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

所以我正在做一个项目，我需要做的是我有一些文本，文本中有很多单词，然后是一个url，这是一个图像。首先我需要做的是，检测该url是网站还是图像，然后如果它是图像，我需要用

标记显示图像，如果它是网站，则用

标记回显url。到目前为止，我有一个脚本来检测它是url还是图像，但我仍然需要在文本中回显图像或url。以下是脚本：

<?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";
        }
    }
}
?>

---

函数是一个例程，返回可在程序中使用的值。 不要使用函数来输出stuf。将函数重写为：

<?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
    }
}

非常简单

if(isset($valid_image_type[$type])){
    $ech = '<img src="'.$url.'"/>';  
} else {
    $ech = '<a href=".'$url'.">".'$url'."<a>';
}
echo $ech;

if（isset（$valid\u image\u type[$type]））{
$ech=''；
}否则{
$ech=“$url.”；
}
echo$ech；
好的，所以我设法解决了它，我的解决方案是

       ?>    
       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[$type])){
                return true;
            } else {
                return false;
            }
        }
    }
    function detectLink($string) {
        $content_array = explode(" ", $string);
        $output = '';
        foreach($content_array as $content) {
            if(substr($content, 0, 7) == "http://" || substr($content, 0, 4) == "www.") {
                if (detectImage($content)===true) {
                    $content = '<img src="'.$content.'">';
                } else {
                    $content = '<a href="'.$content.'">'.$content.'</a>';
                }
            }
            $output .= " " . $content;
        }
        $output = trim($output);
        return $output;
    }
?>

？>
函数detectImage（$url）{
$url\u headers=get\u headers（$url，1）；
if（isset（$url\u头['Content-Type']））{
$type=strtolower（$url_头['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\u image\u type[$type]））{
返回true；
}否则{
返回false；
}
}
}
函数detectLink（$string）{
$content\u数组=分解（“，$string）；
$output=''；
foreach（$content\u数组作为$content）{
如果（substr（$content，0，7）=“http:/”| | substr（$content，0，4）=“www.”）{
if（detectImage（$content）==true）{
$content=''；
}否则{
$content=''；
}
}
$output.=''.$content；
}
$output=trim（$output）；
返回$output；
}
?>

任何人都可以随意使用
 但是，如果字符串中也有文本，比如“hellp”，那么脚本只对图像有效，对文本无效。因此，首先，您必须在字符串中找到http://并且从那里到第一个“”空间是您要查找的URL，然后使用您的函数
       ?>    
       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[$type])){
                return true;
            } else {
                return false;
            }
        }
    }
    function detectLink($string) {
        $content_array = explode(" ", $string);
        $output = '';
        foreach($content_array as $content) {
            if(substr($content, 0, 7) == "http://" || substr($content, 0, 4) == "www.") {
                if (detectImage($content)===true) {
                    $content = '<img src="'.$content.'">';
                } else {
                    $content = '<a href="'.$content.'">'.$content.'</a>';
                }
            }
            $output .= " " . $content;
        }
        $output = trim($output);
        return $output;
    }
?>