从字符串中提取一行的一部分。

I have a string in Ruby like this:

我有一个像这样的Ruby字符串:

var = <<EOS

***************************
Test Spec:  somespecname 
Test Case:  get new status
EOS

I want to retrieve the name following the "Test Spec:" part of the string. So in this case I want the result to be "somespecname".

我想根据“Test Spec:”检索字符串的一部分。在这种情况下，我希望结果是"somespecname"

How can I do it in Ruby?

用Ruby怎么做呢?

4 个解决方案

#1

Looking at the OP's example, I assume that there are cases when the string to be retrieved may include non-word characters. This regex will capture them correctly as well.

查看OP的示例，我假设在某些情况下，要检索的字符串可能包含非单词字符。这个正则表达式也可以正确地捕获它们。

var.match(/^Test Spec:\s*(.*?)\s*$/)[1]

#2

Do this:

这样做:

var.match(/Spec:\s*(\w+)/)[1] # => "somespecname"

#3

You can divide the string into array elements using split for easier reference than regex or match.

可以使用split将字符串分割为数组元素，以便比regex或match更容易引用。

  var = "\n\n\nTest Spec:  somespecname \nTest Case:  get new status"

  var.strip! 
  # => "Test Spec:  somespecname \nTest Case:  get new status" 
  new_var = var.split("\n")
  # => ["Test Spec:  somespecname ", "Test Case:  get new status"]
  test_spec_element = new_var[0]
  # => "Test Spec:  somespecname "
  desired_string = test_spec_element.split(":")[0]
  # => " somespecname "

#4

You could use the String#[] and String#strip methods:

您可以使用String#[]和String#strip方法:

var[/Test Spec:(.*)$/, 1].strip
# => "somespecname"

Update: An alternative expression using lookbehind:

更新:使用lookbehind的另一个表达式:

var[/(?<=Test Spec:).*$/].strip

#1