从多个链接中获取数据

Question

我有一个PHP代码，它将从“level0 nav-1 active parent”的类名检索数据。有没有一种方法可以提供一系列链接，并为每个链接的循环数组使用稍微不同的类名，而无需为10个链接重复相同的代码？

是这样的： 第一连杆（https://www.postme.com.my/men-1.html） - 使用类（ “0级NAV-1活性母体”） 二（https://www.postme.com.my/women.html） - 使用类（ “0级NAV-2活性母体”） 三（https://www.postme.com.my/children.html） - 使用类（“level0 nav-3 active parent”）

注意递增的nav-＃？

这是PHP代码：

<?php 
header('Content-Type: text/html; charset=utf-8'); 
$grep = new DoMDocument(); 
@$grep->loadHTMLFile("https://www.postme.com.my/men-1.html"); 

$finder = new DomXPath($grep); 
$classCat = "level0 nav-1 active parent"; 

$nodesCat = $finder->query("//*[contains(@class, '$classCat')]"); 

$i = 0; 

    foreach ($nodesCat as $node) { 
    $span = $node->childNodes; 
    $replace = str_replace("Items 1-12 of", "",$span->item(1)->nodeValue); 

    echo $replace. " : "; 
    } 

    // Check another link using class name of "level0 nav-2 active parent" 
    //repeat code 

    @$grep->loadHTMLFile("https://www.postme.com.my/women.html"); 

$finder = new DomXPath($grep); 
$classCat = "level0 nav-2 active parent"; 

$nodesCat = $finder->query("//*[contains(@class, '$classCat')]"); 

$i = 0; 

    foreach ($nodesCat as $node) { 
    $span = $node->childNodes; 
    $replace = $span->item(1)->nodeValue; 

    echo $replace. " : "; 
    } 
//check another link with class name "level0 nav-3 active parent". 
//notice the incrementing nav-#? 
//I don't want to make the code long just because each link is using a slightly different class name to refer to the data. 
?> 

感谢

Answer 1

我会做的就是让这些链接（<li>），这是<ul id="nav">的父。然后从那里。提取值。例如：

$dom = new DOMDocument(); 
@$dom->loadHTMLFile('https://www.postme.com.my/men-1.html'); 
$xpath = new DOMXpath($dom); 

$categories = $xpath->query('//ul[@id="nav"]/li'); 

foreach($categories as $category) { 
    echo $xpath->query('./a/span', $category)->item(0)->nodeValue . '<br/>'; 
}