I want to download the zip file from eBay. Using downloadfile api.
response = RestClient.post(url,xml,headers)
This call return the content of zip file that is not exractable in xml I think. So i want to download zip file as it is from eBay.
My code is:
headers = {
"X-EBAY-SOA-OPERATION-NAME"=>"downloadFile",
"X-EBAY-SOA-SECURITY-TOKEN" => access_token_lister,
"X-EBAY-API-SITEID"=>"0",
"Content-Type"=>"application/zip"
}
url = 'https://storage.ebay.com/FileTransferService'
xml = '<?xml version="1.0"
encoding="utf-8"?> <downloadFileRequest xmlns="ebay.com/marketplace/services">; <fileReferenceId>6637191637</fileReferenceId> <taskReferenceId>6474385857</taskReferenceId> </downloadFileRequest>'
The documentation for the API used above can be found here : http://developer.ebay.com/devzone/file-transfer/CallRef/downloadFile.html
You will have to pass the content of the response to a library like rubyzip. Then use it to extract your files from the zipped file. This you can do by first writing the file to disk, and then reading it using rubyzip.
Specifically look at this part of the documentation of rubyzip - Reading a Zip file to accomplish what you are trying to achieve.
Related
I wants to read a excel file existing on Live URL of another website.
When I hit that URL in browser file is downloading. While in my rails app it is giving below error
No such file or directory # rb_sysopen - http://www.carsa.jp/admin/data.xlsx (Errno::ENOENT)
My Rails app code is as below
data = Roo::Excelx.new('http://www.carsa.jp/admin/data.xlsx')
header = data.row(1)
puts header
Note: If I download file and place it within my application it is working fine but the requirement is to read it from the third-party website in a scheduled job as per the above script.
data = Roo::Excelx.new('lib/data.xlsx')
header = data.row(1)
puts header
Try using Roo::Spreadsheet.open instead of Roo::Excelx.new. According to the Roo Readme:
Roo::Spreadsheet.open can accept both paths and File instances.
This should do the trick:
Roo::Spreadsheet.open('http://www.carsa.jp/admin/data.xlsx')
i am having one xml file and i need to fetch value of tags postnumer and regarding every postnumber all the valid number values...i am using xml.dom for reading xml file in python what is the code to fetch the value...how can i do this? please help
xml code:
-<post>
<PostNumber>180</postNumber>
-<validList>
<validNumber>208</validNumber>
<validNumber>209</validNumber>
<validNumber>210</validNumber>
</validlist>
</post>
-<post>
<postNumber>1023832</postNumber>
-<validlist>
<validNumber>264</validNumber>
<validNumber>401</validNumber>
</validlist>
</post>
I am using spreadsheet gem to generate .xls file. After writing it to a file, I am trying to send to client browser for download.
Below is the code in rails
workbook = Spreadsheet::Workbook.new
# Constructed the data
file = "/path/to/file/sheet.xls"
workbook.write file
send_file file
This file when opened contains expected data in ideal format.
Below is the code in js:
CustomRestService.custom_post("report",{report_data: angular.toJson($scope.report_data)},"download_xls",{}).then (data)->
if data
hiddenElement = document.createElement('a')
angular.element(document.body).append(hiddenElement)
hiddenElement.href = 'data:attachment/xls,' + encodeURI(data)
hiddenElement.target = '_blank'
hiddenElement.download = "report.xls"
hiddenElement.click()
hiddenElement.remove()
But the file getting downloaded in browser contains junk data. I tried multiple solutions like below:
Using send_data, instead of send_file
Generated xls data and wrote to StringIO object to directly download
Constructed Blob object in js, with type as "application/vnd.ms-excel" and trying to download it.
All attempts failed, as I am missing something. All suggestions are welcome.
filename = "/path/to/file/sheet.xls"
tempfile = Tempfile.new(filename)
workbook = Spreadsheet::Workbook.new
...
workbook.write(tempfile.path)
send_file tempfile.path, :filename => filename
I am developing an application for crawling the web using crawler4j and Jsoup. I need to parse a webpage using JSoup and check if it has zip files, pdf/doc and mp3/mov file available as a resource for download.
For zip files i did the following and it works:
Elements zip = doc.select("a[href\$=.zip]")
println "No of zip files is " + zip.size()
This code correctly tells me how many zip files are there in a page. I am not sure how to count all audio files or document files using JSoup. Any help is appreciated. Thanks.
Using the same approach I suspect it would be something like this:
Elements docs = doc.select("a[href\$=.doc]")
println "No of doc files is " + docs.size()
Elements mp3s = doc.select("a[href\$=.mp3]")
println "No of mp3 files is " + mp3s.size()
Really it's just a selector where the href attribute ends in some file extension.
So I am trying to send an email containing a .zip file. The .zip file is located at another url on another server. I am able to retrieve the file, attach it and send it. However when I get the attachment from the email. It will not open as it says cannot open *.zip.zip I have tried removing the .zip in the name but then the archive manager cannot open it either.
Any ideas?
Code is below.
http = Net::HTTP.new('www.somedomaim.org')
http.start() { |http|
req = Net::HTTP::Get.new("/path/to/file.zip")
response = http.request(req)
tempfile = Tempfile.new('files')
File.open(tempfile.path, 'w') do |f|
f.write response.body
end
attachments["files.zip"] = File.read(tempfile.path)
mail to: someone#somewhere.com, subject: "Sending zip file"
}
[SOLVED]
The solution is rather simple.
attachments['files.zip'] = open('http://somedomain.com/path/to/file.zip').read
attachments needs to receive the content of the file. .read returns the content of the file. My issue was that i was placing the entire zip file in the content of a new file. The above solution will place just the content of the zip into a new file.
Hope this helps someone someday. Thanks for all the suggestions.
The solution is rather simple.
attachments['files.zip'] = open('http://somedomain.com/path/to/file.zip').read
attachments needs to receive the content of the file. .read returns the content of the file. My issue was that i was placing the entire zip file in the content of a new file. The above solution will place just the content of the zip into a new file.