Pregunta

I would like to get the URL from a Google websearch as follows:

library(httr)
search.term="httr+package+daterange:%3A2456294-2456659"
url.name=paste0("https://www.google.com/search?q=",search.term)
url.get=GET(url.name)
url.content=content(url.get)

Then the attempt to get the links out of the result fails:

links <- xpathApply(url.content, "//h3//a[@href]", function(x) xmlAttrs(x)[[1]])
Error in UseMethod("xpathApply") : 
no applicable method for 'xpathApply' applied to an object of class "XMLDocumentContent"

What is the best method of getting the links out of the url.content?

¿Fue útil?

Solución

Try content() with as="text" to prevent it from returning an object of class XMLDocumentContent:

library(httr)
search.term="httr+package+daterange:%3A2456294-2456659"
url.name=paste0("https://www.google.com/search?q=",search.term)
url.get=GET(url.name)
url.content=content(url.get, as="text")
links <- xpathSApply(htmlParse(url.content), "//a/@href")
head(links,3)
# href 
# "https://www.google.com/webhp?tab=ww" 
# href 
# "https://www.google.com/search?q=httr%2Bpackage%2Bdaterange::2456294-2456659&um=1&ie=UTF-8&hl=en&tbm=isch&source=og&sa=N&tab=wi" 
# href 
# "https://maps.google.com/maps?q=httr%2Bpackage%2Bdaterange::2456294-2456659&um=1&ie=UTF-8&hl=en&sa=N&tab=wl" 

Update:

As Jake points out in a comment, this also works:

url.get=GET(url.name)
links <- xpathSApply(htmlParse(url.get), "//a/@href")
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