php 获取所有 div 的内容,其中 class 是 X
php get all div's content which class is X
我正在尝试使用 php 获取 div 内容,其中 div 的 class 是 item ajax-product-负载。我的代码是:
$doc = new DOMDocument();
libxml_use_internal_errors(true);
$doc->loadHTML($html); // loads your HTML
$xpath = new DOMXPath($doc);
// returns a list of all links with rel=nofollow
$nlist = $xpath->query("//div[@class='item ajax-product-load']");
print_r($nlist);
我 return 有多少 div 可用 class item ajax-product-load。但我需要获取 div 的内容。这段代码的结果:
DOMNodeList Object
(
[length] => 17
)
$xpath->query(...)
returns a DOMNodelist
object. You need to retrieve an item with $item = $nlist->item(0);
and then get the content with $content = $item->textContent;
.
$nlist = $xpath->query("//div[@class='item ajax-product-load']");
$item = $nlist->item(0);
$content = $item->textContent;
print_r($content);
我正在尝试使用 php 获取 div 内容,其中 div 的 class 是 item ajax-product-负载。我的代码是:
$doc = new DOMDocument();
libxml_use_internal_errors(true);
$doc->loadHTML($html); // loads your HTML
$xpath = new DOMXPath($doc);
// returns a list of all links with rel=nofollow
$nlist = $xpath->query("//div[@class='item ajax-product-load']");
print_r($nlist);
我 return 有多少 div 可用 class item ajax-product-load。但我需要获取 div 的内容。这段代码的结果:
DOMNodeList Object
(
[length] => 17
)
$xpath->query(...)
returns a DOMNodelist
object. You need to retrieve an item with $item = $nlist->item(0);
and then get the content with $content = $item->textContent;
.
$nlist = $xpath->query("//div[@class='item ajax-product-load']");
$item = $nlist->item(0);
$content = $item->textContent;
print_r($content);