DECODE在ORACLE

Question

我不太理解這個DECODE函數將如何自行解決，特別是在尖（==>）參數：

DECODE 
(
    SH.RESPCODE, 0, SH.AMOUNT-DEVICE_FEE, 
    102, SH.NEW_AMOUNT-DEVICE_FEE, 
    ==>AMOUNT-DEVICE_FEE, 
    0 
) 

預先感謝什麼該參數將解決任何啓示至。

Answer 1

在僞代碼，這將是等效於下面的IF語句

IF(sh.respcode = 0) 
THEN 
    RETURN sh.amount-device_fee 
ELSIF(sh.respcode = 102) 
THEN 
    RETURN sh.new_amount-device_fee 
ELSIF(sh.respcode = amount-device_fee) 
THEN 
    RETURN 0; 
END IF; 

Answer 2

DECODE基本上相當於CASE表達式。

所以這

DECODE 
(
    SH.RESPCODE, 0, SH.AMOUNT-DEVICE_FEE, 
    102, SH.NEW_AMOUNT-DEVICE_FEE, 
    ==>AMOUNT-DEVICE_FEE, 
    0 
) 

相當於該

CASE SH.RESPCODE 
    WHEN 0 THEN SH.AMOUNT - DEVICE_FEE 
    WHEN 102 THEN SH.NEW_AMOUNT - DEVICE_FEE 
    WHEN AMOUNT - DEVICE_FEE THEN 0 
END 

Answer 3

它說：

If the value of SH.RESPCODE is 0, then return SH.AMOUNT-DEVICE_FEE. 
Otherwise, if it's equal to 102, then return SH.NEW_AMOUNT-DEVICE_FEE. 
Otherwise, if it's equal to AMOUNT-DEVICE_FEE, then return 0 

作爲一個case語句將是：

Case SH.RESPCODE 
    when 0     then SH.AMOUNT-DEVICE_FEE 
    when 102    then rSH.NEW_AMOUNT-DEVICE_FEE 
    when AMOUNT-DEVICE_FEE then 0 
    else null 
end 

有關於兩者之間的待遇NULL的一些細微的差別。

Answer 4

關於DECODE相當於CASE的意見很好。有一件事我發現是有幫助的就是格式化DECODE調用，使其更具可讀性，如：

DECODE(SH.RESPCODE, 
     0,     SH.AMOUNT-DEVICE_FEE, 
     102,    SH.NEW_AMOUNT-DEVICE_FEE, 
     AMOUNT-DEVICE_FEE, 0) 

分享和享受。