Fizz嗡嗡聲SQL邏輯錯誤

Question

-1到100在SQL中顯示'Fizz'如果數字可以被3整除'Buzz'如果可以被5整除並且'FizzBu​​zz'被兩個整除。下面是我的代碼

select 
Case when Remainder (rownum,3)=0 then 'Fizz' 
    when Remainder (rownum,5)=0 then 'Buzz' 
    when (remainder (rownum,3)=0 and remainder(ROWNUM,5) = 0) then 'FizzBuzz' 
    else rownum end 
from DUAL 
Connect by level <=100; 

它給了我錯誤 - ORA-00932：不一致的數據類型：預期CHAR了NUMBER 00932. 00000 - 「不一致的數據類型：有望％S得了％的」 *原因：
*行動： 錯誤在行：5列：18

Answer 1

您需要通過要麼ROWNUM轉換爲字符串在ELSE情況：

• CAST(ROWNUM AS VARCHAR2(3))
• ''||ROWNUM
• TO_CHAR(ROWNUM)

您還需要這樣一種情況，即它可以被3和5整除以排在列表的頂部（否則前面的例子將優先）。就像這樣：

SELECT CASE 
     WHEN remainder (rownum,15)=0 THEN 'FizzBuzz' 
     WHEN Remainder (rownum,3)=0 THEN 'Fizz' 
     WHEN Remainder (rownum,5)=0 THEN 'Buzz' 
     ELSE TO_CHAR(rownum) 
     END 
FROM DUAL 
CONNECT BY LEVEL <= 100; 

不過，可以縮短代碼有很多（從一個答案我張貼Code Golf拍攝）：

SELECT NVL(
     DECODE(MOD(LEVEL,3),0,'Fizz')||DECODE(MOD(LEVEL,5),0,'Buzz'), 
     ''||LEVEL 
     ) 
FROM DUAL 
CONNECT BY LEVEL<101 

Answer 2

這是在你的其他聲明。 Rownum是NUMBER，而Fizz，Buzz和FizzBu​​zz是CHAR。

Answer 3

select 
Case when Remainder (rownum,3)=0 then 'Fizz' 
    when Remainder (rownum,5)=0 then 'Buzz' 
    when (remainder (rownum,3)=0 and remainder(ROWNUM,5) = 0) then 'FizzBuzz' 
    else ''||rownum end 
from DUAL 
Connect by level <=100; 

或

select 
Case when Remainder (rownum,3)=0 then 'Fizz' 
    when Remainder (rownum,5)=0 then 'Buzz' 
    when (remainder (rownum,3)=0 and remainder(ROWNUM,5) = 0) then 'FizzBuzz' 
    else to_char(rownum) end 
from DUAL 
Connect by level <=100;