Question

所以我有这个php变量：

$keywords = array("google"=>"http://google.com","stackoverflow"=>"http://stackoverflow.com");    
$content = "<p>Hello Stackover flow</p>
    <p><img src='IMG_URL' alt='stackoverflow logo'/></p>
    <p>Let's go to <a href='http://stackoverflow.com'>stackoverflow</a></p>
    <p>To use stackoverflow and google</p>";

如何将其转换为：

$content = "<p>Hello Stackover flow</p>
<p><img src='IMG_URL' alt='stackoverflow logo'/></p>
<p>Let's go to <a href='http://stackoverflow.com'>stackoverflow</a></p>
<p>To use <a href='http://stackoverflow.com'>stackoverflow</a> and <a href="http://google.com">google</a></p>";

Answer 1

建议您使用array_map()＆amp; implode()。例如：

$keywords = array("google"=>"http://google.com","stackoverflow"=>"http://stackoverflow.com");
$content = "<p>Hello Stackover flow</p>
<p><img src='IMG_URL' alt='stackoverflow logo'/></p>
<p>Let's go to <a href='http://stackoverflow.com'>stackoverflow</a></p>
<p>To use stackoverflow and google</p>";

$anchors = array_map(function($v, $k){
    return '<a href="'.$v.'">'.$k.'</a>';
}, $keywords, array_keys($keywords));

$anchors = '<p> To use '.implode(' and ', $anchors).'</p>';

echo preg_replace("/<p>To use (.*?)<\/p>/", $anchors, $content);

<强>参考：

preg_replace only string，不在a或img标签中

1 个答案: