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在 PostgreSQL 函数中参数化间隔日期类型的最佳方法是什么?我有以下功能

CREATE OR REPLACE FUNCTION testing(
min_time integer
) 
RETURNS void AS
$BODY$
BEGIN
 EXECUTE format('CREATE TABLE foo AS
 SELECT
     gid,
     now() + concat(%s, ' hours')::interval as x,

  FROM foobar
  limit 3  ',min_time  );

 END;
$BODY$
LANGUAGE plpgsql;

每次我尝试执行该功能时,我都会收到以下错误ERROR: syntax error at or near "hours"

4

1 回答 1

0

单引号必须在字符串中转义:

CREATE OR REPLACE FUNCTION testing(min_time integer) RETURNS void
   LANGUAGE plpgsql AS
$BODY$
BEGIN
   EXECUTE format(
              'CREATE TABLE foo AS
               SELECT gid,
                      now() + ''%s hours''::interval as x
               FROM foobar
               limit 3',
              min_time
           );
END;
$BODY$;

但实际上,您不需要动态 SQL,静态 SQL 可能是更好的选择:

CREATE OR REPLACE FUNCTION testing(min_time integer) RETURNS void
   LANGUAGE plpgsql AS
$BODY$
BEGIN
   CREATE TABLE foo (
      gid bigint NOT NULL,
      x timestamp with time zone NOT NULL
   );

   INSERT INTO foo
   SELECT gid,
          now() + min_time * '1 hour'::interval
   FROM foobar
   limit 3;
END;
$BODY$;
于 2021-12-15T09:50:31.400 回答