我通过DOMDocument从外部xml文件加载25000个项目(每个站点地图5000个),循环中循环显示5个站点地图需要大约15-20秒,这很多
我很担心我在循环中做错了什么。
如果有什么东西导致加载花了这么长时间,你能检查代码吗?
我不知道。
代码:
$resultHTML = '';
$sitemaps = [
'0' => 'http://example.com/sitemap_part1.xml',
'1' => 'http://example.com/sitemap_part2.xml',
'2' => 'http://example.com/sitemap_part3.xml',
'3' => 'http://example.com/sitemap_part4.xml',
'4' => 'http://example.com/sitemap_part5.xml',
];
foreach ( $sitemaps as $sm ) :
$DomDocument = new DOMDocument();
$DomDocument->preserveWhiteSpace = false;
$DomDocument->load($sm);
$DomNodeList = $DomDocument->getElementsByTagName('loc');
foreach($DomNodeList as $url) :
//$i++;
$resultHTML .= '<div class="xml-item">';
$resultHTML .= $url->nodeValue;
$resultHTML .= '</div>';
endforeach;
endforeach;
echo $resultHTML;
答案 0 :(得分:2)
这是一个未经测试的示例,小文件缓存如何工作。 你应该添加一些错误处理,但我认为它会起作用。
已更新: file_put_contents( $filepath, $resultHTML );
$resultHTML = '';
$chacheDir = "cache";// path/to/your/cachedir
$cacheTime = 24 * 60 * 60;// 24 hours
$sitemaps = [
'0' => 'http://example.com/sitemap_part1.xml',
'1' => 'http://example.com/sitemap_part2.xml',
'2' => 'http://example.com/sitemap_part3.xml',
'3' => 'http://example.com/sitemap_part4.xml',
'4' => 'http://example.com/sitemap_part5.xml',
];
foreach ( $sitemaps as $sm ) :
$filepath = $chacheDir.'/'.md5( $sm );
// check if cached file exists, and if it's too old already
if( file_exists( $filepath ) && ( ( time() - filemtime( $filepath ) ) <= $cacheTime ) ) {
// read from cache
$resultHTML .= file_get_contents( $filepath );
} else {
//create cache file
$DomDocument = new DOMDocument();
$DomDocument->preserveWhiteSpace = false;
//$DomDocument->load($sitemap_url);
$DomDocument->load( $sm );
$DomNodeList = $DomDocument->getElementsByTagName( 'loc' );
foreach ( $DomNodeList as $url ) :
//$i++;
$resultHTML .= '<div class="xml-item">';
$resultHTML .= $url->nodeValue;
$resultHTML .= '</div>';
endforeach;
file_put_contents( $filepath, $resultHTML );
}
endforeach;
echo $resultHTML;