【问题标题】:COUNT LIKE to find a string [closed]计数喜欢找到一个字符串[关闭]
【发布时间】:2021-01-29 06:24:53
【问题描述】:

在甲骨文中。我想计算每个字符串出现的次数。以4/10、4/11、4/8为参考。

STRING
Upstream Channel 4/10.0/0
Upstream Channel 4/10.1/0
Upstream Channel 4/11.0/0
Upstream Channel 4/11.1/0
Upstream Channel 4/8.0/0
Upstream Channel 4/8.1/0
Upstream Channel 4/8.2/0
Upstream Channel 4/8.3/0
STRING COUNT
Upstream Channel %4/10.% 2
Upstream Channel %4/11.% 2
Upstream Channel %4/8.% 4

【问题讨论】:

  • 您的结果包含前缀字符串。但是,您的问题仅集中在数字表达式上。这使得问题不清楚。

标签: sql oracle substr


【解决方案1】:

您可以在这里使用REGEXP_SUBSTR 来隔离两个数字版本:

SELECT
    REGEXP_SUBSTR(string, '\s(\d+/\d+)', 1, 1, NULL, 1) AS STRING,
    COUNT(*) AS COUNT
FROM yourTable
GROUP BY REGEXP_SUBSTR(string, '\s(\d+/\d+)', 1, 1, NULL, 1);

Demo

【讨论】:

    【解决方案2】:

    不确定确切的要求是什么。但是您可以将substrinstrcount 一起使用,如下所示:

    select substr(str, 1, instr(str,'.')) as str_trunc,
           count(*) as occurance
      from your_Table t
     group by substr(str, 1, instr(str,'.'))
    

    如果你只想找到4/10, 4/11 and 4/8的计数,那么你可以使用如下:

    select temp.str, count(*)
      from (select '4/10' as val, 'Upstream Channel %4/10.%' as str from dual
            union all select '4/11' as val, 'Upstream Channel %4/11.%' as str from dual
            union all select '4/18' as val, 'Upstream Channel %4/8.%' as str from dual
           ) temp
      left join your_table t on t.str like '%' || temp.val || '%'
     group by temp.str
    

    【讨论】:

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