mysql 重复单号统计

任务：增加重复件统计分析：统计展示选择时间范围内重复1次、重复2次、重复3次、重复4次、重复5次及以上的数据量

17、统计出现的重复次数

增加重复件统计分析：统计展示选择时间范围内重复1次、重复2次、重复3次、重复4次、重复5次及以上的数据量。

1，建表

DROP TABLE IF EXISTS repeat_num;
CREATE TABLE repeat_num(id INT NOT NULL AUTO_INCREMENT,create_date DATE,content VARCHAR(500),nums VARCHAR(500),PRIMARY KEY(id)
);

2，插入数据

INSERT INTO repeat_num VALUES
(1, '2024-01-01', '投诉', '1'),
(2, '2024-01-02', '举报', '1,2'),
(3, '2024-01-03', '申告', '1,2,3'),
(4, '2024-01-04', '提醒', '1,2,3,4'),
(5, '2024-01-05', '告知', '1,2,3,4,5'),
(6, '2024-01-06', '预警', '1,2,3,4,5,6'),
(7, '2024-01-07', '投诉', '2,3'),
(8, '2024-01-08', '举报', '3'),
(9, '2024-01-09', '举报', '4,5,6'),
(10, '2024-01-10', '举报', '2,5,1,3'),
(11, '2024-01-11', '举报', '7,8,9'),
(12, '2024-01-12', '申诉', '1,2,3,4,5,6,8'),
(13, '2024-01-13', '申诉', '2,6'),
(14, '2024-01-14', '申诉', '1,3,4,5,6'),
(15, '2024-01-15', '申诉', '1,3,5,6,7,8,9'),
(16, '2024-01-16', '申诉', '5,7,8,9,10,11,13'),
(17, '2024-01-17', '申诉', '7,8,9,10'),
(18, '2024-01-18', '提醒', '7,8,9,10,14'),
(19, '2024-01-19', '告知', '5,7,8,9,10'),
(20, '2024-01-20', '告知', '4,5,6,7,8,9,11,12'),
(21, '2024-01-21', '预警', '3,5'),
(22, '2024-01-22', '预警', '8,9,11');

3，sql

1，先结算长度：

 -- 先计算原来的长度，然后用replace 替换到目标值，得到的长度，相减就是目标长度。 如果目标是多字符串的
SELECT LENGTH('1,2,3,4,5,6') - 
LENGTH(REPLACE('1,2,3,4,5,6', ',', '')) AS count_commas;

多字符间隔处理：

-- 多字符的，计算结果要除以自身长度
SELECT  (LENGTH('best wish for you! best wish for you!') - 
LENGTH( REPLACE ('best wish for you! best wish for you', 'for', '')))/ 
LENGTH('for')  AS repeat_num;-- 向下取整
SELECT  FLOOR((LENGTH('best wish for you! best wish for you!') - 
LENGTH( REPLACE ('best wish for you! best wish for you', 'for', '')))/ 
LENGTH('for'))  AS repeat_num;

2，计算重复次数：

 SELECT *, (LENGTH(nums) - LENGTH(REPLACE(nums, ',', ''))  + 1) 
AS  repeat_count FROM repeat_num ;

3，进行统计：

-- 统计：
SELECT repeat_count, SUM(repeat_count) repeat_sum
FROM (SELECT *, (LENGTH(nums) - LENGTH(REPLACE(nums, ',', ''))  + 1) AS  repeat_count FROM repeat_num  
) AS temp GROUP BY repeat_count;

4，统计以上的：

-- 统计 重复1次、重复2次、重复3次、重复4次、重复5次及以上 -- 以上的，又要聚合统计，得单独写
SELECT  repeat_count, SUM(repeat_count) repeat_sumFROM (SELECT *, (LENGTH(nums) - LENGTH(REPLACE(nums, ',', ''))  + 1) AS  repeat_count FROM repeat_num  ) AS temp WHERE repeat_count <= 5GROUP BY repeat_countUNION  SELECT  6 AS repeat_count, SUM(repeat_count) repeat_sumFROM (SELECT *, (LENGTH(nums) - LENGTH(REPLACE(nums, ',', ''))  + 1) AS  repeat_count FROM repeat_num  ) AS temp WHERE repeat_count > 5;

2，用函数的方式：

知识点：

SUBSTRING(s,n,len)带有len参数的格式，从字符串s返回一个长度同 len字符相同的子字符串，起始于位置n，也可能对n使用一个负值。假若这样，则子字符串的位置起始于字符串结尾的n字符，即倒数第n个字符，而不是字符串的开头位置。

【例 1】使用SUBSTRING函数获取指定位置处的子字符串，输入语句如下：

 SELECT SUBSTRING('breakfast', 5) AS coll,SUBSTRING('breakfast', 5,3) AS co12,
SUBSTRING('lunch', -3) AS co13,
SUBSTRING('lunch', -5, 3)AS col4;

SUBSTRING('breakfast',5)返回从第5个位置开始到字符串结尾的子字符串，结果为“kfast”; SUBSTRING(breakfast',5,3)返回从第5个位置开始长度为3的子字符串，结果为“kfa”;

SUBSTRING("unch',-3)返回从结尾开始第3个位置到字符串结尾的子字符串，结果为“nch”; SUBSTRING('lunch',-5,3)返回从结尾开始第5个位置，即字符串开头起，长度为3的子字符串，结果为“lun”。

用SUBSTRING去处理，不断去遍历，判断值符合条件的，就记录下。

函数：

DELIMITER //CREATE FUNCTION count_by_symbols(str VARCHAR(255), symbol VARCHAR(10))RETURNS INTBEGINDECLARE cnt INT DEFAULT 0;DECLARE idx INT DEFAULT 1;DECLARE len INT;SET len = CHAR_LENGTH(str);WHILE idx <= len DOIF SUBSTRING(str, idx, 1) = symbol THENSET cnt = cnt + 1;END IF;SET idx = idx + 1;END WHILE;RETURN cnt;END //DELIMITER ;

使用

 SELECT  count_by_symbols('a|lin|yan@sd|fd|ds|f', "|") AS count_symbol_num;

如果新建函数保存

报错

This function has none of DETERMINISTIC, NO SQL, or READS SQL DATA in its declaration and binary logging is enabled (you *might* want to use the less safe log_bin_trust_function_creators variable)

SET GLOBAL log_bin_trust_function_creators = 1;

调用：

SELECT *, (count_by_symbols(nums, ',')   + 1) AS  repeat_num FROM repeat_num ;

统计：

SELECT  repeat_count, SUM(repeat_count) repeat_sumFROM (SELECT *, (count_by_symbols(nums, ',')   + 1) AS  repeat_count FROM repeat_num  ) AS temp WHERE repeat_count <= 5GROUP BY repeat_countUNION  SELECT  6 AS repeat_count, SUM(repeat_count) repeat_sumFROM (SELECT *, (count_by_symbols(nums, ',')   + 1) AS  repeat_count FROM repeat_num  ) AS temp WHERE repeat_count > 5;