Question

我真的很困惑，我可以通过以下查询得到以下结果。当(LR.DATE_TO - LR.DATE_FROM)计算为0时，我想将0.5作为结果，如果它为null，我想将0作为结果，否则我想使用从（LR.DATE_TO - LR.DATE_FROM）返回的确切差异。

请，帮助

    INSERT INTO LEAVE_TAKEN_COUNT (NUMDAYS,EMPLOYEE_SERIAL_ID)
    SELECT NVL((LR.DATE_TO - LR.DATE_FROM), 0) as NumDays , T.ID
    FROM TBL_EMPLOYEE T 
    LEFT JOIN TBL_LEAVE_REQUEST LR on T.ID = LR.EMPLOYEE_SERIAL_ID
    WHERE T.DATE_PERMANENT IS NOT NULL
    ORDER BY T.ID;

Answer 1

使用CASE：

INSERT INTO LEAVE_TAKEN_COUNT (NUMDAYS,EMPLOYEE_SERIAL_ID)
    SELECT CASE WHEN LR.DATE_TO - LR.DATE_FROM = 0 THEN 0.5
                WHEN (LR.DATE_TO - LR.DATE_FROM) IS NULL THEN 0
                ELSE LR.DATE_TO - LR.DATE_FROM
            END as NumDays , T.ID
    FROM TBL_EMPLOYEE T 
    LEFT JOIN TBL_LEAVE_REQUEST LR on T.ID = LR.EMPLOYEE_SERIAL_ID
    WHERE T.DATE_PERMANENT IS NOT NULL
    ORDER BY T.ID;

Answer 2

使用case：

SELECT (CASE WHEN LR.DATE_TO IS NULL OR LR.DATE_FROM IS NULL THEN 0
             WHEN LR.DATE_TO = LR.DATE_FROM THEN 0.5
             ELSE LR.DATE_TO - LR.DATE_FROM
        END) as NumDays, . . .

如果检查结果为0并返回0.5，请选择oracle

2 个答案: