您如何解析网页并提取所有 href 链接? [英] How do you parse a web page and extract all the href links?
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问题描述
我想在 Groovy 中解析一个网页并提取所有的 href 链接和相关的文本.
如果页面包含这些链接:
<a href="http://www.google.com">Google</a><br/><a href="http://www.apple.com">Apple</a>
输出将是:
谷歌,http://www.google.com<br/>苹果,http://www.apple.com
我正在寻找一个 Groovy 答案.又名.简单的方法!
解决方案
假设 XHTML 格式正确,提取 xml,收集所有标签,找到 'a' 标签,并打印出 href 和文本.
>
input = """<a href = "http://www.hjsoft.com/">John</a><a href = "http://www.google.com/">Google</a><a href = "http://www.stackoverflow.com/">StackOverflow</a></body></html>"""doc = new XmlSlurper().parseText(input)doc.depthFirst().collect { it }.findAll { it.name() == "a" }.each {println "${it.text()}, ${it.@href.text()}"}
I want to parse a web page in Groovy and extract all of the href links and the associated text with it.
If the page contained these links:
<a href="http://www.google.com">Google</a><br />
<a href="http://www.apple.com">Apple</a>
the output would be:
Google, http://www.google.com<br />
Apple, http://www.apple.com
I'm looking for a Groovy answer. AKA. The easy way!
解决方案
Assuming well-formed XHTML, slurp the xml, collect up all the tags, find the 'a' tags, and print out the href and text.
input = """<html><body>
<a href = "http://www.hjsoft.com/">John</a>
<a href = "http://www.google.com/">Google</a>
<a href = "http://www.stackoverflow.com/">StackOverflow</a>
</body></html>"""
doc = new XmlSlurper().parseText(input)
doc.depthFirst().collect { it }.findAll { it.name() == "a" }.each {
println "${it.text()}, ${it.@href.text()}"
}
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