如何在“if”条件中使用“if”条件？答案

【问题标题】：How do I use 'if' condition inside a 'if' condition?如何在“if”条件中使用“if”条件？
【发布时间】：2016-03-08 05:28:00
【问题描述】：

我正在尝试在“if”条件中使用“if”条件来创建函数。我知道语法看起来有点像下面，但我无法在我的代码中做到这一点。

IF sales > (quota + 200) THEN
    bonus := (sales - quota)/4;
ELSE
    IF sales > quota THEN
        bonus := 50;
    ELSE
        bonus := 0;
    END IF;
END IF;

以下是我要完成的程序。请帮帮我。它的功能是在给定的通话次数中查找电话账单金额和计划类型

CREATE OR REPLACE  FUNCTION BILL(NUM_OF_CALLS NUMBER, PLAN_TYPE NUMBER) RETURN NUMBER IS :

    BILL_AMT NUMBER;
    MIN_1 NUMBER :=150;
    MIN_2 NUMBER :=1000;    

    BEGIN
    IF PLAN_TYPE:=150 THEN 
        IF NUM_OF_CALLS<150 THEN 
            BILL_AMT:=MIN_1;
            ELSIF NUM_OF_CALLS BETWEEN 151 AND 250 THEN
            BILL_AMT:=MIN_1+(NUM_OF_CALLS-150);
            ELSIF NUM_OF_CALLS BETWEEN 251 AND 400 THEN
            BILL_AMT:=MIN_1+(100*1)+(NUM_OF_CALLS-250)*0.5;
            ELSIF NUM_OF_CALLS>400 THEN
            BILL_AMT:=MIN_1+(100*1)+(150*0.5)+(NUM_OF_CALLS-400)*0.3;
        END IF;
        ELSE PLAN_TYPE:=500 THEN
            IF NUM_OF_CALLS<1000 THEN 
            BILL_AMT:=MIN_2;
            ELSIF NUM_OF_CALLS BETWEEN 1001 AND 1500 THEN
            BILL_AMT:=MIN_2+(NUM_OF_CALLS-1000)*0.50;
            ELSIF NUM_OF_CALLS>1500 THEN
            BILL_AMT:=MIN_2+(500*0.50)+(NUM_OF_CALLS-1500)*0.25;
        END IF;
    END IF;

    RETURN BILL_AMT;

END;

以下是错误

SQL> 显示错误 FUNCTION BILL 错误：

LINE/COL ERROR

11/14    PLS-00103: Encountered the symbol "=" when expecting one of the
         following:
         . ( * @ % & = - + < / > at in is mod remainder not rem then
         <an exponent (**)> <> or != or ~= >= <= <> and or like like2
         like4 likec between || multiset member submultiset
         The symbol "* was inserted before "=" to continue.

22/22    PLS-00103: Encountered the symbol "THEN" when expecting one of
         the following:
         * & = - + ; < / > at in is mod remainder not rem
         <an exponent (**)> <> or != or ~= >= <= <> and or like like2
         like4 likec between || multiset member submultiset

31/6     PLS-00103: Encountered the symbol "IF" when expecting one of the
         following:
         ; <an identifier> <a double-quoted delimited-identifier>
         current delete exists prior <a single-quoted SQL string>

【问题讨论】：

这似乎是一种 Pascal 语言，但是 := 里面如果是错误的
是Oracle PL SQL...
你遇到了什么错误？
Ed：发布错误
@ClaudioDaffra，是的，PL/SQL 至少在语法层面受到 Ada 的严重影响，Ada 借鉴了 Pascal 以及其他影响。

标签： oracle function if-statement plsql oracle11g

【解决方案1】：

您的答案直接在错误输出中。

PLSQL 使用:= 进行赋值，使用= 进行比较。您正在混合这些运算符并导致这些错误。

仔细阅读错误，它们准确地描述了问题。

提示：一次添加少量逻辑，以便您了解问题的根源。我们每个人都会在某个时间点学到这一点:)

