我得到的错误: enter image description here
错误被插入'插入'程序中的陈述。目标表具有与insert语句中相同的列数和数据类型。
我的程序是:
$sql = "SELECT Email , FirstName, LastName,Contact FROM tblUser where UserID=sessionID";
$result = mysqli_query($conn,$sql);
while($row = mysqli_fetch_array ($result)) {
$arr = array(
$row["Email"],
$row["FirstName"],
$row["LastName"],
$row["Contact"]
);
array_push($json, $arr);
}
$jsonstring = json_encode($json);
echo $jsonstring;
存储过程中调用的函数具有以下逻辑:
create or replace procedure updatetax
is
date_30 date:=sysdate-(365*30);
date_50 date:=sysdate-(365*50);
/*employees less than 30 years of age*/
cursor c1 is
select eid
from employee
where dateofbirth>date_30
and enddate is null;
/*employees between the age of 30 and 50*/
cursor c2 is
select eid
from employee
where dateofbirth between date_50 and date_30;
/*employees greater than 50 years of age*/
cursor c3 is
select eid
from employee
where dateofbirth<date_50;
r1 employee.eid%type;
r2 employee.eid%type;
r3 employee.eid%type;
begin
open c1;
for r1 in c1
loop
insert into emptax
values(r1.eid,gettaxlessthan30(select salary from empsalary where eid=r1.eid),sysdate);
end loop;
close c1;
commit;
open c2;
for r2 in c2
loop
insert into emptax
values(r2.eid,gettaxbetween30and50(select salary from empsalary where eid=r2.eid),sysdate);
end loop;
commit;
open c3;
close c2;
for r3 in c3
loop
insert into emptax
values(r3.eid,gettaxgreaterthan50(select salary from empsalary where eid=r3.eid),sysdate);
end loop;
commit;
close c3;
end;
/
答案 0 :(得分:0)
已解决 - 从一开始就重新编写了程序并且工作正常。