PHP 将类似名称的变量转换为 json

Question

我通过 url 查询字符串变量获取，例如：

myserver_state=1&myserver_running=2&myserver_mem=3

目前我正在添加到现有的 json，例如：

{

   "key1": "1",

   "key2": "2",

   "key3": "3",

   "myserver_state": "1",

   "myserver_running": "2",

   "myserver_mem": "3"

}

我真的想要这样：

{

   "key1": "1",

   "key2": "2",

   "key3": "3",

   "myserver": {

      "state": "1",

      "running": "2",

      "mem": "3"

   }

}

我用它来加载它们：

        $formdata = array(

          'state'=> $_POST['state'],

          'uassip'=> $_POST['uassip'],

          'uassipport'=> $_POST['uassipport'],

          'c_uacminrtpport'=> $_POST['c_uacminrtpport'],

          'c_uacmaxrtpport'=> $_POST['c_uacmaxrtpport'],

          'c_cps'=> $_POST['c_cps'],

          'c_totalcalls'=> $_POST['c_totalcalls'],

          'c_maxchannels'=> $_POST['c_maxchannels'],

          'c_duration'=> $_POST['c_duration'],

          'c_to'=> $_POST['c_to'],

          'c_uacxml'=> $_POST['c_uacxml']

        );

        echo "fromdata: <br>"; echo var_dump($formdata) .  "<br><hr>";

        if(file_put_contents('testconfig.json', json_encode($formdata) )) echo 'OK';

        else echo 'Unable to save data in "testconfig.json"';

非常感谢！

编辑：

我尝试过以下评论：

status.php?server1[当前状态]=10

这实际上可以：

    "c_uacxml": "telnyx-uac-invite-ok.xml",

    "server1": {

        "current_state": "10"

    }

}

这很棒，但是，如果我想添加这样的元素：status.php?server1[current_mem]=1

这实际上取代了整个server1

    "c_uacxml": "telnyx-uac-invite-ok.xml",

    "server1": {

        "current_mem": "10"

    }

}

我失去了已经存在的 current_state

Answer 1

只需在 URL 中使用多维数组，例如：test.php?key1=1&key2=2&myserver[state]=1&myserver[running]=2&myserver[mem]=3如此简单的脚本<?phpecho '<pre>';echo json_encode($_GET, JSON_PRETTY_PRINT);会给你{&nbsp; &nbsp; "key1": "1",&nbsp; &nbsp; "key2": "2",&nbsp; &nbsp; "myserver": {&nbsp; &nbsp; &nbsp; &nbsp; "state": "1",&nbsp; &nbsp; &nbsp; &nbsp; "running": "2",&nbsp; &nbsp; &nbsp; &nbsp; "mem": "3"&nbsp; &nbsp; }}当然，如果需要，您也可以使用具有相同命名规则的 POST 请求。

Answer 2

为了创建嵌套的 JSON 对象，您需要在数组中创建数组。例如$example = [&nbsp; &nbsp; 'key1' => 'foo',&nbsp; &nbsp; 'key2' => 'bar',&nbsp; &nbsp; 'key3' => [&nbsp; &nbsp; &nbsp; &nbsp; 'subkey1' => 'foo',&nbsp; &nbsp; &nbsp; &nbsp; 'subkey2' => 'bar',&nbsp; &nbsp; ],];当运行它时json_encode()，它会导致{&nbsp; "key1": "foo",&nbsp; "key2": "bar",&nbsp; "key3": {&nbsp; &nbsp; "subkey1": "foo",&nbsp; &nbsp; "subkey2": "bar"&nbsp; }}也没有必要像这样加载表单数据 –$formdata = [&nbsp; &nbsp; 'state' => $_POST['state'],&nbsp; &nbsp; 'uassip' => $_POST['uassip'],&nbsp; &nbsp; 'uassipport' => $_POST['uassipport'],&nbsp; &nbsp; 'c_uacminrtpport' => $_POST['c_uacminrtpport'],&nbsp; &nbsp; 'c_uacmaxrtpport' => $_POST['c_uacmaxrtpport'],&nbsp; &nbsp; 'c_cps' => $_POST['c_cps'],&nbsp; &nbsp; 'c_totalcalls' => $_POST['c_totalcalls'],&nbsp; &nbsp; 'c_maxchannels' => $_POST['c_maxchannels'],&nbsp; &nbsp; 'c_duration' => $_POST['c_duration'],&nbsp; &nbsp; 'c_to' => $_POST['c_to'],&nbsp; &nbsp; 'c_uacxml' => $_POST['c_uacxml'],];因为$_POST已经包含您正在尝试重新创建的结构。您只需将发布数据分配给新变量即可。另一方面，我强烈建议您查看 PSR PHP 标准，它们将极大地帮助提高代码可读性和代码结构:) https://www.php-fig.org/psr/