Question

我努力让我的头衔变得有意义哈哈。我有这个JSON：

[{
    "0": {
        "id": 130427,
        "created_at": 1512521776301,
        "updated_at": 1512549188911,
        "category": 0,
        "platform": 6,
        "date": 1513987200000,
        "region": 8,
        "y": 2017,
        "m": 12,
        "human": "2017-Dec-23",
        "game": 76663
    },
    "2": {
        "id": 131795,
        "created_at": 1514172411633,
        "updated_at": 1514190849639,
        "category": 0,
        "platform": 39,
        "date": 1513987200000,
        "region": 8,
        "y": 2017,
        "m": 12,
        "human": "2017-Dec-23",
        "game": 78658
    }
}]

正如您所看到的，全局json中JSON对象的位置充当对象的名称，我不想这样做。这就是我想要的：

[{
        "id": 130427,
        "created_at": 1512521776301,
        "updated_at": 1512549188911,
        "category": 0,
        "platform": 6,
        "date": 1513987200000,
        "region": 8,
        "y": 2017,
        "m": 12,
        "human": "2017-Dec-23",
        "game": 76663
    },
    {
        "id": 131795,
        "created_at": 1514172411633,
        "updated_at": 1514190849639,
        "category": 0,
        "platform": 39,
        "date": 1513987200000,
        "region": 8,
        "y": 2017,
        "m": 12,
        "human": "2017-Dec-23",
        "game": 78658
    }
]

我想要没有名字的物品。这是我正在使用的代码：

$json = file_get_contents('./releases.json');
$data = json_decode($json, TRUE);
$region = isset($_GET['region']) ? $_GET['region'] : null;

# if region is not null: ?region=8
if ($region) {
    $region_filter = function($v) use ($region) {
        // 8 == Worldwide
        if ($v['region'] == $region || $v['region'] == 8) {
            return true;
        } else {
            return false;
        }
    };
    $data = array_filter($data['data'], $region_filter);
}

header('Content-Type: application/json');
echo json_encode(array($data)); // good

谢谢

Answer 1

您需要使用array_values()重新索引数组。

PHP的json_encode()函数只会生成一个数组，如果所有数组键都是数字键并且没有任何间隙，例如0,1,2,3等问题是array_filter()可以删除某些键并留下空白，这会导致json_encode()包含您在示例中显示的键。您可以在调用array_values()之前使用json_encode()重新索引数组来解决此问题。

以下是一个例子：

<?php

// numeric array keys with no gaps
$a = ['a', 'b', 'c'];
echo json_encode($a);
// ["a","b","c"]

// filter out the 'b' element to introduce a gap in the keys
$a = array_filter($a, function ($v) {
    return $v !== 'b';
});
echo json_encode($a);
// {"0":"a","2":"c"}

// re-index the array to remove gaps
$a = array_values($a);
echo json_encode($a);
// ["a","c"]

在php中删除基于索引的json对象名称

1 个答案: