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List files on HTTP/FTP server in R

I'm trying to get list of files on HTTP/FTP server from R!, so that in next step I will be able to download them (or select some of files which meet my criteria to download).

I know that it is possible to use external program in web browser (download manager) which will allow me to select files to download from current web page/ftp. However, I wish to have everything scripted, so that it will be easier for me to reproduce.

I thought about calling Python from R! (since it seems much easier), but I tried to do that entirely in R!

I wrote following lines

require("RCurl") 
result <- getURL("http://server",verbose=TRUE,ftp.use.epsv=TRUE, dirlistonly = TRUE)

Result variable is character type:

typeof(result)
[1] "character"

Sample content is as follows:

Interesting file_20150629.txt20 Aug-2015 09:31 289K\nInteresting file_20150630.txt20 Aug-2015 09:31 293K\nInteresting file_20150701.txt20 Aug-2015 09:31 301K\nInteresting file_20150702.txt20 Aug-2015 09:31 304K\nInteresting file_20150703.txt20 Aug-2015 09:31 301K\nInteresting file_20150704.txt20 Aug-2015 09:31 300K\nInteresting file_20150705.txt20 Aug-2015 09:31 300K\nInteresting file_20150706.txt20 Aug-2015 09:31 305K\nInteresting file_20150707.txt20 Aug-2015 09:31 305K\nInteresting file_20150708.txt20 Aug-2015 09:31 301K\nInteresting file_20150709.txt20 Aug-2015 09:31 294K\n

\n\n\n"

So now, I'm trying to parse result content:

myFiles <- strsplit(result,'<a[^>]* href=\\"([^"]*.txt)\\"')[[1]]

hoping that I will match txt file (since it's in brackets: ()). but it matches:

">Interesting file_20150706.txt</a></td><td align=\"right\">20 Aug-2015 09:31  </td><td align=\"right\">305K</td></tr>\n<tr><td valign=\"top\"><img src=\"/apacheIcons/text.gif\" alt=\"[TXT]\"></td><td>

instead.

What is wrong (I tested my expression on https://regex101.com/) or (maybe this question is more appropriate) there is much easier way to obtain list of files with specific extension on the server in R! ?

Solution

  • You really shouldn't use regex on html. The XML package makes this pretty simple. We can use getHTMLLinks() to gather any links we want. 

    library(XML)
getHTMLLinks(result)
#  [1] "Interesting file_20150629.txt"   "Interesting file_20150630.txt"  
#  [3] "Interesting file_20150701.txt"   "Interesting file_20150702.txt"  
#  [5] "Interesting file_20150703.txt"   "Interesting file_20150704.txt"  
#  [7] "Interesting file_20150705.txt"   "Interesting file_20150706.txt"  
#  [9] "Interesting file_20150707.txt"   "Interesting file_20150708.txt"  
# [11] "Interesting file_20150709.txt"

    That will get all /@href links contained in //a. To grab only the ones that contain .txt, you can use a different XPath query from the default.

    getHTMLLinks(result, xpQuery = "//a/@href[contains(., '.txt')]")

    Or even more precisely, to get those files that end with .txt, you can do

    getHTMLLinks(
    result,
    xpQuery = "//a/@href['.txt'=substring(., string-length(.) - 3)]"
)