PHP Xpath：获取所有包含needle的href值
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PHP Xpath : get all href values that contain needle
提问by MattW
Working with PHP Xpath trying to quickly pull certain links within a html page.
使用 PHP Xpath 尝试快速拉取 html 页面中的某些链接。
The following will find all href links on mypage.html:
$nodes = $x->query("//a[@href]");
以下将找到 mypage.html 上的所有 href 链接：
$nodes = $x->query("//a[@href]");
Whereas the following will find all href links where the descriptionmatches my needle:
$nodes = $x->query("//a[contains(@href,'click me')]");
而以下将找到描述与我的针匹配的所有 href 链接：
$nodes = $x->query("//a[contains(@href,'click me')]");
What I am trying to achieve is matching on the href itself, more specific finding url's that contain certain parameters. Is that possible within a Xpath query or should I just start manipulating the output from the first Xpath query?
我想要实现的是匹配 href 本身，更具体的查找包含某些参数的 url。这在 Xpath 查询中是可能的，还是我应该开始操作第一个 Xpath 查询的输出？
回答by Gordon
Not sure I understand the question correctly, but the second XPath expression already does what you are describing. It does not match against the text node of the A element, but the href attribute:
不确定我是否正确理解了这个问题，但是第二个 XPath 表达式已经完成了您所描述的操作。它不匹配 A 元素的文本节点，而是匹配 href 属性：
$html = <<< HTML
<ul>
<li>
<a href="http://example.com/page?foo=bar">Description</a>
</li>
<li>
<a href="http://example.com/page?lang=de">Description</a>
</li>
</ul>
HTML;
$xml = simplexml_load_string($html);
$list = $xml->xpath("//a[contains(@href,'foo')]");
Outputs:
输出：
array(1) {
[0]=>
object(SimpleXMLElement)#2 (2) {
["@attributes"]=>
array(1) {
["href"]=>
string(31) "http://example.com/page?foo=bar"
}
[0]=>
string(11) "Description"
}
}
As you can see, the returned NodeList contains only the A element with href containing foo (which I understand is what you are looking for). It contans the entire element, because the XPath translates to Fetch all A elements with href attribute containing foo. You would then access the attribute with
如您所见，返回的 NodeList 仅包含带有包含 foo 的 href 的 A 元素（我理解这就是您要查找的内容）。它包含整个元素，因为 XPath 转换为使用包含 foo 的 href 属性获取所有 A 元素。然后，您将访问该属性
echo $list[0]['href'] // gives "http://example.com/page?foo=bar"
If you only want to return the attribute itself, you'd have to do
如果您只想返回属性本身，则必须执行
//a[contains(@href,'foo')]/@href
Note that in SimpleXml, this would return a SimpleXml element though:
请注意，在 SimpleXml 中，这将返回一个 SimpleXml 元素：
array(1) {
[0]=>
object(SimpleXMLElement)#3 (1) {
["@attributes"]=>
array(1) {
["href"]=>
string(31) "http://example.com/page?foo=bar"
}
}
}
but you can output the URL now by
但您现在可以通过以下方式输出 URL
echo $list[0] // gives "http://example.com/page?foo=bar"
