Question

大家好我在下面显示了很多数据集（在$ code变量中）。我想知道我如何输出图像的所有值alt =＆＃34; ...＆＃34; ？例如，我想得到：音乐俱乐部（太阳）音乐09

从：

        <img src="http://www.example.com/teststorage/episodes/11224/201.jpg" alt="Music Club ( sun ) Music 09" />

数据样本集：

    <div class="col-lg-3 col-md-3 col-sm-6 col-xs-12 item">
        <div class="portfolio-item">
            <a href="http://www.example.com/en/test/1234/11224/Music 09" title="Music Club ( sun ) Music 09" class="portfolio-item-link" >
                <span class="portfolio-item-hover"></span>
                <span class="fullscreen"><i class="icon-play"></i></span>
                <img src="http://www.example.com/teststorage/episodes/11224/201.jpg" alt="Music Club ( sun ) Music 09" />
            </a>
            <div class="portfolio-item-title">
                <a href="http://www.example.com/en/test/1234/11224/Music 09" title="Music Club ( sun ) Music 09" class="portfolio-item-link"><h4>Music 09</h4></a>
            </div>
            <div class="clearfix"></div>
        </div>
    </div>

Answer 1

如果您有像下面这样的HTML代码，那么您可以这样做

$re = '/(alt)=("[^"]*")/'; 
$str = '<div class="col-lg-3 col-md-3 col-sm-6 col-xs-12 item">\n        <div class="portfolio-item">\n            <a href="http://www.example.com/en/test/1234/11224/Music 09" title="Music Club ( sun ) Music 09" class="portfolio-item-link" >\n                <span class="portfolio-item-hover"></span>\n                <span class="fullscreen"><i class="icon-play"></i></span>\n                <img src="http://www.example.com/teststorage/episodes/11224/201.jpg" alt="Music Club ( sun ) Music 09" />\n            </a>\n            <div class="portfolio-item-title">\n                <a href="http://www.example.com/en/test/1234/11224/Music 09" title="Music Club ( sun ) Music 09" class="portfolio-item-link"><h4>Music 09</h4></a>\n            </div>\n            <div class="clearfix"></div>\n        </div>\n    </div>'; 

preg_match_all($re, $str, $matches);

输出：

    [0] => Array
        (
            [0] => alt="Music Club ( sun ) Music 09"
        )

    [1] => Array
        (
            [0] => alt
        )

    [2] => Array
        (
            [0] => "Music Club ( sun ) Music 09"
        )

Answer 2

如果您真的想使用preg_match_all，可以使用以下代码。

<?php
$str = <<<END
<div class="col-lg-3 col-md-3 col-sm-6 col-xs-12 item">
        <div class="portfolio-item">
            <a href="http://www.example.com/en/test/1234/11224/Music 09" title="Music Club ( sun ) Music 09" class="portfolio-item-link" >
                <span class="portfolio-item-hover"></span>
                <span class="fullscreen"><i class="icon-play"></i></span>
                <img src="http://www.example.com/teststorage/episodes/11224/201.jpg" alt="Music Club ( sun ) Music 09" />
            </a>
            <div class="portfolio-item-title">
                <a href="http://www.example.com/en/test/1234/11224/Music 09" title="Music Club ( sun ) Music 09" class="portfolio-item-link"><h4>Music 09</h4></a>
            </div>
            <div class="clearfix"></div>
        </div>
    </div>
END;

preg_match_all('/<img(.*?)alt=\"(.*?)\"(.*?)>/si', $str, $out, PREG_SET_ORDER);

//see first output
var_dump($out);

/*
array(1) {
  [0]=>
  array(4) {
    [0]=>
    string(105) "<img src="http://www.example.com/teststorage/episodes/11224/201.jpg" alt="Music Club ( sun ) Music 09" />"
    [1]=>
    string(65) " src="http://www.example.com/teststorage/episodes/11224/201.jpg" "
    [2]=>
    string(27) "Music Club ( sun ) Music 09"
    [3]=>
    string(2) " /"
  }
}
*/

//clean array
$alt = array();

foreach($out as $val) {
  $alt[] =  $val[2];
}

//see cleaned output
var_dump($alt);
/*
array(1) {
  [0]=>
  string(27) "Music Club ( sun ) Music 09"
}
*/
?>

如果你想做正确的事情，我会调查simple_html_dom。你可以这样做：

<?php
// Create DOM from URL or file
$html = file_get_html('http://www.example.com/page_i_want_to_spider.php');

// Find all images
foreach($html->find('img') as $element)
       echo $element->alt . '<br>';
?>

Answer 3

使用DOMDocument：

$dom = new DOMDocument();
@$dom->loadHTML($html);

$imgs= $dom->getElementsByTagName('img');

foreach ($imgs as $img) {
    if ($img->hasAttribute('alt')) echo $img->getAttribute('alt') . '<br/>';
}

如何在php中获取图像alt =“...”的值？

3 个答案: