Handling htmlParse error (failed to load HTTP resource)

Question

I'm trying to web-scrape a page. However, from time to time my loop doesn`t work, because the parser "fail to load HTTP resource" . The problem is that the page doesn't load in my browser, so it's not a problem with the code.

However, it's quite annoying to have to restart the process after creating an exception to each page that I find an error. I wonder if there is a way to put an if condition. I'm thinking of something like: if an error occurrs, then restart the loop at the next step.

I looke the the help page for htmlParse, and found that there is an error argument, but couldn`t understand how to use it. Any ideas for my if condition?

Below is a reproducible example:

if(require(RCurl) == F) install.packages('RCurl')
if(require(XML) == F) install.packages('XML')
if(require(seqinr) == F) install.packages('seqinr')

for  (i in 575:585){
    currentPage <- i  # define pagina inicial da busca
# Link que ser? procurado

link <- paste("http://www.cnj.jus.br/improbidade_adm/visualizar_condenacao.php?seq_condenacao=",
             currentPage,
             sep='')

doc <- htmlParse(link, encoding = "UTF-8") #this will preserve characters
    tables <- readHTMLTable(doc, stringsAsFactors = FALSE)
    if(length(tables) != 0) {
    tabela2 <- as.data.frame(tables[10])

    tabela2[,1]  <- gsub( "\\n", " ", tabela2[,1] )
    tabela2[,2]  <- gsub( "\\n", " ", tabela2[,2] )
    tabela2[,2]  <- gsub( "\\t", " ", tabela2[,2] )

    listofTabelas[[i]] <- tabela2

  tabela1 <- do.call("rbind", listofTabelas)
  names(tabela1) <- c("Variaveis", "status")

    }
}

jlhoward · Accepted Answer · 2014-12-03T16:10:11.107

You might be better off using the httr package.

library(httr)
library(XML)

url <- "http://www.cnj.jus.br/improbidade_adm/visualizar_condenacao.php"
for  (i in 575:585){
  response<- GET(url,path="/",query=c(seq_condenacao=as.character(i)))
  if (response$status_code!=200){ # HTTP request failed!!
    # do some stuff...
    print(paste("Failure:",i,"Status:",response$status_code))
    next
  }
  doc <- htmlParse(response, encoding = "UTF-8")
  # do some other stuff
  print(paste("Success:",i,"Status:",response$status_code))
}
# [1] "Success: 575 Status: 200"
# [1] "Success: 576 Status: 200"
# [1] "Success: 577 Status: 200"
# [1] "Success: 578 Status: 200"
# [1] "Success: 579 Status: 200"
# [1] "Success: 580 Status: 200"
# [1] "Success: 581 Status: 200"
# [1] "Success: 582 Status: 200"
# [1] "Success: 583 Status: 200"
# [1] "Success: 584 Status: 200"
# [1] "Success: 585 Status: 200"

what is z in doc <- htmlParse(content(z), encoding = "UTF-8") — jay_phate, Dec 03 '14 at 09:59
Sorry, it should be `htmlParse(response,...)`. The answer has been edited. — jlhoward, Dec 03 '14 at 16:10

Handling htmlParse error (failed to load HTTP resource)

1 Answers1