怎么比较两个list中相同的值

Posted 2023-05-01

或者得到这两个list中相同值个数！请高手指点一下。谢谢

int count=0;
for(int i=0;i<list1.size();i++)
for(int j=0;j<list2.size();j++)
//判断值是否相同
if(list1.get(i).equal(list2.get(j)))
count=count+1;//相同count加一



System.out.println("相同值的个数＝"+count); 参考技术A import java.util.List;
import java.util.Arrays;

public class EqualList
static String a[]="a","b","c";
static String b[]="c","a","b";
static List list1;
static List list2;
static int count=0;
public static void main(String args[])
list1= Arrays.asList(a);
list2= Arrays.asList(b);
for(int i=0;i<list1.size();i++)
for(int j=0;j<list2.size();j++)

if(list1.get(i).equals(list2.get(j)))

count++;
break;


System.out.println("相同数字的个数"+count);


其实就是加个break的问题，呵呵！

比较两个不同列中相同字段的值

我需要将一列中的两个部分并排放在一个表中，具有相同的描述和不同的“外部ID”。这是输入表的样子：

+--+-------+-----------+
|ID|OUTERID|DESCRIPTION|
+--+-------+-----------+
|1 |1      |some desc 1|
+--+-------+-----------+
|2 |1      |some desc 2|
+--+-------+-----------+
|3 |1      |some desc 3|
+--+-------+-----------+
|4 |2      |some desc 2|
+--+-------+-----------+
|5 |2      |some desc 3|
+--+-------+-----------+

以下是我要从我的选择中实现的目标：

+-------------+-------------+
|DESCRIPTION_1|DESCRIPTION_2|
+-------------+-------------+
|some desc 1  |NULL         |
+-------------+-------------+
|some desc 2  |some desc 2  |
+-------------+-------------+
|some desc 3  |some desc 3  |
+-------------+-------------+

我尝试使用简单的JOIN，但正如您所看到的，我需要显示不匹配的值。

SELECT a.DESCRIPTION AS DESCRIPTION_1,
       b.DESCRIPTION AS DESCRIPTION_2
  FROM My_Table a
  JOIN My_Table b on a.DESCRIPTION = b.DESCRIPTION
WHERE a.OUTERID = 1
  AND b.OUTERID = 2

答案

如果我理解正确，您可以使用完整的外部联接：

SELECT a.DESCRIPTION AS DESCRIPTION_1,
       b.DESCRIPTION AS DESCRIPTION_2
FROM (SELECT a.*
      FROM My_Table a
      WHERE a.OUTERID = 1
     ) a FULL JOIN
     (SELECT b.*
      FROM My_Table b 
      WHERE b.OUTERID = 2
     ) b
     ON a.DESCRIPTION = b.DESCRIPTION;

另一答案

你需要dense_rank()并做条件聚合：

select max(case when OUTERID = 1 then DESCRIPTION  end) as DESCRIPTION_1,
       max(case when OUTERID = 2 then DESCRIPTION  end) as DESCRIPTION_2
from (select t.*,
             dense_rank() over (order by DESCRIPTION) as seq
      from table t
     ) t
group by seq;

另一答案

使用cte :(虽然，我不知道为什么你需要提供两个列而不是仅仅识别'dupes'）

表：

Create Table #tbl
(
ID Int,
OUTERID Int,
DESCRIPTION VarChar(20)
)
Insert Into #tbl Values
(1,1,'some desc 1'),
(2,1,'some desc 2'),
(3,1,'some desc 3'),
(4,2,'some desc 2'),
(5,2,'some desc 3')

CTE

With cte As
(
Select Description, Count(Distinct OuterID) As recs From #tbl
Group By Description
)
Select 
      Case When recs = 2 Or recs = 1 Then Description End As Description_1, 
      Case When recs = 2 Then Description End As Description_2
From cte