Json_decode返回NULL

Question

我使用id从api获取一些信息，然后让用户添加一些额外的信息，然后将它们作为json发送到php。但在php中，json_decode返回null。

这是我的jquery： -

$('#btn').click(function(){
         var imdbid=$('#tst').val();
            var url = "https://www.omdbapi.com/?i="+imdbid+"&plot=full&r=json"
            $.ajax({
            url:url,
            dataType:'json',
            success:function (json) {
                 var data=json
                 var foo=$('form').serializeJSON();
                 var merged=$.extend(data,foo);
                 $.ajax({
                     url:"process.php",
                     type:"POST",
                     data:merged,
                     success:function(data){
                        console.log(data);
                     }
                         })
                }

            })
            })   

And this is my php 

if(isset($_POST['imadbID']) || isset($_POST['comment'])|| isset($_POST['Title']) || isset($_POST['Year'])
|| isset($_POST['released']) || isset($_POST['Runtime'])|| isset($_POST['Actors'])||isset($_POST['Country']) || isset($_POST['Language'])
|| isset($_POST['Plot']) || isset($_POST['Poster']) || isset($_POST['Director']) || isset($_POST['Writer']) || isset($_POST['imdbRating']) || isset($_POST['tst'])){
    $imdbid=json_decode($_POST['imdbID']); $comment=json_decode($_POST['comment']);
    $title=json_decode($_POST['Title']);   $year=json_decode($_POST['Year']);   $released=json_decode($_POST['Released']);
$runtime=json_decode($_POST['Runtime']);   $actor=json_decode($_POST['Actors']);  $country=json_decode($_POST['Country']);
$lang=json_decode($_POST['Language']);   $plot=json_decode($_POST['Plot']);   $poster=json_decode($_POST['Poster']);
$director=json_decode($_POST['Director']);  $writer=json_decode($_POST['Writer']);  $rating=json_decode($_POST['imdbRating']);
 $test=json_decode($_POST['tst']);
    }
    else {
        echo"error";
    }

try {
var_dump($title);var_dump($runtime);var_dump($plot);var_dump($released);var_dump($poster);var_dump($lang);var_dump($imdbid);var_dump($country);var_dump($rating);       
$sql=$conn->prepare("INSERT INTO films (title,length,description,releasedate,bannerpath,language,imdbid,country,rating)
VALUES (:t,:r,:p,:re,:po,:s,:i,:c,:ra)");
$sql->execute(array(":t"=>$title, ":r"=>$runtime, ":p"=>$plot, ":re"=>$released, ":po"=>$poster,":s"=>$lang,":i"=>$imdbid, ":c"=>$country, ":ra"=>$rating));

}
    catch(PDOException $e) {
    echo 'ERROR: ' . $e->getMessage();
}   

我已经尝试了var_dump（），只有":ra"=>$rating不是Null。但其他一切都是Null.Been寻找解决方案，但似乎找不到。

当我console.log合并时，显示

object {Title: "Mother's Day", Year: "2016", Rated: "PG-13", Released: "29 Apr 2016", Runtime: "118 min"…}