如何获得谷歌搜索结果

Question

我使用了以下代码:

library(XML)
library(RCurl)
getGoogleURL <- function(search.term, domain = '.co.uk', quotes=TRUE) 
    {
    search.term <- gsub(' ', '%20', search.term)
    if(quotes) search.term <- paste('%22', search.term, '%22', sep='') 
        getGoogleURL <- paste('http://www.google', domain, '/search?q=',
        search.term, sep='')
    }

    getGoogleLinks <- function(google.url) 
    {
       doc <- getURL(google.url, httpheader = c("User-Agent" = "R(2.10.0)"))
       html <- htmlTreeParse(doc, useInternalNodes = TRUE, error=function(...){})
       nodes <- getNodeSet(html, "//a[@href][@class='l']")
       return(sapply(nodes, function(x) x <- xmlAttrs(x)[[1]]))
    }

search.term <- "cran"
quotes <- "FALSE"
search.url <- getGoogleURL(search.term=search.term, quotes=quotes)

links <- getGoogleLinks(search.url)

我想找到我的搜索产生的所有链接,我得到以下结果:

> links
list()

我怎样才能获得链接？另外我想获得谷歌搜索结果的头条新闻和总结如何才能获得它？最后是否有办法获取ChillingEffects.org结果中的链接？

Answer 1

如果查看html变量,可以看到搜索结果链接全部嵌套在<h3 class="r">标记中.

尝试将您的getGoogleLinks功能更改为:

getGoogleLinks <- function(google.url) {
   doc <- getURL(google.url, httpheader = c("User-Agent" = "R
                                             (2.10.0)"))
   html <- htmlTreeParse(doc, useInternalNodes = TRUE, error=function
                          (...){})
   nodes <- getNodeSet(html, "//h3[@class='r']//a")
   return(sapply(nodes, function(x) x <- xmlAttrs(x)[["href"]]))
}

Answer 2

我创建了这个函数来读取公司名称列表,然后获得每个公司名称的顶级网站结果.它会让你开始,然后你可以根据需要调整它.

#libraries.
library(URLencode)
library(rvest)

#load data
d <-read.csv("P:\\needWebsites.csv")
c <- as.character(d$Company.Name)

# Function for getting website.
getWebsite <- function(name)
{
    url = URLencode(paste0("https://www.google.com/search?q=",name))

    page <- read_html(url)

    results <- page %>% 
      html_nodes("cite") %>% # Get all notes of type cite. You can change this to grab other node types.
      html_text()

    result <- results[1]

    return(as.character(result)) # Return results if you want to see them all.
}

# Apply the function to a list of company names.
websites <- data.frame(Website = sapply(c,getWebsite))]

Answer 3

这里的其他解决方案对我不起作用，这是我对 @Bryce-Chamberlain 问题的看法，它在 2019 年 8 月对我有用，它还回答了另一个封闭的问题：R 中的公司名称到 URL

# install.packages("rvest")

get_first_google_link <- function(name, root = TRUE) {
  url = URLencode(paste0("https://www.google.com/search?q=",name))
  page <- xml2::read_html(url)
  # extract all links
  nodes <- rvest::html_nodes(page, "a")
  links <- rvest::html_attr(nodes,"href")
  # extract first link of the search results
  link <- links[startsWith(links, "/url?q=")][1]
  # clean it
  link <- sub("^/url\\?q\\=(.*?)\\&sa.*$","\\1", link)
  # get root if relevant
  if(root) link <- sub("^(https?://.*?/).*$", "\\1", link)
  link
}

companies <- data.frame(company = c("apple acres llc","abbvie inc","apple inc"))
companies <- transform(companies, url = sapply(company,get_first_google_link))
companies
#>           company                            url
#> 1 apple acres llc https://www.appleacresllc.com/
#> 2      abbvie inc        https://www.abbvie.com/
#> 3       apple inc         https://www.apple.com/

^{由reprex 包(v0.2.1)于 2019 年 8 月 10 日创建}