Question

我的表：

id  name    year
1   Carl    1923
2   Carl    1924
3   Carl    1927
4   Carl    1939
5   Carl    1990
6   Carl    1992
7   Carl    1993
8   Suki    1962
9   Suki    1972
10  Suki    2002
11  Suki    2003

我想计算卡尔和苏基之间最大差距。所以我期待的结果是：

Carl    51
Suki    30

提前致谢。

Answer 1

您真的需要lag()或lead()功能。如果没有，一种方法是相关的子查询：

select name, max(year - prev_year) as max_diff
from (select t.*,
             (select t2.year
              from t t2
              where t2.name = t.name and t2.year < t.year
              order by year desc
              limit 1
             ) as prev_year
      from t
     ) t
group by name;

如果您知道id没有间隙（至少在名称中），则可以使用性能更高的join查询：

select t.name, max(t.year - tprev.year) as max_diff
from t join
     t tprev
     on t.year = tprev.year + 1 and t.name = tprev.name
group by t.name;

Answer 2

抱歉，我后来才意识到这不是一个PHP问题。可能它可以帮助将来的某个人你如何做到以下几点：

通过SQL
创建变量$ name =“”，$ year = 0，$ diff = -1
遍历每个$行并检查 if $row['name'] is != $name if so then do $year=$row['year'] and $diff=0, else check if($diff<($row['year']-($year+$diff))) #update diff $diff=$row['year']-($year+$diff) $year=$row['year']
对于打印，您可以在$ row ['name']为！= $ name时打印以前的值或者，以{key：value} JSON格式保存它们并最后打印！

计算多个值的最大单个间隙

2 个答案: