Question

我们假设我有一本餐桌杂志：

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)
);

另一个表格问题：

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INSERTS：

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');

现在只应在“杂志”中插入一行，如果它尚不存在的话。但是，如果它存在，则表'issue'需要获取已存在的行的'magazine_id'以便建立引用。

我该怎么做？

提前谢谢！

Answer 1

你怎么知道一本杂志是否已经在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。

最简单的方法是在客户端应用程序中检查杂志是否存在，如下所示（伪代码）：

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，那么您可以将此伪代码实现为数据库中的pl / pgsql函数，而只是select insert_issue(?, ?, ?, ?, ?, ?)。

Answer 2

如果您使用PostgreSQL 9.1或更高版本，则可以执行以下操作：

WITH ref_key (id) AS (
    WITH ins (id) AS (
       INSERT INTO magazine (longname,shotname,issn)
       VALUES ('a long name','ee','1111-2222')
       RETURNING id
    )
    SELECT id 
      FROM magazine 
      LEFT JOIN ins USING id
     WHERE issn = '1111-2222'
)
INSERT INTO INTO issue (fk_magazine_id,number,year,volume)
SELECT id,'8','1982','6'
  FROM ref_key;

可写CTE的FTW！

Answer 3

我不确定你是否可以用SQL做到这一点。我知道Oracle可以用于触发器，但我不认为SQL能够。如果我错了，有人会纠正我。

检查行是否已存在，如果是，则告诉引用的表id

3 个答案: