Question

我有一个包含多个对象的JSON数组，并尝试使用json_decode创建一个关联数组。

示例数据

$json='[{   
         type: "cool",
         category: "power",
         name: "Robert Downey Jr.",
         character: "Tony Stark / Iron Man",
         bio: "cool kid"
     },
       {
         type: "cool",
         category: "power",
         name: "Chris Hemsworth",
         character: "Thor",
         bio: "cool kid"
     },
     {
         type: "NotCool",
         category: "nothing",
         name: "Alexis Denisof",
         character: "The Other",
         bio: "cool kid"
     }]';

这就是我在做的事情：

$data = json_decode($json, true);

给了我一个NULL的结果。我做错了什么？

（我是PHP的新手。）

Answer 1

您的JSON字符串无效：还需要引用密钥。使用JSONlint网站检查JSON有效性。

Answer 2

这不是有效的JSON。需要引用对象中的键，并使用双引号（"）。

应该是：

$json='[{
     "type": "cool",
     "category": "power",
     "name": "Robert Downey Jr.",
     "character": "Tony Stark / Iron Man",
     "bio": "cool kid"
},
{
     "type": "cool",
     "category": "power",
     "name": "Chris Hemsworth",
     "character": "Thor",
     "bio": "cool kid"
},
{
     "type": "NotCool",
     "category": "nothing",
     "name": "Alexis Denisof",
     "character": "The Other",
     "bio": "cool kid"
}]';

Answer 3

你需要在属性名称周围加双引号，所以它应该是

JSON

[{   
    "type" : "cool",
    "category" : "power",
    "name" : "Robert Downey Jr.",
    "character" : "Tony Stark / Iron Man",
    "bio" : "cool kid"
}]

试试

PHP

echo json_encode(array("name" => "Tony Stark"));

你会看到有效的json

Answer 4

创建验证Json试试这个

<?php
$json='[
    {
        "type": "cool",
        "category": "power",
        "name": "Robert Downey Jr.",
        "character": "Tony Stark / Iron Man",
        "bio": "cool kid"
    },
    {
        "type": "cool",
        "category": "power",
        "name": "Chris Hemsworth",
        "character": "Thor",
        "bio": "cool kid"
    },
    {
        "type": "NotCool",
        "category": "nothing",
        "name": "Alexis Denisof",
        "character": "The Other",
        "bio": "cool kid"
    }
]';
$data = json_decode($json, true);
echo "<pre>" ;
print_r($data);
?>

多个对象数组中的json_decode

4 个答案: