Question

我创建了以下代码来显示一个空白页面，一个外部网站，但我不得不删除一些节点和创建一段代码所需的每个节点，如果它很大，它几乎不可行。项目

我的怀疑：

有没有办法放入一个我们想要消除的内容（页脚，标题，headerContent等）？

是否有更智能的方法来清理而不是删除元素，只显示我想要的内容（TABLE1）？

        # Create a DOM parser object
        $dom = new DOMDocument();
        libxml_use_internal_errors(true);
        $dom->loadHTMLFile('http://www.sptrans.com.br/sac/solicitacoes.aspx');
        $data = $dom -> getElementByid('TABELA1');


        $xpath = new DOMXPath($dom);
        foreach($xpath->query('//div[contains(attribute::id, "novidadeDestaque")]') as $e ) {
            // Delete this node
            $e->parentNode->removeChild($e);
        }

        $xpath = new DOMXPath($dom);
        foreach($xpath->query('//div[contains(attribute::id, "headerLvl1")]') as $e ) {
            // Delete this node
            $e->parentNode->removeChild($e);
        }

        $xpath = new DOMXPath($dom);
        foreach($xpath->query('//div[contains(attribute::id, "headerContent")]') as $e ) {
            // Delete this node
            $e->parentNode->removeChild($e);
        }

        $xpath = new DOMXPath($dom);
        foreach($xpath->query('//div[contains(attribute::id, "novo_menu")]') as $e ) {
            // Delete this node
            $e->parentNode->removeChild($e);
        }
        $xpath = new DOMXPath($dom);
        foreach($xpath->query('//div[contains(attribute::id, "footer")]') as $e ) {
            // Delete this node
            $e->parentNode->removeChild($e);
        }
        $xpath = new DOMXPath($dom);
        foreach($xpath->query('//div[contains(attribute::id, "header")]') as $e ) {
            // Delete this node
            $e->parentNode->removeChild($e);
        }       
        $xpath = new DOMXPath($dom);
        foreach($xpath->query('//div[contains(attribute::id, "pageNovidades")]') as $e ) {
            // Delete this node
            $e->parentNode->removeChild($e);
        }   

            echo $dom->saveHTML();                          
                ?>
</body>

Answer 1

要创建短代码例程以消除所需的元素，您可以使用数组：

$xpath = new DOMXPath($dom);
$idToDelete = [ 'novidadeDestaque', 'headerLvl1', ... ];

foreach( $idToDelete as $id )
{
    foreach($xpath->query('//div[contains(attribute::id, "'.$id.'")]') as $e ) {
        $e->parentNode->removeChild($e);
    }
}

请注意，您不需要为每次搜索创建新的DOMXPath对象：每个DOMDocument对象只能创建一次。

要仅显示您想要的内容，您可以使用以下语法：

$table = $dom->GetElementById( 'MyTable' );
echo $dom->saveHTML( $table );

要使用只有所需表格的完整HTML ，您可以创建新的DOMDocument并使用importNode添加您的表格：

$src = new DOMDocument();
$dst = new DOMDocument();

$src->loadHTML( $html );
$dst->loadHTML( '<html><head><title>Untitled</title></head><body></body></html>' );

$table    = $src->GetElementById( 'MyTable' );
$imported = $dst->importNode( $table );

$dst->getElementsByTagName( 'body' )->item(0)->appendChild( $imported );

$dst->saveHTML();

详细了解DOMDocument::importNode

使用PHP改进使用DOM解析html代码

1 个答案: