Question

CREATE DEFINER=`root`@`localhost` PROCEDURE `apply_Leave_SP`(
in leavetypeid int ,
in empid int,
in reason varchar(100),
in startdate date,
in enddate date,
in startsession int,
in endsession int,
in compoffid int,
in mangerid int

)
BEGIN

declare leavestatus int(10) default 0;
declare optionalyHolidays int(10) default 0;
declare listofholidays int(10) default 0;
declare totalhours int (10) default 0;
declare hours int (10) default 0;
declare satsun int (10) default 0;
declare manger_id int(10) default 0;
declare paidleave int(10) default 0;   
declare days int(10) default 0;
declare leaveappliedid int(10) default 0;
declare lossofpay int(10) default 0;
declare casualleave int (10) default 0;
declare sickleave int (10) default 0;


  select count(holiday_id) into optionalyHolidays from Parabola.holidays where holiday_type_id=2 and 
                          DATEDIFF(startdate,curdate())>=20;

    select count(*) into listofholidays from Parabola.holidays where date between startdate and enddate;                      

 SELECT 
COUNT(*) AS total into hours
FROM 
(   SELECT ADDDATE(startdate, INTERVAL @i:=@i+1 DAY) AS DAY
    FROM (
        SELECT a.a
        FROM (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS a
        CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS b
        CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS c
    ) a
    JOIN (SELECT @i := -1) r1
    WHERE 
    @i < DATEDIFF(enddate, startdate)

) AS dateTable
WHERE WEEKDAY(dateTable.Day) IN (5,6);

       SELECT  datediff(enddate,startdate) into totalhours; 
       set hours=(totalhours-listofholidays-satsun)*8;


    if(leavetypeid=1)then
     set leavestatus=5;
     else
      set leavestatus=1;
       end if;    

       if(optionalyHolidays>=0) then
       set leavestatus=5;
       end if;

insert into leave_applied(leave_type_id,hours,employee_id,
created_at,updated_at,start_date,end_date,start_date_session,
end_date_session,reason,
status,reminder_count,personal_calendar_event_id,system_calendar_event_id)

values(leavetypeid,hours,empid,curdate(),curdate(),startdate,enddate,
startsession,endsession,
reason,leavestatus,1,'sdasdas','sadeew');



Select LAST_INSERT_ID() into leaveappliedid ;


        if((leaveappliedid=1 or 4) and days >paidleave) then
            set lossofpay=days-paidleave;


        else if((leaveappliedid=1 or 4) and days> @CL) then
           set sickleave=days-@CL;
           set casualleave=@CL;


        else     if((leaveappliedid=1 or 4)  and days>@SL) then 
            set casualleave=days-@SL;
            set sickleave=@SL;

            else
            set mangerid=10;

            end if; 

        end;

    insert into leave_actual (leave_applied_id,
     leave_type_id,hours,start_date,
     end_date,created_at,updated_at) values(leaveappliedid,leavetypeid,
     hours,start_date,end_date,curdate(),curdate());     


insert into leave_approval (leave_applied_id,
         manager_id,status ,created_at,updated_at) values
         (leaveappliedid,mangerid,leavestatus,curdate(),curdate());   

  END

这个我的代码，当我尝试应用else如果条件然后有语法错误即将到来，只有如果没有错误即将到来但我必须应用否则如果条件请建议我在哪里做错了它显示其他如果条件

Answer 1

if的语法，if if是这样的：

IF condition THEN
    ... ...
ELSEIF condition THEN
    ... ...
END IF;

请在此处查看manual。

IF search_condition THEN statement_list
    [ELSEIF search_condition THEN statement_list] ...
    [ELSE statement_list]
END IF

如何应用if，否则如果在mysql存储过程中有条件

1 个答案: