Question

我使用了select in if例如：

select IF (LEAVETYPE = 1 , (select test1,test2 from TableName),(select test3,test4 from TableName where id = 2) 
FROM TableName

I Wanne在IF中使用了选择

Answer 1

这个问题已经被问到了。你可以在这里找到答案：https://stackoverflow.com/a/63480/8713889

你应该使用CASE。来自mysql doc的示例：

import java.util.*;

public class BreakingMaps{


    public static void main(String[] args){

            int count = Integer.MAX_VALUE>>5;
            System.out.println(count + " objects tested");
            HashMap<Long, String> set = new HashMap<>(count);
            for(long i = 0; i<count; i++){
                Long l = i;
                set.put(l, l.toString());
            }

            Random r = new Random();
            for(int i = 0; i<1000; i++){
                long k = r.nextInt()%count;
                k = k<0?-k:k;
                System.out.println(set.get(k));
            }

        }

}

参考：https://dev.mysql.com/doc/refman/5.7/en/case.html

如何使用select in if

1 个答案: