FasterCSV：读取远程CSV文件 [英] FasterCSV: Read Remote CSV Files

问题描述

我似乎无法让这个工作。我想从不同的Web服务器中提取一个CSV文件，以便读入我的应用程序。这就是我想说的：

  url ='http://www.testing.com/test。 csv'
 records = FasterCSV.read（url，：headers => true，：header_converters =>：symbol）
  

但这不行。我试过谷歌，所有我想出了这个摘录：实用Ruby Gems < a>

所以，我尝试修改它如下：

  require'open-uri'
 url ='http://www.testing.com/test.csv'
 csv_url = open（url）
 records = FasterCSV.read（csv_url，： headers => true，：header_converters =>：symbol）
  

无法将Tempfile转换为字符串错误（来自FasterCSV gem）。

任何人都可以告诉我如何

p

  require'open-uri'
 url = 'http://www.testing.com/test.csv'
 open（url）do | f | 
 f.each_line do | line | 
 FasterCSV.parse（line）do | row | 
＃您的代码在这里
 end 
 end 
 end 
  

http://www.ruby-doc.org/core/classes/OpenURI .html
http://fastercsv.rubyforge.org/

I can't seem to get this to work. I want to pull a CSV file from a different webserver to read in my application. This is how I'd like to call it:

url = 'http://www.testing.com/test.csv'
records = FasterCSV.read(url, :headers => true, :header_converters => :symbol)

But that doesn't work. I tried Googling, and all I came up with was this excerpt: Practical Ruby Gems

So, I tried modifying it as follows:

require 'open-uri'
url = 'http://www.testing.com/test.csv'
csv_url = open(url)
records = FasterCSV.read(csv_url, :headers => true, :header_converters => :symbol)

... and I get a can't convert Tempfile into String error (coming from the FasterCSV gem).

Can anyone tell me how to make this work?

解决方案

require 'open-uri'
url = 'http://www.testing.com/test.csv'
open(url) do |f|
  f.each_line do |line|
    FasterCSV.parse(line) do |row|
      # Your code here
    end
  end
end

http://www.ruby-doc.org/core/classes/OpenURI.html http://fastercsv.rubyforge.org/

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