小编Cha*_*997的帖子

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

我在jclipse中安装了jOOQ,为mySQL生成了类,但是我仍然有问题要写一些基本的查询.

我试图通过返回生成的密钥来组合插入查询,但编译器抛出错误

表:tblCategory列:category_id,parent_id,name,rem,uipos

Result<TblcategoryRecord> result= create.insertInto(Tblcategory.TBLCATEGORY, 
    Tblcategory.PARENT_ID, Tblcategory.NAME, Tblcategory.REM, Tblcategory.UIPOS)
        .values(node.getParentid())
        .values(node.getName())
        .values(node.getRem())
        .values(node.getUipos())
        .returning(Tblcategory.CATEGORY_ID)
        .fetch();

还尝试了其他不同的方法如何正确地做到这一点？

谢谢charis

这是嵌套集模型的"添加节点"SQL查询

LOCK TABLE mytestdb.tbltree WRITE;

SELECT @myRight := rgt FROM mytestdb.tbltree
WHERE name = 'apples';

UPDATE mytestdb.tbltree SET rgt = rgt + 2 WHERE rgt > @myRight;
UPDATE mytestdb.tbltree SET lft = lft + 2 WHERE lft > @myRight;

INSERT INTO mytestdb.tbltree(name, lft, rgt) 
VALUES('beans', @myRight + 1, @myRight + 2);

UNLOCK TABLES;   

并映射到jOOQ

Record record = create.select(Tbltree.RGT)
    .from(Tbltree.TBLTREE)
    .where(Tbltree.NAME.equal("apples"))
    .fetchOne();

int myright = record.getValue(Tbltree.RGT);

create.update(Tbltree.TBLTREE).set(Tbltree.RGT, Tbltree.RGT.add(2))
      .where(Tbltree.RGT.greaterThan(myright)).execute();

create.update(Tbltree.TBLTREE).set(Tbltree.LFT, Tbltree.LFT.add(2))
      .where(Tbltree.LFT.greaterThan(myright)).execute();

TbltreeRecord record2 = (TbltreeRecord) create 
    .insertInto(Tbltree.TBLTREE, Tbltree.NAME, Tbltree.LFT, Tbltree.RGT) …