树状结构的阶层数的sql

前提有一张组织表.

CREATE TABLE organization_

(

organizationid bigint NOT NULL,

companyid bigint,

parentorganizationid bigint,

leftorganizationid bigint,

rightorganizationid bigint,

"name" character varying(100),

type_ character varying(75),

recursable boolean,

regionid bigint,

countryid bigint,

statusid integer,

comments text,

CONSTRAINT organization__pkey PRIMARY KEY (organizationid)

)

树节点的阶层的sql如下:(阶层数从1开始)

SELECT Children.organizationid , COUNT(Parents.organizationid) AS level

FROM organization_ Parents, organization_ Children

WHERE Children.leftorganizationid BETWEEN Parents.leftorganizationid AND Parents.rightorganizationid

GROUP BY Children.organizationid;

结果例子:

13260,1

13290,1

10852,2

15300，3