Question

我有一个我正在阅读的文件，其中包含以下信息：

  Organization: QC
  Company: Luxury Mortgage Corp. (0020)
  Folio: 3366326
  Doc Code: QCMAIL_STMT         
  Sequence: 3
  Pages: 7
  Method: SCAN            
  User: LAS             
  Received: 01/20/2016

我正在尝试从文件中提取行，只使用Folio，Sequence，Pages和User。

然而，当我这样做时，它会显示如下：

Folio: 
Sequence: 
Page: 
User: 
Folio: 
Sequence: 
Page: 
User: 
Folio: 
Sequence: 
Page: 
User: 
Folio: 
Sequence: 
Page: 
User: 
Folio: 3366326
Sequence: 
Page: 
User: 
Folio: 
Sequence: 
Page: 
User: 
Folio: 
Sequence: 3
Page: 
User: 
Folio: 
Sequence: 
Page: 7
User: 
Folio: 
Sequence: 
Page: 
User: 
Folio: 
Sequence: 
Page: 
User: LAS

我需要的是它出现的喜欢这个：

Folio: 3366326
Sequence: 3
Page: 7
User: LAS

来源：

#!/usr/local/bin/ruby

require 'colored'

class UncommittedDocs

  #attr_accessor :file

 # def initialize(username)
  #  @file = file
 # end

  def pull_info
    File.open("/home/qc/tep/bin/ruby/uncomitted.txt", "r").each_line do |line|
      puts "Folio: #{line.split(" ")[1] if line.include?("Folio")}"
      puts "Sequence: #{line.split(" ")[1] if line.include?("Sequence")}"
      puts "Page: #{line.split(" ")[1] if line.include?("Pages")}"
      puts "User: #{line.split(" ")[1] if line.include?("User")}"
    end
  end
end

test = UncommittedDocs.new#"/home/qc/tep/bin/ruby/uncommittedddocstest.txt")
test.pull_info

Answer 1

你的病情是错误的。

使用此：

puts "Folio: #{line.split(" ")[1] if line.include?("Folio")}"

而不是：

puts "Folio: #{line.split(" ")[1]}" if line.include?("Folio")

Answer 2

当您真正想要打印符合特定条件的行并跳过其余部分时，似乎您正在使用jasmine.getEnv().addReporter({ jasmineDone: function () { var HTMLReport = require('jasmine-xml2html-converter'); // Call custom report for html output testConfig = { reportTitle: 'Test Execution Report', outputPath: browser.reportPath, seleniumServer: 'default', applicationUrl: browser.baseUrl, testBrowser: browser.browserName + ' v.' + browser.browserVersion }; new HTMLReport().from(browser.reportPath + 'junitresults.xml', testConfig); console.log("... aaaannnnd... done."); } });等进行大量工作：

line.split

如果您要删除前导空格，则以下更改将适应：

class UncommittedDocs
  MATCH_LINE_EXPR = /^(Folio|Sequence|Page|User):/

  def pull_info
    File.open("/home/qc/tep/bin/ruby/uncomitted.txt", "r").each_line do |line|
      puts line if line =~ MATCH_LINE_EXPR
    end
  end
end

test = UncommittedDocs.new
test.pull_info

这只是在正则表达式的开头添加MATCH_LINE_EXPR = /^\s*(Folio|Sequence|Page|User):/ def pull_info File.open("/home/qc/tep/bin/ruby/uncomitted.txt", "r").each_line do |line| puts line.lstrip if line =~ MATCH_LINE_EXPR end end（以匹配零个或多个前导空格字符）并将\s*更改为puts line以从匹配的行中去除任何前导空格。

要在puts line.lstrip行后添加空白行，请根据您的评论添加另一个仅在此情况下执行的User:。由于我们已经在捕获组1中捕获puts，因此这很简单：

Folio|Sequence|Page|User

Answer 3

将if移出字符串插值，以便它适用于puts次调用：

puts "Folio: #{line.split(" ")[1]}" if line.include?("Folio")

你应该没事。也就是说，直到您遇到包含关键字的数据，例如

  User: Mr. Folio Pages

要避免此问题，请使用更严格的条件。您对文件格式的了解越多，就越能定制它。如果您没有任何规范，但只有上述示例文件内容可以使用，那么请查看仍然强大的内容，例如

puts "Folio: #{line.split(" ")[1]}" if line.strip.start_with?("Folio:")

Answer 4

如果您的输入文件是YAML（看起来可能是这样），请使用Ruby's built-in YAML support：

require 'yaml'
require 'active_support/core_ext/hash/slice'  # For comfortably dissecting a Hash
                                 # See https://stackoverflow.com/a/25206082/674064

# ...

  def pull_info
    data = YAML.load_file('/home/qc/tep/bin/ruby/uncomitted.txt')
    puts data.slice('Folio', 'Sequence', 'Pages', 'User').to_yaml
  end

# ...

Answer 5

IO.foreach('/path/to/file.txt') { |l| puts l if l =~ /[Folio|Sequence|Pages|User]: \w/ }

从文件中读取行并将它们缩合在一起

5 个答案: