SQL中的IF/ELSE條件

Question

我正在嘗試爲PLSQL執行IF/ELSE語句，但我不確定是否以正確的方式執行，因爲它總是有錯誤：缺少右括號如果（TO_NUMBER （SUBSTR（ATTR_VALUE，6，2））！= TO_NUMBER（SUBSTR（ATTR_VALUE，1，2）））那麼 但括號對我來說似乎是平衡的。

SELECT * 
FROM 
(
IF (TO_NUMBER(SUBSTR(ATTR_VALUE, 6, 2)) != TO_NUMBER(SUBSTR(ATTR_VALUE, 1, 2))) THEN 
    SELECT ID, DATE, ATTR_VALUE, (TO_NUMBER(SUBSTR(ATTR_VALUE, 6, 4))-TO_NUMBER(SUBSTR(ATTR_VALUE, 1, 4))-48) DIFF 
ELSE 
    SELECT ID, DATE, ATTR_VALUE, (TO_NUMBER(SUBSTR(ATTR_VALUE, 6, 4))-TO_NUMBER(SUBSTR(ATTR_VALUE, 1, 4))) DIFF 
END IF 
FROM Table A 
ORDER BY TIME 
) 
WHERE DIFF>26 

請在這個問題上協助我。

Answer 1

您應該能夠使用CASE語句來得到結果：

SELECT * 
FROM 
(
    SELECT ID, 
    DATE, 
    ATTR_VALUE, 
    CASE 
     WHEN TO_NUMBER(SUBSTR(ATTR_VALUE, 6, 2)) != TO_NUMBER(SUBSTR(ATTR_VALUE, 1, 2)) 
     THEN (TO_NUMBER(SUBSTR(ATTR_VALUE, 6, 4))-TO_NUMBER(SUBSTR(ATTR_VALUE, 1, 4))-48) 
     ELSE (TO_NUMBER(SUBSTR(ATTR_VALUE, 6, 4))-TO_NUMBER(SUBSTR(ATTR_VALUE, 1, 4))) 
    END DIFF 
    FROM Table A 
    ORDER BY TIME 
) 
WHERE DIFF>26 

Answer 2

，如下圖所示，您可以修改您的查詢上線的東西，

SELECT * 
    FROM (SELECT ID, 
       DATE, 
       ATTR_VALUE, 
       decode(TO_NUMBER(SUBSTR(ATTR_VALUE, 6, 2)), 
         TO_NUMBER(SUBSTR(ATTR_VALUE, 1, 2)), 
         TO_NUMBER(SUBSTR(ATTR_VALUE, 6, 4)) - 
         TO_NUMBER(SUBSTR(ATTR_VALUE, 1, 4)), 
         TO_NUMBER(SUBSTR(ATTR_VALUE, 6, 4)) - 
         TO_NUMBER(SUBSTR(ATTR_VALUE, 1, 4))) DIFF 
      FROM A) 
WHERE diff > 26