javascript - PHP:检测文本中的url，检查url是图像还是网站，然后回显图像

Question

所以我正在开发一个项目，我需要做的是我有一些文本，在该文本中它有很多单词，然后是一个图像的网址。我首先需要做的是，检测该网址是网站还是图像，然后如果是图像，我需要使用 <img> 显示图像。标签，如果是网站，则回显带有 <a href> 的 url标签。到目前为止，我有一个脚本来检测它是网址还是图像，但我仍然需要在文本中回显图像或网址。这是脚本:

<?php
function detectImage($url) {
    $url_headers=get_headers($url, 1);
    if(isset($url_headers['Content-Type'])){
        $type=strtolower($url_headers['Content-Type']);
        $valid_image_type=array();
        $valid_image_type['image/png']='';
        $valid_image_type['image/jpg']='';
        $valid_image_type['image/jpeg']='';
        $valid_image_type['image/jpe']='';
        $valid_image_type['image/gif']='';
        $valid_image_type['image/tif']='';
        $valid_image_type['image/tiff']='';
        $valid_image_type['image/svg']='';
        $valid_image_type['image/ico']='';
        $valid_image_type['image/icon']='';
        $valid_image_type['image/x-icon']='';
        if(isset($valid_image_type[$type])){
            echo "url is image";
        } else {
            echo "url is website";
        }
    }
}
?>

Answer 1

函数是一个返回可在程序中使用的值的例程。不要使用函数来输出内容。将您的函数重写为:

<?php
function isValidImage($url) {
    $url_headers=get_headers($url, 1);
    if(isset($url_headers['Content-Type'])){
        $type=strtolower($url_headers['Content-Type']);
        $valid_image_type=array();
        $valid_image_type['image/png']='';
        $valid_image_type['image/jpg']='';
        $valid_image_type['image/jpeg']='';
        $valid_image_type['image/jpe']='';
        $valid_image_type['image/gif']='';
        $valid_image_type['image/tif']='';
        $valid_image_type['image/tiff']='';
        $valid_image_type['image/svg']='';
        $valid_image_type['image/ico']='';
        $valid_image_type['image/icon']='';
        $valid_image_type['image/x-icon']='';
        if(isset($valid_image_type[$type])){
            return true; // Its an image
        }
        return false;// Its an URL
    }
}

然后在您的逻辑中使用该函数:

<?php
$urls = [
    'http://www.google.be',
    'http://hearstcommerce.ca/customcontent/members/premium/sample.jpg',
];

foreach($urls as $url) {
   if (isValidImage($url) {
      echo '<img src="'.$url.'" />';
   }else{
      echo '<a href="'.$url.'">'.$url.'</a>';
   }
}