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如何使用参数创建SQL函数？

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2

以下代码根据输入的内容返回一个字段，即字符串，其中显示“28” .

SELECT data.id, LTRIM(SYS_CONNECT_BY_PATH(name, ', '),',') conc_names
  FROM (
     SELECT id, name, ROW_NUMBER() OVER (order by name) rownumber, COUNT(*) OVER () cnt
      FROM (
         SELECT es.EVENT_ID as id, s.SERVICE_NAME as name FROM DT_SERVICES s
            JOIN DT_EVENT_SERVICE es ON s.SERVICE_ID = es.SERVICE_ID
         WHERE es.EVENT_ID = 28
      )
) data
WHERE rownumber = cnt
START WITH rownumber = 1
CONNECT BY PRIOR rownumber = rownumber-1;

如何创建一个SQL函数，以便我可以传入任何数字（而不是28），函数将返回该选择的结果是什么？

我尝试创建一个，但我不断收到编译错误 .

当前SQL创建函数

create or replace function "DT_SERVICE_STRING" (id in VARCHAR2)
  return VARCHAR2 is
begin
  DECLARE 
     result VARCHAR(200);
  SELECT data.id, LTRIM(SYS_CONNECT_BY_PATH(name, ', '),',') conc_names
    INTO result
    FROM (
      SELECT id, name, ROW_NUMBER() OVER (order by name) rownumber, COUNT(*) OVER () cnt
      FROM (
        SELECT es.EVENT_ID as id, s.SERVICE_NAME as name FROM DT_SERVICES s
        JOIN DT_EVENT_SERVICE es ON s.SERVICE_ID = es.SERVICE_ID
        WHERE es.EVENT_ID = id
      )
  ) data
  WHERE rownumber = cnt
  START WITH rownumber = 1
  CONNECT BY PRIOR rownumber = rownumber-1;

  return result;
end;​

错误：
编译失败，第7行（15:22:21）PLS-00103：遇到以下其中一项时遇到符号"SELECT"：begin function pragma procedure子类型当前游标删除存在之前

sql oracle plsql oracle-apex

2 回答

  • 4

    假设你想要一个可以从SQL语句调用的PL / SQL函数（你不能在SQL中定义一个函数），听起来你想要类似的东西

    CREATE OR REPLACE FUNCTION get_conc_names( p_event_id IN dt_event_service.event_id%type )
  RETURN VARCHAR2
IS
  l_conc_names VARCHAR2(32676); 
  -- You may want a smaller variable if you know the result will be smaller
BEGIN
  SELECT LTRIM(SYS_CONNECT_BY_PATH(name, ', '),',') conc_names
    INTO l_conc_names
    FROM (
      SELECT id, name, ROW_NUMBER() OVER (order by name) rownumber, COUNT(*) OVER () cnt
        FROM (SELECT es.EVENT_ID as id, s.SERVICE_NAME as name 
                FROM DT_SERVICES s
                     JOIN DT_EVENT_SERVICE es ON s.SERVICE_ID = es.SERVICE_ID
               WHERE es.EVENT_ID = p_event_id )
          ) data
   WHERE rownumber = cnt
   START WITH rownumber = 1
 CONNECT BY PRIOR rownumber = rownumber-1;
  RETURN l_conc_names;
END;

    基于您刚刚发布的代码，您似乎只需要摆脱 DECLARE 并在 BEGIN 之前和 IS 之后移动局部变量 result 的声明 .

    回复于
  • 0

    我认为分析函数需要在内部内联视图中，然后外部内联视图可以选择它们 - 这就是我总是这样做的 . 试试这个：

    CREATE OR REPLACE FUNCTION get_conc_names( p_event_id IN dt_event_service.event_id%type ) 
  RETURN VARCHAR2 
IS 
  l_conc_names VARCHAR2(32676);  
  -- You may want a smaller variable if you know the result will be smaller 
BEGIN 
  SELECT LTRIM(SYS_CONNECT_BY_PATH(name, ', '),',') conc_names 
    INTO l_conc_names 
    FROM ( 
      SELECT id, name, rownumber, cnt 
        FROM (SELECT es.EVENT_ID as id
                    ,s.SERVICE_NAME as name
                    ,ROW_NUMBER() OVER (order by name) as rownumber
                    ,COUNT(*) OVER () as cnt
                FROM DT_SERVICES s 
                     JOIN DT_EVENT_SERVICE es ON s.SERVICE_ID = es.SERVICE_ID 
               WHERE es.EVENT_ID = p_event_id ) 
             ) data 
    WHERE rownumber = cnt 
    START WITH rownumber = 1 
    CONNECT BY PRIOR rownumber = rownumber-1; 
  RETURN l_conc_names; 
END;
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