php - 我想从这个 JSON URL 回显 IMDB ID、标题、质量和嵌入 URL

Question

现在我有了这个 JSON URL https://vidsrc.me/movies/latest/page-1.json我想为每个数组回显 IMDB ID、标题、质量和嵌入 URL，因为这听起来像一个多维数组。但每次我尝试这样做。我收到以下错误之一:- 未定义索引- undefined offset 所以我想遍历它并回显每个项目并在每个项目之间换行，这是代码

<?php
$url = file_get_contents('https://vidsrc.me/movies/latest/page-1.json');
$decode = json_decode($url, TRUE);
$res = $decode->result;
foreach($decode as $value){
    $imdb    = $res->imdb_id;
    $title   = $res->title;
    $quality = $res->quality;
    $url     = $res->embed_url;
    echo $imdb;
    echo "<br>";
    echo $tite;
    echo "<br>";
    echo $quality;
    echo "<br>";
    echo $url;
}
?>

Answer 1

json_decode()返回你的数组，但你像对象一样处理它，你应该将它作为数组处理:

<?php
$url = file_get_contents('https://vidsrc.me/movies/latest/page-1.json');
$decode = json_decode($url, TRUE);
foreach($decode['result'] as $value){
    $imdb    = $value['imdb_id'];
    $title   = $value['title'];
    $quality = $value['quality'];
    $url     = $value['embed_url'];
    echo $imdb;
    echo "<br>";
    echo $title;
    echo "<br>";
    echo $quality;
    echo "<br>";
    echo $url;
}
?>