13

如何使用 jOOQ 编写以下 SQL?

SELECT *
FROM food_db_schema.tblCategory AS t1
LEFT OUTER JOIN food_db_schema.tblCategory AS t2 ON t1.category_id = t2.parent_id
WHERE t2.parent_id IS NULL
AND t1.heartbeat = "ALIVE";

数据库是 mySQL

4

2 回答 2

9

flesk 的回答很好地描述了如何使用 jOOQ 1.x 完成此操作。使用别名的自连接或多或少等同于使用别名的常规连接,如手册中所述:

https://www.jooq.org/doc/latest/manual/sql-building/table-expressions/aliased-tables/

在即将到来的 2.0 版本中,别名将变得不那么冗长,并且更加类型安全。因此 flesk 的解决方案可以简化为:

// Type-safe table aliasing:
TblCategory t1 = TBLCATEGORY.as("t1");
TblCategory t2 = TBLCATEGORY.as("t2");

Record record = create.select()
                      .from(t1)
                       // t1 and t2 give access to aliased fields:
                      .leftOuterJoin(t2).on(t1.CATEGORY_ID.equal(t2.PARENT_ID))
                      .where(t2.PARENT_ID.isNull())
                      .and(t1.HEARTBEAT.equal("ALIVE"));

我还在这里描述了一个更复杂的自加入示例:

http://blog.jooq.org/jooq-meta-a-hard-core-sql-proof-of-concept/

于 2011-11-20T13:46:37.400 回答
6

也许

SELECT *
FROM food_db_schema.tblCategory AS t1
WHERE t1.category_id IS NULL
AND t1.heartbeat = "ALIVE";

,但你确定t2.parent_id都应该是 NULL 和等于t1.category_id

编辑:

然后像

Table<TblCategoryRecord> t1 = TBLCATEGORY.as("t1");
Table<TblCategoryRecord> t2 = TBLCATEGORY.as("t2");

Field<Integer> t1CategoryId = t1.getField(TblCategory.CATEGORY_ID);
Field<String> t1Heartbeat = t1.getField(TblCategory.HEARTBEAT);
Field<Integer> t2ParentId = t2.getField(TblCategory.PARENT_ID);

Record record = create.select().from(t1)
      .leftOuterJoin(t2).on(t1CategoryId.equal(t2ParentId))
      .where(t2ParentId.isNull())
      .and(t1Heartbeat.equal("ALIVE"));

取决于生成的类、属性和元模型对象的名称。

于 2011-11-19T21:43:57.387 回答