JavaScript:如何反对字符之间的推送字符串？

Question

我正在尝试从如下所示的网站上抓取数据。它在一个 td 中有 4 行信息，我需要帮助抓取我希望最终结果在 JSON 文件中如下所示:

{
 "mainTitle" : "Main Title"
 "subTitle" : "Sub Title here"
 "address1" : "Address Part One"
 "address2" : "Address Part Two"
 "postalCode" : "666666"
 "telNo" : "12345678"
}

HTML:

<div class="contentpage">
  <table>
    <thead></thead>
    <tbody>
      <tr>Main Title</tr>
      <tr>
        <td>
          <strong>Sub Title here</strong>
          <br>
          "Address Part One"
          <br>
          "Address Part Two"
          <br>
          "666666"
          <br>
          "Tel No: 12345678"
        </td>
      </tr>  
    </tbody>
  </table>
</div>

到目前为止，我只能废弃主标题和副标题，我的 JavaScript:

function parseInfo ($) {
  const result = {}
  const enterRow = $('.contentpage > table').first().find('tr)

  result.mainTitle = enterRow.eq(0).find('td').text().trim()
  result.subTitle = enterRow.eq(1).find('td > strong').text().trim()

  return result
}

Answer 1

您可以通过用换行符分割文本来获取这些值。

function parseInfo () {
  const result = {};
  const enterRow = $('.contentpage > table').first().find('tr');
  result.mainTitle = enterRow.eq(0).text().trim();
  result.subTitle = enterRow.eq(1).find('td > strong').text().trim();
  let elements = enterRow.eq(1).find('td').text().trim().split('\n');
  let filteredElements = elements.filter(val => val.replace(/\s/g, '').length).map(val => val.trim());
  result.addressOne = filteredElements[1];
  result.addressTwo = filteredElements[2];
  result.addressCode = filteredElements[3];
  result.telNumber = filteredElements[4];
  return result;
}

我的 fiddle :https://jsfiddle.net/01zmmgy5/1/