Question

我有这样的功能：

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;

当我在我的程序中使用它来调用它时：

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)

如果你帮助我发现缺失部分，我将不胜感激谢谢

Answer 1

不解决您报告的错误。但是要简化你的功能：

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是组合RETURNS SETOF record和OUT参数的现代，更优雅，等效的形式。
IF exists ...在这里买不到任何东西。运行查询;如果没有找到，则不返回任何内容。相同的结果是成本的一半。

Answer 2

从这段代码：

soci::rowset<sim_mob::SinglePath> rs =
  (sql.prepare << "get_path_set_1(:pathset_id_in)",soci::use(testStr));

您似乎正在尝试准备一个只包含函数调用的查询，甚至没有SELECT。

这在SQL中无效。您想要准备此查询：

 SELECT * FROM get_path_set_1(:pathset_id_in)

此表单（select * from function(...)）也是必需的，因为该函数返回的结果集包含多个列，而不是标量值。

同样在Erwin提及的情况下，OUT和SETOF RECORD在这种情况下很奇怪，我会在使用RETURNS TABLE时提出建议。

SOCI无法准备声明

2 个答案: