c++ - SOCI 无法准备声明

Question

我有这样一个函数:

CREATE OR REPLACE FUNCTION get_path_set_1(IN pathset_id_in character varying, OUT id character varying, OUT pathset_id character varying, OUT utility double precision)
  RETURNS SETOF record AS
$BODY$

    begin
        if exists(SELECT 1 FROM "PathSet_Scaled_HITS_distinctODs" WHERE "ID" = $1) then
            return query SELECT "ID", "PATHSET_ID", "UTILITY"
            FROM "SinglePath_Scaled_HITS_distinctODs"
            where "PATHSET_ID" = $1;
        end if; 
    end;
$BODY$
  LANGUAGE plpgsql VOLATILE
  COST 100
  ROWS 1000;
ALTER FUNCTION get_path_set_1(character varying)
  OWNER TO postgres;

当我在我的程序中调用它时:

std::string testStr("43046,75502");// or std::string testStr("'43046,75502'");
soci::rowset<sim_mob::SinglePath> rs = (sql.prepare << "get_path_set_1(:pathset_id_in)",soci::use(testStr));

我得到以下异常:

terminate called after throwing an instance of 'soci::postgresql_soci_error'
  what():  Cannot prepare statement. ERROR:  syntax error at or near "get_path_set_1"
LINE 1: get_path_set_1($1)

如果你帮我检测丢失的部分，我将不胜感激谢谢

Answer 1

这不会解决您报告的错误。但是简化你的功能:

CREATE OR REPLACE FUNCTION get_path_set_1(pathset_id_in varchar)
  RETURNS TABLE(id varchar, pathset_id varchar, utility double precision) AS
$func$
BEGIN
   RETURN QUERY
   SELECT "ID", "PATHSET_ID", "UTILITY"
   FROM   "SinglePath_Scaled_HITS_distinctODs"
   WHERE  "PATHSET_ID" = $1;
END
$func$  LANGUAGE plpgsql;

• RETURNS TABLE是现代的、更优雅的、等效的组合形式 RETURNS SETOF record和 OUT参数。

• IF exists ...在这里不给你买任何东西。运行查询；如果未找到任何内容，则不返回任何内容。用一半的成本获得相同的结果。