一文详解小白也能懂的SQL高效去重技巧

SELECT
  *,
  ROW_NUMBER() OVER (
    PARTITION BY order_number, product, craft, trade_name 
    ORDER BY create_time DESC
  ) AS rn
FROM client_product

SELECT *
FROM (
  -- 步骤1：给所有记录标记组内序号
  SELECT *,
    ROW_NUMBER() OVER (
      PARTITION BY order_number, product, craft, trade_name 
      ORDER BY create_time DESC
    ) AS rn
  FROM client_product
  WHERE 
    production_order_number IS NOT NULL  -- 排除生产订单号为空
    AND order_number IS NOT NULL         -- 排除订单号为空
    AND craft != ''                      -- 排除工艺为空
    AND del_flag = '0'                   -- 只取未删除记录
    AND deliver_status != '0'            -- 排除未交付状态
) AS ranked
-- 步骤2：只取每组最新记录
WHERE rn = 1

SELECT DISTINCT column1 [, column2, ...]  
FROM table_name;  

SELECT DISTINCT address FROM student; -- 获取不重复的地址  

SELECT column1 [, aggregate_function(column2), ...]  
FROM table_name  
GROUP BY column1 [, column2, ...];  

SELECT MIN(id), address FROM student GROUP BY address; -- 按地址去重，返回每组最小 id  

SELECT *  
FROM (  
    SELECT *, ROW_NUMBER() OVER (PARTITION BY column1 ORDER BY column2) AS rn  
    FROM table_name  
) AS t  
WHERE rn = 1;  

SELECT id, name, address  
FROM (  
    SELECT *, ROW_NUMBER() OVER (PARTITION BY address ORDER BY id ASC) AS rn  
    FROM student  
) AS a  
WHERE a.rn = 1; -- 按地址去重，保留每组 id 最小的记录  

SELECT *  
FROM table_name  
WHERE id IN (SELECT MAX(id) FROM table_name GROUP BY column1);  

SELECT * FROM student WHERE id IN (SELECT MAX(id) FROM student GROUP BY address); -- 按地址去重，取每组最大 id 的记录  

SELECT a.*  
FROM table_name a  
WHERE NOT EXISTS (  
    SELECT 1 FROM table_name b  
    WHERE a.column1 = b.column1 AND a.id < b.id  
);  

SELECT a.* FROM student a WHERE NOT EXISTS (SELECT 1 FROM student b WHERE a.address = b.address AND a.id < b.id); -- 按地址去重，保留每组 id 最大的记录  

SELECT column1 [, column2, ...]  
FROM table_name1  
UNION  
SELECT column1 [, column2, ...]  
FROM table_name2;  

SELECT address FROM student UNION SELECT address FROM teacher; -- 合并两表地址并去重  

SELECT a.*  
FROM table_name a  
INNER JOIN (  
    SELECT column1, MAX(id) AS max_id  
    FROM table_name  
    GROUP BY column1  
) b ON a.column1 = b.column1 AND a.id = b.max_id;  

SELECT a.* FROM student a  
INNER JOIN (SELECT address, MAX(id) AS max_id FROM student GROUP BY address) b  
ON a.address = b.address AND a.id = b.max_id; -- 按地址去重，取每组最大 id 的记录