Question

我失去了一些头发。我发现有几个类似问题的在线主题，但不完全是我遇到的问题，所以我仍然有点卡住。

我正在编写一个非常复杂的查询（至少对我来说很复杂：））

在我的年龄表中，我有一个age_start和age_end字段。

例如：

age_start   |   age_end
------------------------
0      -    64
19     -     23
0      -     19
25     -     29
65     -     100
...

我想要实现的是一个在这些范围之间读取的查询。

到目前为止，我提出的最好的方法是将这两个字段设置为数组值，并在我的MySQL查询中执行以下操作：

输入年龄：[33]

将年龄数据输入数组。然后查询如下：

...WHERE $age >= $age_start AND age <= $age_end...

会回来：

age_start   |   age_end
------------------------
0       -     64
25      -    29

有没有人有任何想法或建议或更好的方法？

Answer 1

SELECT *
  FROM ages
 WHERE 32 BETWEEN age_start AND age_end

Answer 2

SELECT
                        CASE
                            WHEN age < 13 THEN 'Under 13'
                            WHEN age BETWEEN 13 and 15 THEN '13 - 15'
                            WHEN age BETWEEN 16 and 20 THEN '16 - 20'
                            WHEN age BETWEEN 21 and 25 THEN '21 - 25'
                            WHEN age BETWEEN 26 and 30 THEN '26 - 30'
                            WHEN age BETWEEN 31 and 35 THEN '31 - 35'
                            WHEN age BETWEEN 36 and 40 THEN '36 - 40'
                            WHEN age BETWEEN 41 and 45 THEN '41 - 45'
                            WHEN age BETWEEN 46 and 50 THEN '46 - 50'
                            WHEN age BETWEEN 51 and 55 THEN '51 - 55'
                            WHEN age BETWEEN 51 and 55 THEN '56 - 60'
                            WHEN age BETWEEN 51 and 55 THEN '61 - 65'
                            WHEN age BETWEEN 51 and 55 THEN '66 - 70'
                            WHEN age BETWEEN 51 and 55 THEN '71 - 75'
                            WHEN age BETWEEN 51 and 55 THEN '76 - 80'
                            WHEN age BETWEEN 51 and 55 THEN '81 - 85'
                            WHEN age BETWEEN 51 and 55 THEN '86 - 90'
                            WHEN age BETWEEN 51 and 55 THEN '91 - 95'
                            WHEN age BETWEEN 51 and 55 THEN '96 - 100'
                            WHEN age >= 100 THEN 'Over 100'
                            WHEN age IS NULL THEN 'Not Filled In (NULL)'
                        END as age_range,
                        COUNT(ids) AS age_count,group_concat(ids) as chart_ids

                        FROM (SELECT 
                                TIMESTAMPDIFF(YEAR, `TABLE_FIELD_OF_DATE_OF_BIRTH`, CURDATE()) AS age, user_id as ids 
                           FROM 
                                jw_user 
                            WHERE 
                                is_active=1 
                            AND 
                                is_deleted = 0 
                        ) as derived

                    GROUP BY age_range

                    ORDER BY age_range

，这将是惊人的:)。

基于两个年龄字段的MySQL中的年龄范围查询

2 个答案: