假设我有一本桌子杂志:
CREATE TABLE magazine
(
magazine_id integer NOT NULL DEFAULT nextval(('public.magazine_magazine_id_seq'::text)::regclass),
longname character varying(1000),
shortname character varying(200),
issn character varying(9),
CONSTRAINT pk_magazine PRIMARY KEY (magazine_id)
);
还有另一个餐桌问题:
CREATE TABLE issue
(
issue_id integer NOT NULL DEFAULT nextval(('public.issue_issue_id_seq'::text)::regclass),
number integer,
year integer,
volume integer,
fk_magazine_id integer,
CONSTRAINT pk_issue PRIMARY KEY (issue_id),
CONSTRAINT fk_magazine_id FOREIGN KEY (fk_magazine_id)
REFERENCES magazine (magazine_id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION
);
当前插入:
INSERT INTO magazine (longname,shotname,issn)
VALUES ('a long name','ee','1111-2222');
INSERT INTO issue (fk_magazine_id,number,year,volume)
VALUES (currval('magazine_magazine_id_seq'),'8','1982','6');
现在,如果一个行不存在的话,应该只插入一个行。但是,如果存在,表“问题”需要获得已经存在的行的“杂志”ID,以便建立引用。
我该怎么做?
提前通知!
最佳答案
你怎么知道杂志是否已经在magazine
表中?issn
列是否定义了杂志?如果是,那么它应该是主键,或者至少是unique
。
最简单的方法是检查你的客户端应用程序中的杂志的存在,像这样(在伪代码中):
function insert_issue(longname, shotname, issn, number,year,volume) {
/* extensive comments for newbies */
start_transaction();
q_get_magazine_id = prepare_query(
'select magazine_id from magazine where issn=?'
);
magazine_id = execute_query(q_get_magazine_id, issn);
/* if magazine_id is null now then there’s no magazine with this issn */
/* and we have to add it */
if ( magazine_id == NULL ) {
q_insert_magazine = prepare_query(
'insert into magazine (longname, shotname, issn)
values (?,?,?) returning magazine_id'
);
magazine_id = execute_query(q_insert_magazine, longname, shortname, issn);
/* we have tried to add a new magazine; */
/* if we failed (magazine_id==NULL) then somebody else just added it */
if ( magazine_id == NULL ) {
/* other, parerelly connected client just inserted this magazine, */
/* this is unlikely but possible */
rollback();
start_transaction();
magazine_id = execute_query(q_get_magazine_id, issn);
}
}
/* now magazine_id is an id of magazine, */
/* added if it was not in a database before, new otherwise */
q_insert_issue = prepare_query(
'insert into issue (fk_magazine_id,number,year,volume)
values (?,?,?,?)'
);
execute_query(q_insert_issue, magazine_id, number, year, volume);
/* we have inserted a new issue referencing old, */
/* or if it was needed new, magazine */
if ( ! commit() ) {
rollback();
raise "Unable to insert an issue";
}
}
如果您只需要在一个查询中执行此操作,那么您可以在数据库中将此伪代码实现为pl/pgsql函数,只需
select insert_issue(?, ?, ?, ?, ?, ?)
。