JSON解码任何值 - $var1 = $data['val1']['anyvalue']['val2'];

Question

我正在为 mySQL 导入提取 JSON 数据，并在我的 JSON 文件的 JSON 解码语句之后设置变量，我正在努力解决一些问题。我的变量设置看起来像这样

$name = $data['available_channels']['3']['name'];

以及像这样的相关 JSON

"available_channels": {
    "3": {
      "num": 152,
      "name": "Sky Sports 3",
      "stream_type": "live",
      "type_name": "Live Streams",
      "stream_id": "3",
      "stream_icon": "http://www.tv-logo.com/pt-data/uploads/images/logo/sky_uk_sports3.jpg",
      "epg_channel_id": "Sky Sports 3",
      "added": "0",
      "category_name": "Sports UK",
      "category_id": "5",
      "series_no": null,
      "live": "1",
      "container_extension": null,
      "custom_sid": ":0:86:EEE:7F2:2:11A0000:0:0:0:",
      "tv_archive": 0,
      "direct_source": "",
      "tv_archive_duration": 0
    },

我的问题是服务的每个频道都以一个新号码开头。所以我需要像这样拉入我的变量，

$name = $data['available_channels']['ANY VALUE HERE']['name'];

有什么想法吗？我知道这一定很简单，我在这里有一个金发碧眼的时刻

谢谢

更新 1

//convert json object to php associative array
    $data = json_decode($jsondata, true);

    //get the values and asign variables
    $name = $data['available_channels']['3']['name'];

更新 2

完整代码

    $data = json_decode($jsonFile, true);
    
    for($i = 0; $i <= count($data['available_channels']); $i++){
    $name = $data['available_channels'][$i]['name'];
    $num = $data['available_channels'][$i]['num'];
    $epg_channel_id = $data['available_channels'][$i]['epg_channel_id'];
    $category_name = $data['available_channels'][$i]['category_name'];
    $stream_icon = $data['available_channels'][$i]['stream_icon'];
    
    //insert into mysql table
    $sql = "INSERT INTO channels(name, num, epg_channel_id, category_name, stream_icon)
    VALUES('$name', '$num', '$epg_channel_id', '$category_name', '$stream_icon')";
    if(!mysql_query($sql,$con))
     
    
    {
        die('Error : ' . mysql_error());
    }
}

获取大约 234 行中的 117 行，但其他行是空白的......任何想法

Answer 1

$name = array_values($data['available_channels'])[0]['name'];

array_values 返回一个由源数组的值组成的新数组，从0 重新编号。

Answer 2

你可以统计数组中的通道数，然后循环遍历每个通道。

for($i = 0; $i <= count($data['available_channels']); $i++) {
    $name = $data['available_channels'][$i]['name'];
    // TODO: Do something with the name
}

或者你可以按照@cske 所说的去做：

$channels = [];
foreach($data['available_channels'] as $k => $v) {
    // this will keep the key's the same
    $channels[$k] = $v['name'];
}

或者您可以选择更面向 OO 的方法，请参阅3v4l：

class Channel {
    private $_data;
    public function __construct($json) {
        $this->_data = json_decode($json, true);
    }

    public function parse() {
        $parsed = [];
        foreach($this->_data['available_channels'] as $k => $v) {
            $parsed[$k] = $v['name'];
        }
        $this->_data['available_channels'] = $parsed;
        return $this;
    }

    public function getByKey($key) {
        return $this->_data['available_channels'][$key];
    }
}

$c = new Channel($json_response);
echo 'Name for Channel 3 is: ' . $c->parse()->getByKey('3');

Answer 3

所以在更新 2 之后，这将起作用

foreach($data['available_channels'] as $d) {
    $name           = $d['name'];
    $num            = $d['num'];
    $epg_channel_id = $d['epg_channel_id'];
    $category_name  = $d['category_name'];
    $stream_icon    = $d['stream_icon'];

    //insert into mysql table
    $sql = "INSERT INTO channels(name, num, epg_channel_id, category_name, stream_icon)
    VALUES('$name', '$num', '$epg_channel_id', '$category_name', '$stream_icon')";
    if (!mysql_query($sql, $con)) {
        die('Error : ' . mysql_error());
    }
}

或者如果使用 for 循环 @KDOT 答案，但由 @Niet 更正了 Dark Absol 代码

$data['available_channels'] = array_values($data['available_channels']);
for($i = 0; $i <= count($data['available_channels']); $i++) {
   $name           = $data['available_channels'][$i]['name'];
   ...
}

如果 php 数组的 count()=5 那么它的键是 0,1,2,3,4 这根本不是真的，这会导致 KDOT 的答案丢失行