Question

我有以下PL / SQL函数：

CREATE OR REPLACE FUNCTION GETOVERUREN(v_user_id IN NUMBER, v_jaar IN NUMBER)
RETURN NUMBER
AS
    v_resultaat number := 0;
    v_min_uren_id number := 0;
    v_max_uren_id number := 0;
BEGIN
    SELECT MIN(UREN_ID) INTO v_min_uren_id FROM UREN WHERE JAAR_NR = v_jaar AND UREN_ID IN (SELECT UREN_ID FROM GEBRUIKER_UREN g WHERE g.USER_ID = v_user_id);
    SELECT MAX(UREN_ID) INTO v_max_uren_id FROM UREN WHERE JAAR_NR = v_jaar AND UREN_ID IN (SELECT UREN_ID FROM GEBRUIKER_UREN g WHERE g.USER_ID = v_user_id);

    DECLARE
        v_subtotaal number := 0;
    BEGIN
    v_max_uren_id := v_max_uren_id +1;
    WHILE v_min_uren_id < v_max_uren_id LOOP
        SELECT SUM(OMAANDAG+ODINSDAG +OWOENSDAG +ODONDERDAG +OVRIJDAG +OZATERDAG +OZONDAG) INTO v_subtotaal FROM UREN WHERE UREN_ID = v_min_uren_id;

                    ----------------------------------------------
                    FIXED: v_min_uren_id := v_min_uren_id +1;(FORGOT TO + THE LOOP ITSELF)
                    ----------------------------------------------

                    v_resultaat := v_resultaat + v_subtotaal;
    END LOOP;
RETURN v_resultaat;
END;
END;

The following sql command should give the following outcome:
SELECT GETOVERUREN(1,2013) FROM UREN; WHERE 1 is the userid and 2013 is the year

GETOVERUREN(1,2013)
-------------------
          10.25

But instead it gives:

GETOVERUREN(1,2013)
-------------------
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
          10.25 
 Up to 157 times (wich is the total of rows i have in my table.)

/ *当我使用函数SQLDeveloper保持运行并且不会停止运行该函数时。我等了半个小时，但没有结果。我做错了什么？ * /

跑步的问题是固定的，现在我得到了很多结果。

我自己就是一个看起来像这样的示例函数：

create or replace FUNCTION TESTING(v_user_id IN NUMBER)
RETURN NUMBER
AS
    v_resultaat number := 0;
BEGIN
    SELECT IS_ADMIN INTO v_resultaat FROM GEBRUIKER WHERE USER_ID = v_user_id;
  return v_resultaat;
END TESTING;

这个函数确实返回一个值，但是它返回值X再次在GEBRUIKER中的行数，这是错误的但我似乎无法弄清楚它是什么。

Answer 1

你必须在循环中移动v_min_uren_id变量的增量（并将其从max更改为min） - 否则，你会得到一个无限循环。试试这个：

CREATE OR REPLACE FUNCTION GETOVERUREN(v_user_id IN NUMBER, v_jaar IN NUMBER) RETURN NUMBER AS v_resultaat number := 0; v_min_uren_id number := 0; v_max_uren_id number := 0; BEGIN SELECT MIN(UREN_ID) INTO v_min_uren_id FROM UREN WHERE JAAR_NR = v_jaar AND UREN_ID IN (SELECT UREN_ID FROM GEBRUIKER_UREN g WHERE g.USER_ID = v_user_id); SELECT MAX(UREN_ID) INTO v_max_uren_id FROM UREN WHERE JAAR_NR = v_jaar AND UREN_ID IN (SELECT UREN_ID FROM GEBRUIKER_UREN g WHERE g.USER_ID = v_user_id); DECLARE v_subtotaal number := 0; BEGIN WHILE v_min_uren_id < v_max_uren_id LOOP v_min_uren_id := v_min_uren_id +1; SELECT SUM(OMAANDAG+ODINSDAG +OWOENSDAG +ODONDERDAG +OVRIJDAG +OZATERDAG +OZONDAG) INTO v_subtotaal FROM UREN WHERE UREN_ID = v_min_uren_id; v_resultaat := v_resultaat + v_subtotaal; END LOOP; RETURN v_resultaat; END; END;

PL / SQL命令继续运行

1 个答案: