会对行排序）

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这是SQLite虚拟表支持中的一个错误还是我们遗漏了什么？

正如

allocateIndexInfo（）

中的注释所述，只有在“ORDERBY子句仅包含当前虚拟表中的列”的情况下，虚拟表才有机会实现排序。在您的查询中，排序列所来自的虚拟表仅用于查找ID列，这种单个值的查找不可用于排序

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使用SQLite 3.7.14，我在查询2的计划中得到了“使用TEMP B-TREE FOR ORDER BY”。xBestIndex返回的值是什么？

问题的原因在于对

orderbyconsumered

输出标志值的处理不当

对于这个带有

JOIN

的查询，我们的

xBestIndex

实现返回了

orderbyconsumered=True

的索引，该索引在条件下匹配

，实际上是不正确的，因为在这种情况下没有进行排序。这忽略了

子句成员的任何进一步的

命令

在修复后，
orderbyconsumered
开始返回
True
仅适用于
nOrderBy
标志大于0的索引，否则返回
False
。
感谢您的回答。我在我的
xBestIndex
实现中发现了一个bug。它意外地为其中一个索引返回了
orderbyconsumered=True
，而根本不需要排序（对于符合条件的索引），SQLite引擎只是暗示我对我这边的所有东西都进行了排序。修复后，问题消失了。