分析内容并提取php中的链接和图像

Question

您可能已经看到 facebook 如何分析内容并提取链接图像和视频

当我们插入这样的东西时

this is an example text which is linked linke this   http://stackoverflow.com/questions/ask
and i like it so much like this picture     http://stackoverflow.com/img.png

变成这个

    this is an example text which is linked linke this   
<a href='http://stackoverflow.com/questions/ask'>
    and i like it so much like this picture   
 <img src='http://stackoverflow.com/img.png'>

现在我想做同样的事情并分析我的内容并产生丰富的内容。

我怎样才能在 php 中做这样的事情？有没有准备好的课程？

Answer 1

这是两个可以正常工作的功能，经过我的测试。

function isImage($url) {
    if (substr($url, 0, 7) == 'http://') {
        $i = str_ireplace('http://', '', $url); // delete first part
        $i = explode('/', $i); // divide link into parts
        $end = end($i); // get the last part
        $ex = explode('.', $end); // get the extension if exists so for 'image.jpg' $ex = jpg
        if (count($ex) === 2) {
            $allowed = array('jpg', 'jpeg', 'png', 'gif'); // allowed image extensions
            foreach ($allowed as $a) {
                if ($ex = $a) {
                    # Is an image
                    return true;
                }
            }
        } else {
            # It's an url
            return false;
        }
    } else {
        # It's not even a link
        return false;
    }
}

function isLink($url) {
    if (substr($url, 0, 7) == 'http://') {
        $i = str_ireplace('http://', '', $url); // delete first part
        $i = explode('/', $url); // divide link into parts
        $end = end($i); // get the last part
        $ex = explode('.', $end); // get the extension if exists so for 'image.jpg' $ex = jpg
        return (count($ex) == 0) ? true : false;
    } else {
        # It's not even a link
        return false;
    }
}

$image = isImage($input); // true
$link = isLink($input); // false
var_dump($image, $link);

示例 (isLink())：

• 'google.com' = false (*)
• 'http://google.com' = true
• 'http://google.com/image.png' = 假

示例 (isImage())：

• 'google.com' = 假
• 'google.com/image.png' = false (*)
• 'http://google.com/image.png' = true
• 'http://google.com/view/model/controller' = false

(*) = 这是因为它检查链接是否有 http:// 作为前缀。如果你不希望这种情况发生，那么下面是函数：

function isImage($url) {
    $i = explode('/', $i); // divide link into parts
    $end = end($i); // get the last part
    $ex = explode('.', $end); // get the extension if exists so for 'image.jpg' $ex = jpg
    if (count($ex) === 2) {
        $allowed = array('jpg', 'jpeg', 'png', 'gif'); // allowed image extensions
        foreach ($allowed as $a) {
            if ($ex = $a) {
                # Is an image
                return true;
            }
        }
    } else {
        # It's an url
        return false;
    }
}

function isLink($url) {
    $i = explode('/', $url); // divide link into parts
    $end = end($i); // get the last part
    $ex = explode('.', $end); // get the extension if exists so for 'image.jpg' $ex = jpg
    return (count($ex) == 0) ? true : false;
}

然后您可以将其用作：

if (isLink($url)) { echo "<a href=\"$url\">$url</a>"; }
if (isImage($url)) { echo "<img src=\"$url\" alt=\"image\">"; }

Answer 2

您可以使用 preg_replace 和适当的正则表达式来替换匹配的模式并添加 HTML 标记。