将 php json 转换为 javascript getJSON

Question

你好 stackoverflow 社区，我为我对 javascript/ajax 的无知而道歉，但我很难将此 php json 转换为 javascript 函数

$json =    file_get_contents('http://videoapi.my.mail.ru/videos/mail/alex.costantin/_myvideo/4375.json');
$json_a = json_decode($json,true);
$url = $json_a[videos][0][url];
$img = $json_a[meta][poster];
echo $url;
echo $img;

提前感谢您提供的任何帮助

Var_dump Json

string(984) "{"version":3,"service":"mail","provider":"ugc","author":{"email":"alex.costantin@mail.ru","name":"alex.costantin","profile":"http://my.mail.ru/mail/alex.costantin"},"meta":{"title":"avg","externalId":"mail/alex.costantin/_myvideo/4375","itemId":4375,"accId":54048083,"poster":"http://videoapi.my.mail.ru/file/sc03/2500725436577747223","duration":7955,"url":"http://my.mail.ru/mail/alex.costantin/video/_myvideo/4375.html","timestamp":1430140403,"viewsCount":13345},"videos":[{"key":"360p","url":"http://cdn28.my.mail.ru/v/54048083.mp4?sign=dab566053f09db40a63a263f17190aeeb09f1d8d&slave[]=s%3Ahttp%3A%2F%2F127.0.0.1%3A5010%2F54048083-v.mp4&p=f&expire_at=1430773200&touch=1430140403","seekSchema":3},{"key":"720p","url":"http://cdn28.my.mail.ru/hv/54048083.mp4?sign=e9ea54e857ca590b171636efae1b80ccdf0bb5bf&slave[]=s%3Ahttp%3A%2F%2F127.0.0.1%3A5010%2F54048083-hv.mp4&p=f&expire_at=1430773200&touch=1430140403","seekSchema":3}],"encoding":true,"flags":16387,"spAccess":3,"region":"200"}" 

Var_dump Json_a

array(10) { ["version"]=> int(3) ["service"]=> string(4) "mail" ["provider"]=> string(3) "ugc" ["author"]=> array(3) { ["email"]=> string(22) "alex.costantin@mail.ru" ["name"]=> string(14) "alex.costantin" ["profile"]=> string(37) "http://my.mail.ru/mail/alex.costantin" } ["meta"]=> array(9) { ["title"]=> string(3) "avg" ["externalId"]=> string(33) "mail/alex.costantin/_myvideo/4375" ["itemId"]=> int(4375) ["accId"]=> int(54048083) ["poster"]=> string(56) "http://videoapi.my.mail.ru/file/sc03/2500725436577747223" ["duration"]=> int(7955) ["url"]=> string(62) "http://my.mail.ru/mail/alex.costantin/video/_myvideo/4375.html" ["timestamp"]=> int(1430140403) ["viewsCount"]=> int(13345) } ["videos"]=> array(2) { [0]=> array(3) { ["key"]=> string(4) "360p" ["url"]=> string(185) "http://cdn28.my.mail.ru/v/54048083.mp4?sign=dab566053f09db40a63a263f17190aeeb09f1d8d&slave[]=s%3Ahttp%3A%2F%2F127.0.0.1%3A5010%2F54048083-v.mp4&p=f&expire_at=1430773200&touch=1430140403" ["seekSchema"]=> int(3) } [1]=> array(3) { ["key"]=> string(4) "720p" ["url"]=> string(187) "http://cdn28.my.mail.ru/hv/54048083.mp4?sign=e9ea54e857ca590b171636efae1b80ccdf0bb5bf&slave[]=s%3Ahttp%3A%2F%2F127.0.0.1%3A5010%2F54048083-hv.mp4&p=f&expire_at=1430773200&touch=1430140403" ["seekSchema"]=> int(3) } } ["encoding"]=> bool(true) ["flags"]=> int(16387) ["spAccess"]=> int(3) ["region"]=> string(3) "200" } 

到目前为止我已经做到了，但没有成功，代码有什么问题？

$.ajax({ type: "GET", url: "http://videoapi.my.mail.ru/videos/mail/alex.costantin/_myvideo/4375.json", async: false, beforeSend: function(x) { if(x && x.overrideMimeType) { x.overrideMimeType("application/j-son;charset=UTF-8"); } }, dataType: "json", success: function(data){ alert(data.meta.poster); }});

