Python读取字符串对象并搜索格式

Question

我正在使用 pyGithib 库查看 github repo 中所有以 *.rb 结尾的文件的内容，使用该库我得到一个这种格式的字符串

desc  'heading \'Test this too\')
                 Rationale: Best Practice
                 this line is also included in description '

config 'xxx' do

          title 'this is a dummy title \'Test this too\' for this block'
          desc  'Demo (test this) description \'Test this too\')
                 Rationale: Best Practice
                 this line is also included in description '
          
          tag benchmark: 'xyz:11'
          tag level: 1
          tag version: '0.0.1'
          tag reference: 'version 2.4'
          tag resource_type: 'A'

使用正则表达式，我如何获取desc 之后的多行值以及config 块之后包含version 的字符串

Answer 1

您可以使用匹配器来执行此操作，如以下问题所示：How to extract a substring using regex

您必须用正确的正则表达式替换正则表达式。

要获取正则表达式，您可以使用https://regexr.com/ 等网站，让您轻松测试与您的正则表达式匹配的内容。

在这种情况下，这样的事情应该可以工作：

String mydata = """config 'xxx' do

      title 'this is a dummy title \'Test this too\' for this block'
      desc  'Demo (test this) description \'Test this too\')
             Rationale: Best Practice
             this line is also included in description '
      
      tag benchmark: 'xyz:11'
      tag level: 1
      tag version: '0.0.1'
      tag reference: 'version 2.4'
      tag resource_type: 'A'""";
Pattern desc_pattern = Pattern.compile("desc  '(.|\n)*'\n\s*\n");
Matcher desc_matcher = desc_pattern.matcher(mydata);

desc = desc_matcher.find()

Pattern version_pattern = Pattern.compile("tag version: '.*'");
Matcher version_matcher = version_pattern.matcher(mydata);

version = version_matcher.find()

然后你可以剪掉前几个字符来得到你想要的字符串。