更新数据精度的Regex

Question

我有一个字符串，它包含以下日期范围格式变体。我需要使用单一的java regex模式来查找和替换小时精度。日期范围是可变的。你能为我想个理由吗？

字符串示例

Published_date：{05/31/16.23:41:24-}

published_date:{05/31/16.23:41:24-06/21/16.23:41:24}

预期结果

published_date:{05/31/16.23:00:00-?}

published_date:{05/31/16.23:00:00-06/21/16.23:00:00}

Answer 1

描述

这个正则表达式会找到类似于日期/时间戳的子字符串，比如05/31/16.23:41:24。它将捕获00的日期和小时部分，并允许您替换分钟和秒。

([0-9]{2}\/[0-9]{2}\/[0-9]{2}\.[0-9]{2}):[0-9]{2}:[0-9]{2}

替换为： $1:00:00

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示例

现场演示

https://regex101.com/r/qK8bL7/1

样本文本

published_date:{05/31/16.23:41:24-?}

published_date:{05/31/16.23:41:24-06/21/16.23:41:24}

置换后

published_date:{05/31/16.23:00:00-?}

published_date:{05/31/16.23:00:00-06/21/16.23:00:00}

解释

NODE                     EXPLANATION
----------------------------------------------------------------------
  (                        group and capture to \1:
----------------------------------------------------------------------
    [0-9]{2}                 any character of: '0' to '9' (2 times)
----------------------------------------------------------------------
    \/                       '/'
----------------------------------------------------------------------
    [0-9]{2}                 any character of: '0' to '9' (2 times)
----------------------------------------------------------------------
    \/                       '/'
----------------------------------------------------------------------
    [0-9]{2}                 any character of: '0' to '9' (2 times)
----------------------------------------------------------------------
    \.                       '.'
----------------------------------------------------------------------
    [0-9]{2}                 any character of: '0' to '9' (2 times)
----------------------------------------------------------------------
  )                        end of \1
----------------------------------------------------------------------
  :                        ':'
----------------------------------------------------------------------
  [0-9]{2}                 any character of: '0' to '9' (2 times)
----------------------------------------------------------------------
  :                        ':'
----------------------------------------------------------------------
  [0-9]{2}                 any character of: '0' to '9' (2 times)
----------------------------------------------------------------------

Answer 2

尝尝这个。":0-9+:0-9+“。

公共类Regex {

public static void main(String ar[]){
    String st = "05/31/16.23:41:24-06/21/16.23:41:24";

    st = st.replaceAll(":[0-9]+:[0-9]+", ":00:00");
    System.out.println(st);

    st = "05/31/16.23:41:24-?";

    st = st.replaceAll(":[0-9]+:[0-9]+", ":00:00");
    System.out.println(st);

}

}