【讨论】：

【解决方案2】：

这样的东西应该适合你：

CREATE OR REPLACE  
FUNCTION BILL(NUM_OF_CALLS NUMBER, PLAN_TYPE NUMBER) 
   RETURN NUMBER 
IS

    BILL_AMT NUMBER;
    MIN_1    NUMBER := 150;
    MIN_2    NUMBER := 1000;    

BEGIN
    IF PLAN_TYPE = 150 
    THEN 
       IF NUM_OF_CALLS<150 
       THEN 
          BILL_AMT:= MIN_1;
       ELSIF NUM_OF_CALLS BETWEEN 151 AND 250 
       THEN
          BILL_AMT:=MIN_1+(NUM_OF_CALLS-150);
       ELSIF NUM_OF_CALLS BETWEEN 251 AND 400 
       THEN
          BILL_AMT:=MIN_1+(100*1)+(NUM_OF_CALLS-250)*0.5;
       ELSIF NUM_OF_CALLS>400 
       THEN
          BILL_AMT:=MIN_1+(100*1)+(150*0.5)+(NUM_OF_CALLS-400)*0.3;
       END IF;
    ELSIF PLAN_TYPE = 500 
    THEN
       IF NUM_OF_CALLS<1000 
       THEN 
          BILL_AMT:=MIN_2;
       ELSIF NUM_OF_CALLS BETWEEN 1001 AND 1500 THEN
          BILL_AMT:=MIN_2+(NUM_OF_CALLS-1000)*0.50;
       ELSIF NUM_OF_CALLS>1500 THEN
          BILL_AMT:=MIN_2+(500*0.50)+(NUM_OF_CALLS-1500)*0.25;
       END IF;
    END IF;

    RETURN BILL_AMT;

END BILL;

希望对你有帮助...

【讨论】：

【解决方案3】：

你不能

   ELSE PLAN_TYPE:=500 THEN

但你可以

   ELSIF PLAN_TYPE =500 THEN

如果我弄清楚了你的逻辑流程

CREATE OR REPLACE  FUNCTION BILL(NUM_OF_CALLS NUMBER, PLAN_TYPE NUMBER) RETURN NUMBER IS 

    BILL_AMT NUMBER;
    MIN_1    NUMBER :=150;
    MIN_2    NUMBER :=1000;    

 BEGIN
    IF PLAN_TYPE = 150 
    THEN 
        IF NUM_OF_CALLS<150 THEN 
            BILL_AMT  :=MIN_1;
        ELSIF NUM_OF_CALLS BETWEEN 151 AND 250 THEN
            BILL_AMT:=MIN_1+(NUM_OF_CALLS-150);
        ELSIF NUM_OF_CALLS BETWEEN 251 AND 400 THEN
            BILL_AMT:=MIN_1+(100*1)+(NUM_OF_CALLS-250)*0.5;
        ELSIF NUM_OF_CALLS>400 THEN
            BILL_AMT:=MIN_1+(100*1)+(150*0.5)+(NUM_OF_CALLS-400)*0.3;
        END IF;
    ELSIF PLAN_TYPE = 500 THEN
        IF NUM_OF_CALLS<1000 THEN 
            BILL_AMT:=MIN_2;
        ELSIF NUM_OF_CALLS BETWEEN 1001 AND 1500 THEN
            BILL_AMT:=MIN_2+(NUM_OF_CALLS-1000)*0.50;
        ELSIF NUM_OF_CALLS>1500 THEN
            BILL_AMT:=MIN_2+(500*0.50)+(NUM_OF_CALLS-1500)*0.25;
        END IF;
    END IF;

    RETURN BILL_AMT;

END;

【讨论】：

我很确定END IF; 关闭了IF NUM_OF_CALLS<150 THEN，下面的ELSE 与IF PLAN_TYPE=150 THEN 相关（尽管它应该是ELSIF 而不是ELSE）跨度>
是的，格式让我失望。但是你仍然不能 ELSE ...我已经相应地更新了我的答案