我写了一个函数在PHP让我这，我希望它对某人有用。缩略图的路径包含在视频页面的链接标记中。这似乎对我有用。

    $video_url = "http://vimeo.com/7811853"  
    $file = fopen($video_url, "r");
    $filedata = stream_get_contents($file);
    $html_content = strpos($filedata,"<link rel=\"videothumbnail");
    $link_string = substr($filedata, $html_content, 128);
    $video_id_array = explode("\"", $link_string);
    $thumbnail_url = $video_id_array[3];
    echo $thumbnail_url;

希望能对大家有所帮助。

Foggson

我创建了一个code depen来为您获取图像。

https://codepen.io/isramv/pen/gOpabXg

HTML

<input type="text" id="vimeoid" placeholder="257314493" value="257314493">
<button id="getVideo">Get Video</button>
<div id="output"></div>

JavaScript:

const videoIdInput = document.getElementById('vimeoid');
const getVideo = document.getElementById('getVideo');
const output = document.getElementById('output');

function getVideoThumbnails(videoid) {
  fetch(`https://vimeo.com/api/v2/video/${videoid}.json`)
  .then(response => {
    return response.text();
  })
  .then(data => {
    const { thumbnail_large, thumbnail_medium, thumbnail_small } = JSON.parse(data)[0];
    const small = `<img src="${thumbnail_small}"/>`;
    const medium = `<img src="${thumbnail_medium}"/>`;
    const large = `<img src="${thumbnail_large}"/>`;
    output.innerHTML = small + medium + large;
  })
  .catch(error => {
    console.log(error);
  });
}

getVideo.addEventListener('click', e => {
  if (!isNaN(videoIdInput.value)) {
    getVideoThumbnails(videoIdInput.value);
  }
});

2020的解决方案:

我写了一个使用Vimeo Oembed API的PHP函数。

/**
 * Get Vimeo.com video thumbnail URL
 *
 * Set the referer parameter if your video is domain restricted.
 * 
 * @param  int    $videoid   Video id
 * @param  URL    $referer   Your website domain
 * @return bool/string       Thumbnail URL or false if can't access the video
 */
function get_vimeo_thumbnail_url( $videoid, $referer=null ){

    // if referer set, create context
    $ctx = null;
    if( isset($referer) ){
        $ctxa = array(
            'http' => array(
                'header' => array("Referer: $referer\r\n"),
                'request_fulluri' => true,
            ),
        );
        $ctx = stream_context_create($ctxa);
    }

    $resp = @file_get_contents("https://vimeo.com/api/oembed.json?url=https://vimeo.com/$videoid", False, $ctx);
    $resp = json_decode($resp, true);

return $resp["thumbnail_url"]??false;
}

用法:

echo get_vimeo_thumbnail_url("1084537");

我写了一个函数在PHP让我这，我希望它对某人有用。缩略图的路径包含在视频页面的链接标记中。这似乎对我有用。

    $video_url = "http://vimeo.com/7811853"  
    $file = fopen($video_url, "r");
    $filedata = stream_get_contents($file);
    $html_content = strpos($filedata,"<link rel=\"videothumbnail");
    $link_string = substr($filedata, $html_content, 128);
    $video_id_array = explode("\"", $link_string);
    $thumbnail_url = $video_id_array[3];
    echo $thumbnail_url;

希望能对大家有所帮助。

Foggson

api/v2似乎已经死了。 为了使用新的API，您需要注册应用程序，base64将client_id和client_secret编码为授权标头。

$.ajax({
    type:'GET',
    url: 'https://api.vimeo.com/videos/' + video_id,
    dataType: 'json',
    headers: {
        'Authorization': 'Basic ' + window.btoa(client_id + ":" + client_secret);
    },
    success: function(data) {
        var thumbnail_src = data.pictures.sizes[2].link;
        $('#thumbImg').attr('src', thumbnail_src);
    }
});

为了安全起见，您可以返回已经从服务器编码的client_id和client_secret。

您应该解析Vimeo API的响应。没有办法用URL调用(如dailymotion或youtube)。

下面是我的PHP解决方案:

/**
 * Gets a vimeo thumbnail url
 * @param mixed $id A vimeo id (ie. 1185346)
 * @return thumbnail's url
*/
function getVimeoThumb($id) {
    $data = file_get_contents("http://vimeo.com/api/v2/video/$id.json");
    $data = json_decode($data);
    return $data[0]->thumbnail_medium;
}