吉林市建设局网站semi是什么意思

今天，面试了一家公司，什么也不说先来三道面试题做做，第三题。

那么，我们就开始做题吧，谁叫我们是打工人呢。

题目是这样的：

DEALER_INFO表在每个经销商申请代理一个品牌的额度时，单独维护对应的基本信息。
CREDIT_LIMIT_INFO表维护每个授信协议的额度金额。                                    
问：要求以相同的证件号码为唯一标识识别为一个客户，取每个客户最近一次维护的基本信息进行报送，同时取这个客户已用额度总和。                    
最终报文数据样式：

---------------------------------------------------------------------------------------------------------------------------------

DROP TABLE IF EXISTS #DEALER_INFO;				
CREATE TABLE #DEALER_INFO (				
DEALER_NUMBER VARCHAR(20),				
DEALER_NAME VARCHAR(50),				
DEALER_ID_NO VARCHAR(50),				
ADDRESS VARCHAR(100),				
STUFF_NUMBER INT,				
MODIFY_DATE VARCHAR(10)				
);				INSERT INTO #DEALER_INFO(DEALER_NUMBER, DEALER_NAME, DEALER_ID_NO, ADDRESS, STUFF_NUMBER, MODIFY_DATE) VALUES ('1001001','TEST_01', 'ID_001', 'BEIJING CHAOYANG' , 100,  '2020-01-01');				
INSERT INTO #DEALER_INFO(DEALER_NUMBER, DEALER_NAME, DEALER_ID_NO, ADDRESS, STUFF_NUMBER, MODIFY_DATE) VALUES ('2001001','TEST_01', 'ID_001', 'BEIJING CHAOYANG' , 200,  '2020-02-01');				
INSERT INTO #DEALER_INFO(DEALER_NUMBER, DEALER_NAME, DEALER_ID_NO, ADDRESS, STUFF_NUMBER, MODIFY_DATE) VALUES ('1002002','TEST_02', 'ID_002', 'SHANGHAI PUDONG'  , 1000, '2020-03-15');				
INSERT INTO #DEALER_INFO(DEALER_NUMBER, DEALER_NAME, DEALER_ID_NO, ADDRESS, STUFF_NUMBER, MODIFY_DATE) VALUES ('3002002','TEST_02', 'ID_002', 'SHANGHAI MINHANG' , 1000, '2020-01-20');				
INSERT INTO #DEALER_INFO(DEALER_NUMBER, DEALER_NAME, DEALER_ID_NO, ADDRESS, STUFF_NUMBER, MODIFY_DATE) VALUES ('2003003','TEST_03', 'ID_003', 'BEIJING HAIDIAN'  , 50,   '2020-02-25');				
INSERT INTO #DEALER_INFO(DEALER_NUMBER, DEALER_NAME, DEALER_ID_NO, ADDRESS, STUFF_NUMBER, MODIFY_DATE) VALUES ('4003003','TEST_03', 'ID_003', 'BEIJING DONGCHENG', 200,  '2020-01-01');  				
INSERT INTO #DEALER_INFO(DEALER_NUMBER, DEALER_NAME, DEALER_ID_NO, ADDRESS, STUFF_NUMBER, MODIFY_DATE) VALUES ('1004004','TEST_04', 'ID_004', 'SHANGHAI HUANGPU' , 100,  '2020-03-01');  				DROP TABLE IF EXISTS #CREDIT_LIMIT_INFO;				
CREATE TABLE #CREDIT_LIMIT_INFO (				
DEALER_NUMBER VARCHAR(20),				
BRAND VARCHAR(50),				
UTILIZED_LIMIT INT				
);				INSERT INTO #CREDIT_LIMIT_INFO (DEALER_NUMBER, BRAND, UTILIZED_LIMIT) VALUES ('1001001','FAW'	  , 1000);			
INSERT INTO #CREDIT_LIMIT_INFO (DEALER_NUMBER, BRAND, UTILIZED_LIMIT) VALUES ('2001001','SKODA'   , 1500);				
INSERT INTO #CREDIT_LIMIT_INFO (DEALER_NUMBER, BRAND, UTILIZED_LIMIT) VALUES ('1002002','FAW'	  , 200 );			
INSERT INTO #CREDIT_LIMIT_INFO (DEALER_NUMBER, BRAND, UTILIZED_LIMIT) VALUES ('3002002','AUDI'	  , 400 );			
INSERT INTO #CREDIT_LIMIT_INFO (DEALER_NUMBER, BRAND, UTILIZED_LIMIT) VALUES ('2003003','SVW'	  , 500 );			
INSERT INTO #CREDIT_LIMIT_INFO (DEALER_NUMBER, BRAND, UTILIZED_LIMIT) VALUES ('4003003','PORCHE'  , 50  );				
INSERT INTO #CREDIT_LIMIT_INFO (DEALER_NUMBER, BRAND, UTILIZED_LIMIT) VALUES ('1004004','FAW'	  , 700 );			SELECT A.DEALER_NAME, A.DEALER_ID_NO, A.ADDRESS, A.STUFF_NUMBER, B.SUM_LIMIT				
FROM				
(				
SELECT * FROM				
(				
select DEALER_NAME, DEALER_ID_NO, ADDRESS, STUFF_NUMBER, MODIFY_DATE, 				
ROW_NUMBER() OVER(PARTITION BY DEALER_ID_NO ORDER BY MODIFY_DATE DESC) AS ROW_NUMBER				
from #DEALER_INFO A				
) T WHERE T.ROW_NUMBER = 1				
) A				
LEFT JOIN				
(				
select A.DEALER_NAME, SUM(B.UTILIZED_LIMIT) AS SUM_LIMIT				
from #DEALER_INFO A INNER JOIN #CREDIT_LIMIT_INFO B  ON A.DEALER_NUMBER = B.DEALER_NUMBER 				
GROUP BY A.DEALER_NAME				
) B				
ON A.DEALER_NAME = B.DEALER_NAME;				

查询结果如下：

结语：这个题目，在百度没有搜到，可能我的搜素能力不行吧，但是我给出了答案，希望给各位一个参考，不当之处请指出。

仁者见仁智者见智。