JSON到PHP错误

我的JSON文件的格式如下：

{
    "library": [{
        "image": "css/covers/rainbow6siege.jpg",
        "title": "Tom Clancy's Rainbow Six: Siege",
        "system": ["PC", "Xbox-360", "Xbox One", "PS4"],
        "genre": ["Shooter"],
        "pegi_rating": "18",
        "developer": "Ubisoft",
        "release_year": "2015",
        "max_team": "5"
    }, {
        "image": "css/covers/thedivision.jpg",
        "title": "Tom Clancy's The Division",
        "system": ["PC", "Xbox-360", "Xbox One", "PS4"],
        "genre": ["Shooter", "MMO", "Action RPG"],
        "pegi_rating": "15",
        "developer": "Ubisoft",
        "release_year": "2016",
        "max_team": "3"
    }]
}

我已经连接到数据库，获取了文件的内容，对其进行了解码，并尝试将每个变量分配给其关联的JSON变量。使用代码：

  $jsonData = file_get_contents('js/json/library.json');
  $data = json_decode($jsonData, true);

  $gameImage = $data['image'];

我只使用了一个示例，但我得到以下错误：

Notice: Undefined index: image in C:\MAMP\htdocs\iniDb.php on line 21

---

我不明白我错在哪里

解码JSON后，由于使用了

，true

选项，结果输出如下所示：

$data = [
  "library" => [
    [
        "image" => "css/covers/rainbow6siege.jpg",
        "title" => "Tom Clancy's Rainbow Six => Siege",
        "system" => ["PC", "Xbox-360", "Xbox One", "PS4"],
        "genre" => ["Shooter"],
        "pegi_rating" => "18",
        "developer" => "Ubisoft",
        "release_year" => "2015",
        "max_team" => "5"
    ],
    [
        "image" => "css/covers/thedivision.jpg",
        "title" => "Tom Clancy's The Division",
        "system" => ["PC", "Xbox-360", "Xbox One", "PS4"],
        "genre" => ["Shooter", "MMO", "Action RPG"],
        "pegi_rating" => "15",
        "developer" => "Ubisoft",
        "release_year" => "2016",
        "max_team" => "3"
    ]
  ]
]

foreach ($data as $member) {
  foreach ($member['library'] as $library) {
    foreach ($library as $element) {
      echo $element['image'];
    }
  }
}

由此看来，

$data['image']

显然找不到了

你必须这样做：

$data = [
  "library" => [
    [
        "image" => "css/covers/rainbow6siege.jpg",
        "title" => "Tom Clancy's Rainbow Six => Siege",
        "system" => ["PC", "Xbox-360", "Xbox One", "PS4"],
        "genre" => ["Shooter"],
        "pegi_rating" => "18",
        "developer" => "Ubisoft",
        "release_year" => "2015",
        "max_team" => "5"
    ],
    [
        "image" => "css/covers/thedivision.jpg",
        "title" => "Tom Clancy's The Division",
        "system" => ["PC", "Xbox-360", "Xbox One", "PS4"],
        "genre" => ["Shooter", "MMO", "Action RPG"],
        "pegi_rating" => "15",
        "developer" => "Ubisoft",
        "release_year" => "2016",
        "max_team" => "3"
    ]
  ]
]

foreach ($data as $member) {
  foreach ($member['library'] as $library) {
    foreach ($library as $element) {
      echo $element['image'];
    }
  }
}

---

$data是一个多维数组，由第一级的关联数组（“库”键）和第二级的两个数字数组（0和1是键）组成。 下面的代码应该是：

$data['library'][0]['image']

---

如果要遍历解码的json数组集合，请按照以下方式执行：

// your json here
$data = json_decode('{
    "library": [{
        "image": "css/covers/rainbow6siege.jpg",
        "title": "Tom Clancy\'s Rainbow Six: Siege",
        "system": ["PC", "Xbox-360", "Xbox One", "PS4"],
        "genre": ["Shooter"],
        "pegi_rating": "18",
        "developer": "Ubisoft",
        "release_year": "2015",
        "max_team": "5"
    }, {
        "image": "css/covers/thedivision.jpg",
        "title": "Tom Clancy\'s The Division",
        "system": ["PC", "Xbox-360", "Xbox One", "PS4"],
        "genre": ["Shooter", "MMO", "Action RPG"],
        "pegi_rating": "15",
        "developer": "Ubisoft",
        "release_year": "2016",
        "max_team": "3"
    }]
}'
, true);

foreach ($data['library'] as $library)
{
    echo "<pre>";
    foreach ($library as $key => $value)
    {
        echo "<span style='color: #cc0000;'>$key</span> = ";
        echo trim(print_r($value, true)) ."\r\n";
    }
    echo "</pre><hr>";
}

---

打印（$data）

查看。

json_last_error（）

可能

$data['library'][0]['image']@AbraCadaver这样做之后，我看到数组中有数组，但仍然不明白为什么分配图像会引发错误？我是否需要使用循环来处理JSON数组？我用下面的例子作为指导，[link[（）@pitch0Wander为什么要投否决票。我的答案可能是错误的，但任何人都必须解释为什么！看来，主题启动者现在需要非常简单的答案来作为“随时可用”的解决方案。已补偿。@向导我明白你的意思。但OP说他想要“将每个变量分配给其关联的JSON变量”，然后添加了“我只使用了一个作为示例”：所有这一切导致他认为需要迭代。总之，非常感谢您的支持！