[英]PHP: get subarray of subarray data from array by key from JSON file
我有一个如下所示的JSON文件,并且我已经想要创建一个以电影名作为键的最终数组,而actor是键存储的值
{
"movies": [{
"title": "Diner",
"cast": [
"Steve Guttenberg",
"Daniel Stern",
"Mickey Rourke",
"Kevin Bacon",
"Tim Daly",
"Ellen Barkin",
"Paul Reiser",
"Kathryn Dowling",
"Michael Tucker",
"Jessica James",
"Colette Blonigan",
"Kelle Kipp",
"Clement Fowler",
"Claudia Cron"
]
},
{
"title": "Footloose",
"cast": [
"Kevin Bacon",
"Lori Singer",
"Dianne Wiest",
"John Lithgow",
"Sarah Jessica Parker",
"Chris Penn",
"Frances Lee McCain",
"Jim Youngs",
"John Laughlin",
"Lynne Marta",
"Douglas Dirkson"
]
}
]
}
Array(
["Diner"]=>Array(Steve Guttenberg","Daniel Stern","Mickey Rourke","Kevin Bacon","Tim Daly","Ellen Barkin","Paul Reiser","Kathryn Dowling","Michael Tucker","Jessica James","Colette Blonigan","Kelle Kipp","Clement Fowler","Claudia Cron")
["Footloose"]=>Array("Kevin Bacon","Lori Singer","Dianne Wiest","John Lithgow","Sarah Jessica Parker","Chris Penn","Frances Lee McCain","Jim Youngs","John Laughlin","Lynne Marta","Douglas Dirkson")
)
$movies = json_decode(file_get_contents("movies.json"),true);
$actors = array();
foreach($movies as $movie){
$key = "cast";
echo $movie->$key;
}
但是,当我运行当前代码时,php给我一个提示“ 试图获取非对象的属性 ”,有人可以解释为什么会发生这种情况以及如何解决它吗? 错误在此行上:
echo $movie->$key;
提前致谢!
首先,您的json是具有movies
属性的对象。 因此,您必须在通过获取movies
属性进行解码时才能获取电影。 然后,如果json_decode
第二个参数为true,则返回关联的数组而不是对象。 如果要获取对象,请这样调用:
$json = json_decode(file_get_contents("movies.json"));
$movies = $json->movies;
最后,您要获得名称为title且值强制转换的数组。
您可以使用以下代码:
$json = json_decode(file_get_contents("movies.json"));
$movies = $json->movies;
$actors = array();
foreach($movies as $movie){
$title = $movie->title;
$actors[$title] = $movie->cast;
}
print_r($actors); //to see ideal output
这未经测试,但是尝试这样的事情。 (请注意,访问$movies
就像一个数组,而不是像第二个参数一样将true
传递给json_decode()
)
$movies = json_decode(file_get_contents("movies.json"), true);
$actors = array();
foreach($movies['movies'] as $movie){
$actors[$movie['title']] = $movie['cast'];
}
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