Thanks,
That was very helpful.
I am using readLines and grep. If grep isn't powerful enough I might end up
using the XML package but I hope that won't be necessary.
Thanks again,
KW
--
On May 14, 2012, at 7:18 PM, J Toll wrote:
> On Mon, May 14, 2012 at 4:17 PM, Keith Weintraub <kw1...@gmail.com> wrote:
>> Folks,
>> I want to scrape a series of web-page sources for strings like the
>> following:
>>
>> "/en/Ships/A-8605507.html"
>> "/en/Ships/Aalborg-8122830.html"
>>
>> which appear in an href inside an <a> tag inside a <div> tag inside a table.
>>
>> In fact all I want is the (exactly) 7-digit number before ".html".
>>
>> The good news is that as far as I can tell the the <a> tag is always on it's
>> own line so some kind of line-by-line grep should suffice once I figure out
>> the following:
>>
>> What is the best package/command to use to get the source of a web page. I
>> tried using something like:
>> if(url.exists("http://www.omegahat.org/RCurl";)) {
>> h = basicTextGatherer()
>> curlPerform(url = "http://www.omegahat.org/RCurl";, writefunction = h$update)
>> # Now read the text that was cumulated during the query response.
>> h$value()
>> }
>>
>> which works except that I get one long streamed html doc without the line
>> breaks.
>
> You could use:
>
> h <- readLines("http://www.omegahat.org/RCurl";)
>
> -- or --
>
> download.file(url = "http://www.omegahat.org/RCurl";, destfile = "tmp.html")
> h = scan("tmp.html", what = "", sep = "\n")
>
> and then use grep or the XML package for processing.
>
> HTH
>
> James
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