Question

我是php的新手，并且遇到了将php array转换为JS object的麻烦。我试图获取给定YouTube视频的信息。我能够在客户端收到信息，但仅限php array。我尝试使用$.parseJSON()，但数据受到退格和冗余字符的污染 JS（使用角度$http）：

$http({
    url: "assets/controllers/youtubeInfo.php",
    method: "POST",
    headers: { 'Content-Type': 'application/x-www-form-urlencoded' },
    data: $.param({ getInfo: videoUrl })
}).success(function(data, status, headers, config) {
    //    console.log(JSON.parse(data));
    console.log(data);
}).error(function(data, status, headers, config) { });

PHP代码：

function get_youtube($url) {
    $youtube = "http://www.youtube.com/oembed?url=".$url."&format=json";

    $curl = curl_init($youtube);
    curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1);
    $return = curl_exec($curl);
    curl_close($curl);
    return json_decode($return, true);
}

$url = $videoKey;

// Display Data 
print_r(get_youtube($url));

这是我的输出：

Array
(
    [title] => StarCraft - OST
    [html] => <iframe width="459" height="344" src="https://www.youtube.com/embed/pNt0iVG2VOA?feature=oembed" frameborder="0" allowfullscreen></iframe>
    [provider_name] => YouTube
    [thumbnail_height] => 360
    [author_url] => https://www.youtube.com/user/JisengSo
    [provider_url] => https://www.youtube.com/
    [type] => video
    [height] => 344
    [thumbnail_url] => https://i.ytimg.com/vi/pNt0iVG2VOA/hqdefault.jpg
    [version] => 1.0
    [author_name] => Jiseng So
    [width] => 459
    [thumbnail_width] => 480
)

Answer 1

要获取JS对象，请不要json_decode输出（或再次编码）：

function get_youtube($url) {
    $youtube = "http://www.youtube.com/oembed?url=".$url."&format=json";

    $curl = curl_init($youtube);
    curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1);
    $return = curl_exec($curl);
    curl_close($curl);
    return $return;    // change here
}

$url = $videoKey;

// Display Data 
echo get_youtube($url);    // change here

Answer 2

以下内容可能会起作用：

PHP代码：

function get_youtube($url) {
    $youtube = "http://www.youtube.com/oembed?url=".$url."&format=json";

    $curl = curl_init($youtube);
    curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1);
    $return = curl_exec($curl);
    curl_close($curl);
    return $return; //If you're only returing it to forward it there's no point decoding it before re-encoding it.
}

$url = $videoKey;

// Display Data 
echo get_youtube($url);

JavaScript的：

$http({
    url: "assets/controllers/youtubeInfo.php",
    method: "POST",
    headers: { 'Content-Type': 'application/x-www-form-urlencoded' },
    data: $.param({ getInfo: videoUrl }),
    responseType: "json", //Tells it to expect JSON in the response 
}).success(function(data, status, headers, config) {
    console.log(data);
}).error(function(data, status, headers, config) { });

注意：据我所知，$ http使用XMLHttpRequest，因此https://developer.mozilla.org/en-US/docs/Web/API/XMLHttpRequest#xmlhttprequest-responsetype适用。

PHP - 将php数组转换为JS对象时遇到麻烦

2 个答案: