node.js – node js cheerio xml

我有以下代码,它可以正常工作：

<troveUrl>http://trove.nla.gov.au/work/23043869</troveUrl>

但是我希望从同一页面获得以下’id’之后的值,并且无法得到它！

<work id="23043869" url="/work/23043869">

这是我目前拥有的代码

var request = require ('request'),
    cheerio = require ('cheerio');
    request('http://api.trove.nla.gov.au/result?key=6k6oagt6ott4ohno&zone=book&q-year1-date=2000&l-advformat=Thesis&l-australian=y&q-term2=&q-term3=&q-term0=&q-field1=title%3A&q-type2=all&q-field0=&q-term1=&q-type3=all&q-field3=subject%3A&q-type0=all&q-field2=creator%3A&q-type1=all&l-availability=y%2Ff&q=+date%3A[2000+TO+2014]&q-year2-date=2014&n=1', function (error, response, html) {
      if (!error && response.statusCode == 200) {
        var $= cheerio.load(html);
        $('troveurl').each(function(i, element){
          var id = $(this);
          console.log(id.text());
        });
      }
    });

任何协助赞赏.

您应该在options对象中传递xmlMode：true,然后您可以将其解析为XML.

然后,您可以使用$(‘tag’).attr(‘attribute’)和$(‘tag’).text()来获取标签和数据,以便在您完成之后获取标签之间的数据.

var request = require('request'),
    cheerio = require('cheerio');
request('http://api.trove.nla.gov.au/result?key=6k6oagt6ott4ohno&zone=book&q-year1-date=2000&l-advformat=Thesis&l-australian=y&q-term2=&q-term3=&q-term0=&q-field1=title%3A&q-type2=all&q-field0=&q-term1=&q-type3=all&q-field3=subject%3A&q-type0=all&q-field2=creator%3A&q-type1=all&l-availability=y%2Ff&q=+date%3A[2000+TO+2014]&q-year2-date=2014&n=1', function(error, response, html) {
    if (!error && response.statusCode == 200) {
        var $= cheerio.load(html, {
          xmlMode: true
        });
        console.log($('work').attr('id'))
    }
});