Regex 从较长的字符串中提取字符串| Ruby

我正在尝试使用Ruby从长字符串中提取一个句子

例如：

您好，文件位于以下目录中：/Volumes/GAW-FS01/08\u Video/02\u Projects/Plus/S/metadata.xml谢谢您的查询

它应该返回：/Volumes/GAW-FS01/08_Video/02_Projects/Plus/S/metadata.xml 开头总是包含“/Volume”，并且以“.xml”结尾

---

谢谢你

你可能在找这个-

long_string = <<file
Hi There the file is in this directory: /Volumes/GAW-FS01/08_Video/02_Projects/Plus/S/metadata.xml Thank you for your inquiry.Hi There the file is in this directory: /Volumes/GAW-def/08_Video/02_Projects/Plus/S/metadata.xml Thank you for your inquiry.Hi There the file is in this directory: /Volumes/GAW-FS01/08_Video/02_Projects/hello/S/metadata.xml Thank you for your inquiry.
file

long_string.split.select{ |word| word.starts_with?("/Volumes") && word.end_with?(".xml") }

---

有很多方法可以做到这一点

string = "Hi There the file is in this directory: /Volumes/GAW-FS01/08_Video/02_Projects/ Plus/S/metadata.xml Thank you for your inquiry." 

时髦的：

"Volume#{string[/Volume(.*?).xml/, 1]}.xml"

更有趣的是，找到字符串的索引并使用范围返回所需的字符串

first = "Volume"
last = ".xml"
string[(string.index(first)) .. (string.index(last) + last.length)]
 => "Volumes/GAW-FS01/08_Video/02_Projects/ Plus/S/metadata.xml "

"Volume#{string.split('Volume').last.split('.xml').first}.xml"
 => "Volumes/GAW-FS01/08_Video/02_Projects/ Plus/S/metadata.xml"

不那么奇怪，将字符串按起点和终点拆分，然后返回所需的字符串

first = "Volume"
last = ".xml"
string[(string.index(first)) .. (string.index(last) + last.length)]
 => "Volumes/GAW-FS01/08_Video/02_Projects/ Plus/S/metadata.xml "

"Volume#{string.split('Volume').last.split('.xml').first}.xml"
 => "Volumes/GAW-FS01/08_Video/02_Projects/ Plus/S/metadata.xml"

最后，你可能想这样做

string[/Volume(.*?).xml/, 0]
=> "Volumes/GAW-FS01/08_Video/02_Projects/ Plus/S/metadata.xml"

---

您注意到第一个和最后一个选项是相同的，尽管传递的int值在0和1之间变化。0捕获用于匹配1不匹配的正则表达式。

欢迎使用SO！请在代码中提供一个尝试。那么基本上

字符串[%r[/Volumes/*/\.xml]]

？