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某些问题。我的变量设置如此

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

和像这样的相关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新号码开头。所以我需要把我的变量拉进去，

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

任何想法？我知道这一定很简单，我在这里有一个金发的时刻

由于

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

现在完整代码

＆＃13;

    $data = json_decode($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());
    }
}

＆＃13;

获取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返回一个新数组，该数组由源数组的值组成，从0重新编号。

Answer 2

您可以计算数组内的通道数，然后遍历每个通道。

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

或者你可以做@cske说的话：

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

或者您可以选择更加OO风格的方法，看它在3v4l工作：

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

因此，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'];
   ...
}

如果一个php数组有count（）= 5然后它的键是0,1,2,3,4那么KDOT的答案会丢失行，那就不是真的

JSON解码任何值 - $ var1 = $ data [＆＃39; val1＆＃39;] [＆＃39; anyvalue＆＃39;] [＆＃39; val2＆＃39;];

3 个答案: