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如何在字符串变量中检索特定网页的整个页面源信息并在 php.ini 中回显它。我是 php 新手，不知道这样做有没有人可以给我它的完整源代码。以下是我的源代码：

<?php
$dom = new DOMDocument;
$dom->loadHtmlFile('http://www.google.com');

$xpath = new DOMXPath($dom);

$elements = $xpath->query('//input[@name="session_id"]');
if ($elements->length) {
    echo "found: ", $elements->item(0)->getAttribute('value');
} else {
    echo "not found";
}
}
?>

我已将上面的代码替换为

只是将网址更改为：'http://www.flipkart.com/professional-android-2-application-development-8126525894/p/itmdytmwpjzyhade?pid=9788126525898&ref=8a47bf68-7558-43ce-a9b2-17c1ac119e84'

但它给出了以下错误： 警告：file_get_contents(http://www.flipkart.com/professional-android-2-application-development-8126525894/p/itmdytmwpjzyhade?pid=9788126525898&ref=8a47bf68-7558-43ce-a9b2-17c1ac119e84 ) [function.file-get-contents]：打开流失败：HTTP 请求失败！在第 2 行的 C:\wamp\www\displaycontentswebpage.php

预期结果：（页面来源）

<title>Professional Android 2 Application Development 8126525894: Book: Reto Meier (9788126525898) | Flipkart.com</title>
<meta http-equiv="Content-Type" content="application/xhtml+xml; charset=utf-8" />
<!--<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7" /> -->
<meta property="fb:page_id" content="102988293558" />
<meta property="fb:admins" content="658873552,1412400758,624500995,100000233612389"/>
<meta name="Keywords" content="professional android 2 application development, buy professional android 2 application development, professional android 2 application development india, professional android 2 application development review, reto meier, 8126525894, 9788126525898" />
<meta name="Description" content="Professional Android 2 Application Development by Reto Meier. Rs.449, Save 25%. Buy Professional Android 2 Application Development, All India Free Home Delivery. 8126525894, 9788126525898 |" />


    <link rel="canonical" href="http://www.flipkart.com/professional-android-2-application-development-8126525894/p/itmdytmwpjzyhade" />
<link rel='shortcut icon' href='http://img5.flixcart.com/www/prod/images/favicon-18354.ico' />................something something..........................




src="http://googleads.g.doubleclick.net/pagead/viewthroughconversion/1017598645/?value=0&amp;label=9tgBCLOv-QIQtaWd5QM&amp;guid=ON&amp;script=0"/>
                </div>
            </noscript></div>

请帮忙。

4

3 回答 3

1

当您将文档加载到$dom变量中时，您可以执行以下操作：

echo htmlspecialchars($dom->saveHTML());

请参阅手册saveHTML

我正在使用htmlspecialchars以便显示 html 而不是渲染。

于 2012-07-13T14:02:52.730 回答
0

这是通过 PHP file_get_contents()( http://php.net/file-get-contents ) 中的单个函数实现的，它将文件的内容作为字符串返回。

// print source to current output
echo file_get_contents( 'http://www.google.com' );

// print content as readble format
echo htmlspecialchars( file_get_contents( 'http://www.google.com' ), ENT_SUBSTITUTE );

如果输入在给定编码中包含无效的代码单元序列，则-ENT_SUBSTITUTE或 -ENT_IGNORE标志是必需的。htmlspecialchars()见http://php.net/htmlspecialchars#refsect1-function.htmlspecialchars-returnvalues

于 2012-07-13T14:03:38.917 回答
0 
$dom = new DOMDocument ('1.0');

@$dom->loadHTMLfile ('https://mp3skull.cr');

$thisi=$dom->saveHTML();

echo htmlentities($thisi);

this will print the html source of the page

于 2015-06-21T16:44:45.710 回答