Answer 1

如果您无法通过 JSONP 获取它，您可以创建处理请求的 PHP 包装器，然后调用您的 PHP url 来获取它。

这是同一页面上的一个粗略示例（当然，如果您将 PHP 文件分开会更好）。

<?php

if($_SERVER['REQUEST_METHOD'] === 'POST' && isset($_POST['json_call'])) {
    echo file_get_contents('http://videoapi.my.mail.ru/videos/mail/alex.costantin/_myvideo/4375.json');
    exit;
}

?>

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script type="text/javascript">
$.ajax({
    url: document.URL,
    dataType: 'JSON',
    type: 'POST',
    data: {json_call : true},
    success: function(response) {
        alert(response.version);
        alert(response.videos[0].key);
    }
});
</script>

Sample Output

Answer 2

我认为这正是您访问变量的方式，您需要在键周围使用引号

<?php
   $json = '{"version":3,"service":"mail","provider":"ugc","author":{"email":"alex.costantin@mail.ru","name":"alex.costantin","profile":"http://my.mail.ru/mail/alex.costantin"},"meta":{"title":"avg","externalId":"mail/alex.costantin/_myvideo/4375","itemId":4375,"accId":54048083,"poster":"http://videoapi.my.mail.ru/file/sc03/2500725436577747223","duration":7955,"url":"http://my.mail.ru/mail/alex.costantin/video/_myvideo/4375.html","timestamp":1430140403,"viewsCount":13345},"videos":[{"key":"360p","url":"http://cdn28.my.mail.ru/v/54048083.mp4?sign=dab566053f09db40a63a263f17190aeeb09f1d8d&slave[]=s%3Ahttp%3A%2F%2F127.0.0.1%3A5010%2F54048083-v.mp4&p=f&expire_at=1430773200&touch=1430140403","seekSchema":3},{"key":"720p","url":"http://cdn28.my.mail.ru/hv/54048083.mp4?sign=e9ea54e857ca590b171636efae1b80ccdf0bb5bf&slave[]=s%3Ahttp%3A%2F%2F127.0.0.1%3A5010%2F54048083-hv.mp4&p=f&expire_at=1430773200&touch=1430140403","seekSchema":3}],"encoding":true,"flags":16387,"spAccess":3,"region":"200"}';

  $json_a = json_decode($json,true);

  var_dump($json_a["meta"]["poster"]); 
  //string(56) "http://videoapi.my.mail.ru/file/sc03/2500725436577747223"
?>

用你的 php 设置一个 js 变量，你可以这样做：

<script>
    var poster = '<?php echo $json_a["meta"]["poster"]; ?>';
</script>

用你的 php 设置一个 js 对象，你可以这样做：

<script>
    var jsonString = '<? phpfile_get_contents('http://videoapi.my.mail.ru/videos/mail/alex.costantin/_myvideo/4375.json'); ?>';
    var jsonObj = JSON.parse(jsonString);
</script>

Answer 3

您可以使用Simple JSON for PHP 库来伪造您的复杂 JSON 并将多个 json 合并在一起而无需对其进行解码。

<?php

  include('../includes/json.php');

  // $json = new json(); // Pure JSON
  $json = new json('callback', 'myCallback'); // JSON with Callback

  $jsonOnly = file_get_contents('http://videoapi.my.mail.ru/videos/mail/alex.costantin/_myvideo/4375.json');

  $json->add('status', '200');

  if(connected){
    $json->add("alex.constantin", $jsonOnly, false);
    $json->add("authorized", true);
    // $json->add("authorized"); can also be used
  }
  else 
    $json->add("authorized", false);

  $json->send();
?>

在您的 HTML 中，您主要可以通过两种方式调用它：

旧版 JS 和 DOM 元素。

回调必须与以下内容“集成”：mycallback({ ... });

function load_script(url) {
  var s = document.createElement('script'); 
  s.src = url;
  document.body.appendChild(s);
}

function load_scripts() {
  load_script('myPhpPage');
}

window.onload=load_scripts;

Ajax 通过 Legacy JS 或 JQuery

回调应该是这样的：{} 并被这样调用：

$.getJSON('http://example.com/MyPHP.php',
data,
function(json) {
  alert(json);
});