2013-09-24 177 views
1

我想運行一個函數來返回兩個日期之間的天差。出於這個原因,我根據日期參數獲取兩個日期(這工作正常)。接下來,我使用參數日期pdate來構造日期,如果它們未在表中找到。日期的這種構造不能按需要工作。以下是代碼:MYSQL STR_TO_DATE不返回想要的結果

`DELIMITER $$ 

USE payrolldb001 $$ 

DROP FUNCTION IF EXISTS NoOfWorkingDays $$ 

CREATE DEFINER=`root`@`localhost` FUNCTION NoOfWorkingDays(P_EmployeeID INT,pdate    DATE) RETURNS INT(10) 
READS SQL DATA 
BEGIN 

DECLARE v_FromDate DATE; 
DECLARE todate DATE; 
DECLARE workingdays INT; 
DECLARE shiftoff INT; 
DECLARE gazzetted INT; 

SET v_FromDate=(SELECT cp.FromDate FROM current_payroll cp 
       INNER JOIN employees e ON e.employeeid=P_EmployeeID AND e.CompanyCode=cp.CompanyCode 
       AND e.SubOfficeCode=cp.SubOfficeCode 
       WHERE pdate BETWEEN cp.FromDate AND cp.ToDate); 
IF (v_FromDate IS NOT NULL) THEN    
SET todate=(SELECT cp.ToDate FROM current_payroll cp 
       INNER JOIN employees e ON e.employeeid=P_EmployeeID AND e.CompanyCode=cp.CompanyCode 
       AND e.SubOfficeCode=cp.SubOfficeCode 
       WHERE pdate BETWEEN cp.FromDate AND cp.ToDate); 
    ELSE 
     SET v_FromDate=STR_TO_DATE(CONCAT(YEAR(pdate),'',MONTH(pdate)-1,'-21'),"%d%m%Y"); 
     SET todate=STR_TO_DATE(CONCAT(YEAR(pdate),'-',MONTH(pdate),'-20'),"%d%m%Y"); 

END IF; 

SET workingdays = DATEDIFF(todate,v_FromDate); 

WHILE v_FromDate<=todate DO 
    SET shiftoff= `isShiftOff`(P_EmployeeID,v_FromDate); 
    IF (shiftoff=0) THEN 
     SET gazzetted = `GazettedHoliday`(v_FromDate,P_Employeeid); 
     SET workingdays = workingdays-gazzetted; 
    ELSE 
     SET workingdays=workingdays-shiftoff; 
    END IF; 

    -- RESET QUERY CACHE; 
    -- FLUSH QUERY CACHE; 
    SET v_FromDate = ADDDATE(v_FromDate,1); 
END WHILE; 

RETURN workingdays ; 

END$$ 

DELIMITER ;` 

我得到的錯誤如下: Incorrect datetime value: '2013-8-21' for function str_to_date 這是在爲v_FromDate轉換。

回答

0

一個解決將是改變日期變量作爲設置:

SET v_FromDate=STR_TO_DATE(CONCAT(YEAR(pdate),'-',MONTH(pdate)-1,'-',21),"%Y-%c-%d");

SET todate=STR_TO_DATE(CONCAT(YEAR(pdate),'-',MONTH(pdate),'-',20),"%Y-%c-%d");