从数据库内容中解析信息

Question

我有一个充满文章的数据库表。在某些情况下，文章底部有一个我想解析以从中获取信息的块。例如，以下是文章表中的两个可能值：

<p>Test test <blockquote class="pull">text quote</blockquote></p>

<p>&nbsp;</p>

<p><span class="italic">italic text</span></p>

<div class="bottom-block"><div class="picture" style="background-image:url('/generator?f=somepicture.jpg');"></div><div class="blurb">Blurb about person<a href="http://website.com">http://website.com</a></div></div>

还有一个例子：

<p>Some content</p>
<div class="bottom-block"><img alt="John Doe" class="picture" src="/assets/images/JOHN_DOE_1.jpg"><div class="blurb"><p>John Doe is a guy from Texas. <a href="http://johnswebsite.com" target="_blank">John's Website</a> and has a large following.</p></div></div>

以上是在数据库中看到的两个值的示例。现在，我希望能够提取某些信息。更准确地说，我想提取 Name、Url、ImageName 和 Blurb

在第一个示例中，在对该值运行查询后，我想看看：

名称： Url：http://website.com ImageName：somepicture.jpg Blurb：Blurb about person<a href="http://website.com">http://website.com</a>

在第二个例子中：

名称：John Doe Url：http://johnswebsite.com ImageName：JOHN_DOE_1.jpg Blurb：<p>John Doe is a guy from Texas. <a href="http://johnswebsite.com" target="_blank">John's Website</a> and has a large following.</p>

我正在玩一个 SQL 查询，它做得不错，但仍然有很多不一致之处。

SELECT id, url, content, TRIM(BOTH '\n' FROM TRIM(TRAILING '</div>\n</div>' FROM TRIM(TRAILING '</div></div>' FROM TRIM(SUBSTRING(content, LOCATE('class="bottom-block"',content)+18))))) as block_extract, TRIM(BOTH '\n' FROM TRIM(TRAILING '</div>\n</div>' FROM TRIM(TRAILING '</div></div>' FROM TRIM(SUBSTRING(content, LOCATE('class="blurb"',content)+12))))) as blurb FROM articles WHERE content LIKE '%bottom-block%' GROUP BY block_extract;

Answer 1

这是一种 DOM 方式：

$results = array();

$fields = array('name', 'img', 'url', 'blurb');

$queries = array('name'  => '//img/@alt',
                 'img'   => '//img[@class = "picture"]/@style |
                             //img/@src |
                             //div[@class = "picture"]/@style',
                 'url'   => '//div[@class = "blurb"]//a/@href',
                 'blurb' => '//div[@class = "blurb"]');

$imgPattern = <<<'EOD'
~
(?|
    .*? background-image:url\( [^)]*? ([^?="\')/]+ \.(?:png|jpe?g|gif) ).*
  | 
    .*? ([^=;/]+)$
)
~ix
EOD;

foreach ($data as $html) {
    $srcDom = new DOMDocument();
    @$srcDom->loadHTML($html);

    $elts = $srcDom->getElementsbyTagName("body")->item(0)->childNodes;

    $tmp['other'] = '';
    foreach ($elts as $elt) {
        if ( $elt->nodeType === XML_ELEMENT_NODE &&
             $elt->hasAttribute('class') &&
             $elt->getAttribute('class') == 'bottom-block' )
            $bbnode = $elt;
        else
            $tmp['other'] .= $srcDom->saveHTML($elt);
    }
    echo htmlspecialchars(print_r($other, true));
    if ( $bbnode ):
        $bbDom = new DOMDocument();
        $bbDom->appendChild($bbDom->importNode($bbnode, true));

        $xpath = new DOMXPath($bbDom);

        foreach($fields as $field) {
            $$field = $xpath->query($queries[$field]);

            if ( $field == 'blurb' ):
                $tmp[$field] = '';
                foreach ($$field->item(0)->childNodes as $child) {
                    $tmp[$field] .= $bbDom->saveHTML($child);
                }
            else:
                $tmp[$field] = ($$field->length) ? $$field->item(0)->nodeValue : '';
            endif;
        }
        $tmp['img'] = preg_replace($imgPattern, '$1', $tmp['img']);
    endif;
    $results[] = $tmp;
}

echo htmlspecialchars(print_r($results, true));

Answer 2

好的，所以我不知道如何使用 SQL 查询来执行此操作，但我将使用 PHP 执行此操作。基本前提是使用五个单独的匹配查询，然后将它们打印出来。匹配查询如下：

1. 底部块内容
2. 图片
3. 网址
4. 简介
5. 名字

这里有一些代码来演示。

// GET THE BOTTOM BLOCK CONTENT
preg_match('~(?<=<div class="bottom-block">).*?(?=</div>$)~ims', $mysql_row, $bottom_block_array);
$string = $bottom_block_array[0];

// GRAB THE IMAGES
preg_match_all('~[A-Z0-9_]+\.(?:jpg|jpeg|gif|png)(?=\'|")~i', $string, $images);
$images = $images[0];

// GRAB THE URLS
preg_match_all('~(?<=href=").*?(?=")~ims', $string, $urls);
$urls = $urls[0];

// GRAB THE BLURBS
preg_match_all('~(?<=<div class="blurb">).*?(?=</div>)~ims', $string, $blurbs);
$blurbs = $blurbs[0];

// GRAB THE NAMES
preg_match_all('~(?<=alt=").*?(?=")~ims', $string, $names);
$names = $names[0];



// LOOP THROUGH AND PRINT OUT ALL OF THE NAMES (OR ONLY ONE, IF DESIRED)
if ($names) {
    foreach ($names AS $name) {print "\nName: ".$name;} // USE THIS IF YOU WANT ALL OF THE NAMES
    // print "\nName: ".$names[0]; // USE THIS IF YOU ONLY WANT ONE POSSIBLE NAME TO SHOW UP
}
else {print "\nName:";}


if ($urls) {
    foreach ($urls AS $url) {print "\nUrl: ".$url;} // PRINT OUT ALL URLS
    // print "\nUrl: ".$urls[0]; // PRINT OUT ONLY ONE URL    
}
else {print "\nUrl:";}


if ($images) {
    foreach ($images AS $image) {print "\nImageName: ".$image;} // PRINT OUT ALL THE IMAGES
    // print "\nImageName: ".$images[0]; // PRINT OUT ONLY ONE IMAGE
}
else {print "\nImageName:";}


if ($blurbs) {
    foreach ($blurbs AS $blurb) {print "\nBlurb: ".$blurb;} // PRINT OUT ALL OF THE BLURBS
    // print "\nBlurb: ".$blurbs[0]; // PRINT OUT ONLY ONE BLURB
}
else {print "\nBlurb:";}


print "\n\n\n\n\n";

Here is a working demo