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planner: update the doc of optimizer hint (#17499)
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@qw4990
qw4990 authored May 31, 2024
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75 changes: 73 additions & 2 deletions optimizer-hints.md
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Expand Up @@ -134,6 +134,10 @@ SELECT /*+ NO_MERGE_JOIN(t1, t2) */ * FROM t1, t2 WHERE t1.id = t2.id;

### INL_JOIN(t1_name [, tl_name ...])

> **注意：**
>
> 部分情况下 `INL_JOIN` Hint 可能无法生效，详情请参阅 [`INL_JOIN` Hint 不生效](#inl_join-hint-不生效)。
`INL_JOIN(t1_name [, tl_name ...])` 提示优化器对指定表使用 Index Nested Loop Join 算法。这个算法可能会在某些场景更快，消耗更少系统资源，有的场景会更慢，消耗更多系统资源。对于外表经过 WHERE 条件过滤后结果集较小（小于 1 万行）的场景，可以尝试使用。例如：

{{< copyable "sql" >}}
Expand Down Expand Up @@ -939,7 +943,74 @@ Warning 信息如下：

在上面的示例中，你需要将 Hint 直接放在 `SELECT` 关键字之后。具体的语法规则参见 [Hint 语法](#语法)部分。

### 排序规则不兼容导致 `INL_JOIN` Hint 不生效
### `INL_JOIN` Hint 不生效

#### 关联表的列上使用内置函数导致 `INL_JOIN` Hint 不生效

在某些情况下，如果在关联表的列上使用了内置函数，优化器可能无法选择 `IndexJoin` 计划，导致 `INL_JOIN` Hint 也无法生效。

例如，以下查询在关联表的列 `tname` 上使用了内置函数 `substr`：

```sql
CREATE TABLE t1 (id varchar(10) primary key, tname varchar(10));
CREATE TABLE t2 (id varchar(10) primary key, tname varchar(10));
EXPLAIN SELECT /*+ INL_JOIN(t1, t2) */ * FROM t1, t2 WHERE t1.id=t2.id and SUBSTR(t1.tname,1,2)=SUBSTR(t2.tname,1,2);
```

查询计划输出结果如下：

```sql
+------------------------------+----------+-----------+---------------+-----------------------------------------------------------------------+
| id | estRows | task | access object | operator info |
+------------------------------+----------+-----------+---------------+-----------------------------------------------------------------------+
| HashJoin_12 | 12500.00 | root | | inner join, equal:[eq(test.t1.id, test.t2.id) eq(Column#5, Column#6)] |
| ├─Projection_17(Build) | 10000.00 | root | | test.t2.id, test.t2.tname, substr(test.t2.tname, 1, 2)->Column#6 |
| │ └─TableReader_19 | 10000.00 | root | | data:TableFullScan_18 |
| │ └─TableFullScan_18 | 10000.00 | cop[tikv] | table:t2 | keep order:false, stats:pseudo |
| └─Projection_14(Probe) | 10000.00 | root | | test.t1.id, test.t1.tname, substr(test.t1.tname, 1, 2)->Column#5 |
| └─TableReader_16 | 10000.00 | root | | data:TableFullScan_15 |
| └─TableFullScan_15 | 10000.00 | cop[tikv] | table:t1 | keep order:false, stats:pseudo |
+------------------------------+----------+-----------+---------------+-----------------------------------------------------------------------+
7 rows in set, 1 warning (0.01 sec)
```

```sql
SHOW WARNINGS;
```

```
+---------+------+------------------------------------------------------------------------------------+
| Level | Code | Message |
+---------+------+------------------------------------------------------------------------------------+
| Warning | 1815 | Optimizer Hint /*+ INL_JOIN(t1, t2) */ or /*+ TIDB_INLJ(t1, t2) */ is inapplicable |
+---------+------+------------------------------------------------------------------------------------+
1 row in set (0.00 sec)
```
从该示例中可以看到，`INL_JOIN` Hint 没有生效。该问题的根本原因是优化器限制导致无法使用 `Projection` 或者 `Selection` 算子作为 `IndexJoin` 的探测 (Probe) 端。
从 TiDB v8.0.0 起，你通过设置 [`tidb_enable_inl_join_inner_multi_pattern`](/system-variables.md#tidb_enable_inl_join_inner_multi_pattern-从-v700-版本开始引入) 为 `ON` 来避免该问题。
```sql
SET @@tidb_enable_inl_join_inner_multi_pattern=ON;
Query OK, 0 rows affected (0.00 sec)
EXPLAIN SELECT /*+ INL_JOIN(t1, t2) */ * FROM t1, t2 WHERE t1.id=t2.id AND SUBSTR(t1.tname,1,2)=SUBSTR(t2.tname,1,2);
+------------------------------+--------------+-----------+---------------+--------------------------------------------------------------------------------------------------------------------------------------------+
| id | estRows | task | access object | operator info |
+------------------------------+--------------+-----------+---------------+--------------------------------------------------------------------------------------------------------------------------------------------+
| IndexJoin_18 | 12500.00 | root | | inner join, inner:Projection_14, outer key:test.t1.id, inner key:test.t2.id, equal cond:eq(Column#5, Column#6), eq(test.t1.id, test.t2.id) |
| ├─Projection_32(Build) | 10000.00 | root | | test.t1.id, test.t1.tname, substr(test.t1.tname, 1, 2)->Column#5 |
| │ └─TableReader_34 | 10000.00 | root | | data:TableFullScan_33 |
| │ └─TableFullScan_33 | 10000.00 | cop[tikv] | table:t1 | keep order:false, stats:pseudo |
| └─Projection_14(Probe) | 100000000.00 | root | | test.t2.id, test.t2.tname, substr(test.t2.tname, 1, 2)->Column#6 |
| └─TableReader_13 | 10000.00 | root | | data:TableRangeScan_12 |
| └─TableRangeScan_12 | 10000.00 | cop[tikv] | table:t2 | range: decided by [eq(test.t2.id, test.t1.id)], keep order:false, stats:pseudo |
+------------------------------+--------------+-----------+---------------+--------------------------------------------------------------------------------------------------------------------------------------------+
7 rows in set (0.00 sec)
```

#### 排序规则不兼容导致 `INL_JOIN` Hint 不生效

如果两个表的 Join key 的排序规则不能兼容，将无法使用 IndexJoin 来执行查询。此时 [`INL_JOIN` Hint](#inl_joint1_name--tl_name-) 将无法生效。例如：

Expand Down Expand Up @@ -976,7 +1047,7 @@ SHOW WARNINGS;
1 row in set (0.00 sec)
```

### 连接顺序导致 `INL_JOIN` Hint 不生效
#### 连接顺序导致 `INL_JOIN` Hint 不生效

[`INL_JOIN(t1, t2)`](#inl_joint1_name--tl_name-) 或 `TIDB_INLJ(t1, t2)` 的语义是让 `t1` 和 `t2` 作为 `IndexJoin` 的内表与其他表进行连接，而不是直接将 `t1` 和 `t2` 进行 `IndexJoin` 连接。例如：

Expand Down

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