我有一个刮刀类，从一个受欢迎的赔率门户网站收集玩家。 确切的名字不存在，网站使用简短形式。例如Rafael - ＆gt; R.幸运的是，它们可以在slugified形式的链接中找到（nadal-rafael）
我制作了一个方法processPlayers，尝试解决这个问题。它适用于更简单的情况，但如果玩家的名字中有连字符或者他/她有两个名字，则会失败。
我写了一些测试来演示html结构和问题。
class Scraper {
/**
* Converts a html string to a cheerio object
* @param {String} html The html string
* @return {Object} The cheerio object
*/
htmlToDom(html) {
return cheerio.load(html)
}
/**
* Gives back the number of parts if the name would slugify
* It takes in account, that the name could contains hyphen
* Leopold von Sacher-Masoch -> leopold-von-sacher-masoch
* @param {Array} a_name The name splitted by space (' ')
* @return {Integer} The length of the name
*/
getNameLength(a_name) {
let name = a_name.length > 1 ? a_name.join(' ') : a_name[0]
return a_name.length + name.split('-').length - 1
}
capitalize(a_name) {
let res = []
a_name.forEach(str => {
res.push(str.substr(0, 1).toUpperCase() + str.substr(1))
})
return res.join(' ')
}
processPlayers(players) {
let link = players('a')
let href = link.attr('href')
let a_players = link.text().split(' - ')
let a_href = href.split('/')
let a_link = a_href[a_href.length - 2].split('-')
let a_player1 = a_players[0].trim().split(' ')
let a_player2 = a_players[1].trim().split(' ')
let a_player1_lastName = a_player1.slice(0, -1)
let a_player2_lastName = a_player2.slice(0, -1)
let a_player1ShortFirstName = [a_player1[a_player1.length - 1]]
let a_player2ShortFirstName = [a_player2[a_player2.length - 1]]
let p1_lnLength = this.getNameLength(a_player1_lastName)
let p1_fnLength = this.getNameLength(a_player1ShortFirstName)
let p2_lnLength = this.getNameLength(a_player2_lastName)
let p2_fnLength = this.getNameLength(a_player2ShortFirstName)
let p1_length = p1_lnLength + p1_fnLength
let p2_length = p2_lnLength + p2_fnLength
let player1FirstName = this.capitalize(a_link.slice(p1_lnLength, p1_length))
let player2FirstName = this.capitalize(a_link.slice(p1_length + p2_lnLength, p1_length + p2_length))
return {
p1: {
firstName: player1FirstName,
lastName: a_player1_lastName.join(' ')
},
p2: {
firstName: player2FirstName,
lastName: a_player2_lastName.join(' ')
}
}
}
}
// test ===========================================================
test('simple case', function() {
let playersCell = `
<td>
<a href="/t/pavlyuchenkova-anastasia-sorribes-tormo-sara/">
<span>Pavlyuchenkova A.</span>
- Sorribes Tormo S.
</a>
</td>
`
const scraper = new Scraper()
const td = scraper.htmlToDom(playersCell)
const players = scraper.processPlayers(td)
equal(players.p1.firstName, 'Anastasia')
equal(players.p1.lastName, 'Pavlyuchenkova')
deepEqual(players.p2, {
firstName: 'Sara',
lastName: 'Sorribes Tormo',
})
});
// =====================================================================
test('hyphen in last name', function() {
let playersCell = `
<td><a href="/t/kudermetova-veronika-duque-marino-mariana/">
<span>Kudermetova V.</span> - Duque-Marino M.</a>
</td>
`
const scraper = new Scraper()
const td = scraper.htmlToDom(playersCell)
const players = scraper.processPlayers(td)
equal(players.p2.firstName, 'Mariana')
equal(players.p2.lastName, 'Duque-Marino')
});
// =====================================================================
test('hyphen in first name', function() {
let playersCell = `
<td>
<a href="/t/tsonga-jo-wilfried-mayer-florian/">
<span>Tsonga J-W.</span>
- Mayer F.
</a>
</td>
`
const scraper = new Scraper()
const td = scraper.htmlToDom(playersCell)
const players = scraper.processPlayers(td)
equal(players.p1.firstName, 'Jo-Wilfried')
equal(players.p1.lastName, 'Tsonga')
});
test('two first names', function() {
let playersCell = `
<td>
<a href="/t/alexandrova-ekaterina-muguruza-blanco-garbine/">
Alexandrova E. - <span>Muguruza</span> B. G.</a>
</td>
`
const scraper = new Scraper()
const td = scraper.htmlToDom(playersCell)
const players = scraper.processPlayers(td)
equal(players.p2.firstName, 'Blanco Garbine')
equal(players.p2.lastName, 'Muguruza')
});<!DOCTYPE html>
<html>
<head>
<meta charset=utf-8 />
<title></title>
<script src="https://wzrd.in/standalone/cheerio@latest"></script>
<link rel="StyleSheet" href="http://code.jquery.com/qunit/qunit-1.12.0.css" type="text/css">
</head>
<body>
<div id="qunit"></div>
<div id="qunit-fixture"></div>
<script src="http://code.jquery.com/qunit/qunit-1.12.0.js" type="text/javascript">
</script>
</body>
</html>
看起来stackoverflow的代码段与qunit有一些问题。所以这里是JsBin链接：
答案 0 :(得分：1)
看起来问题是要在“困难”中正确确定last name和first name。例。
您将文本分为两个全名，OK 然后你用几个单词分割全名，好的。
然后，假设first name始终是最后一个字，而你剩下的就是last name，那么你就犯了一个错误。
事实上，last name是所有不是简短形式的单词，其余的是first name。
我已经用这种方式解决了你的问题。
请在下方找到更新的class Scraper（我已删除了不再使用的功能）：
class Scraper {
/**
* Converts a html string to a cheerio object
* @param {String} html The html string
* @return {Object} The cheerio object
*/
htmlToDom(html) {
return cheerio.load(html)
}
processPlayers(players) {
let href = players('a').attr('href')
let N = players('a').text().trim().split(/\s+-\s+/).map(n => {
let r = new RegExp('.+('+n.replace(/-/g,'\\S+').replace(/\./g,'[^-]+').replace(/\s/g,'.')+').+','i')
let p = {lastName: n.replace(/\s\S+\./g,'')}
href.match(r).map(m => {
r = new RegExp(p.lastName.replace(/\s/g,'.') + '-', 'i')
m = m.replace(r,'').split(/-/).map(i => {return i.substring(0,1).toUpperCase() + i.substring(1)})
p.firstName = n.split(p.lastName + ' ')[1].replace(/\./g,'').split('').map(l => {return m[0] && m[0].indexOf(l) === 0 ? m.shift() : l}).join('')
})
return p
})
return {p1: N[0], p2: N[1]}
}
}
它通过了所有测试。 对不起，如果它不太可读。