【发布时间】:2020-03-19 19:45:40
【问题描述】:
我从未使用过 HTML 或 CSS,但我知道 R,所以我在网上查看了几种抓取方法,并在 Stack 中查看了在 R 中执行此操作的方法。我一直在从职位列表页面中提取公司评级时遇到问题。我得到character(0),在示例网址中,该公司的评级为4.0。
这是我的尝试:
library(rvest)
library(tidyverse)
library(xml2)
#example URL
url<- "https://www.indeed.com/viewjob?jk=a25a91736b1f7042&tk=1e3q54n49heai800&from=serp&vjs=3&advn=8876452989351355&adid=95236293&sjdu=TDSJNe66qIM3gcXFOG94m--bPylNW2vvO3WAHEKN7JhCAD1FQ-2FXD1gQyElsLNkg6gfXO2CD3rQYOYjO9iXITyFdYOp8tCECkHuDmf3Og8qdMmciGFIv2ahigETjLmuY8uXdLjnQTg4__yOXqHJkA"
page<- read_html(url)
page%>%
rvest::html_nodes("span") %>%
rvest::html_nodes(xpath = '//*[contains(concat( " ", @class, " " ), concat( " ", "ratingsContent", " " ))]')%>%
rvest::html_text()
#Output is
#character(0)
#Where as it should have been 4.0!
谁能告诉我如何获得它,如果公司缺少评级,我该如何返回NA?谢谢!
【问题讨论】:
标签: html css r web-scraping rvest