Question

为此JSON代码创建PHP部件的简单方法

    {
        "contacts": [
            {
                    "id": "c200",
                    "name": "Ravi Tamada",
                    "email": "ravi@gmail.com",
                    "address": "xx-xx-xxxx,x - street, x - country",
                    "gender" : "male",
                    "phone": {
                        "mobile": "+91 0000000000",
                        "home": "00 000000",
                        "office": "00 000000"
                    }
            },
            {
                    "id": "c201",
                    "name": "Johnny Depp",
                    "email": "johnny_depp@gmail.com",
                    "address": "xx-xx-xxxx,x - street, x - country",
                    "gender" : "male",
                    "phone": {
                        "mobile": "+91 0000000000",
                        "home": "00 000000",
                        "office": "00 000000"
                    }
            },

    ]
}

我创建了简单的JSON响应但是多对象和数组存在问题。

Answer 1

您可以创建如下数组：

    Error.prepareStackTrace = function (err, stack) { return stack; };

    currentfile = err.stack.shift().getFileName();

    while (err.stack.length) {
        callerfile = err.stack.shift().getFileName();

        if(currentfile !== callerfile) return callerfile;
    }
} catch (err) {}
return undefined;

结果：

$a = array("contacts" => array(
     array(
            "id" => "c200",
            "name" => "Ravi Tamada",
            "email" => "ravi@gmail.com",
            "address" => "xx-xx-xxxx,x - street, x - country",
            "gender"  => "male",
            "phone" => array(
                "mobile" => "+91 0000000000",
                "home" => "00 000000",
                "office" => "00 000000"
            )
        ),
       array(
            "id" => "c201",
            "name" => "Johnny Depp",
            "email" => "johnny_depp@gmail.com",
            "address" => "xx-xx-xxxx,x - street, x - country",
            "gender"  => "male",
            "phone" => array(
                "mobile" => "+91 0000000000",
                "home" => "00 000000",
                "office" => "00 000000"
            )
        ),
    )
);

echo json_encode($a);

Answer 2

这实际上很容易自己做

获取现有的JSON字符串，然后将其放入Map。这将为您创建等效的PHP数据结构。

json_decode

Answer 3

echo json_encode([
    "contacts" => [
        [
            "id" => "c200",
            "name" => "Ravi Tamada",
            "email" => "ravi@gmail.com",
            "address" => "xx-xx-xxxx,x - street, x - country",
            "gender"  => "male",
            "phone" => [
                "mobile" => "+91 0000000000",
                "home" => "00 000000",
                "office" => "00 000000"
            ]
        ],
        [
            "id" => "c201",
            "name" => "Johnny Depp",
            "email" => "johnny_depp@gmail.com",
            "address" => "xx-xx-xxxx,x - street, x - country",
            "gender"  => "male",
            "phone" => [
                "mobile" => "+91 0000000000",
                "home" => "00 000000",
                "office" => "00 000000"
            ]
        ],

    ]
]);

Answer 4

要创建一个从该（或另一个）json生成结构的PHP代码，您应该结合使用json_decode + var_export。

$json = '... your json code ...';
$phpcode = var_export(json_decode($json_string));

在您的特定情况下，它将为您的代码创建结构：

stdClass::__set_state(array(
   'contacts' => 
  array (
    0 => 
    stdClass::__set_state(array(
       'id' => 'c200',
       'name' => 'Ravi Tamada',
       'email' => 'ravi@gmail.com',
       'address' => 'xx-xx-xxxx,x - street, x - country',
       'gender' => 'male',
       'phone' => 
      stdClass::__set_state(array(
         'mobile' => '+91 0000000000',
         'home' => '00 000000',
         'office' => '00 000000',
      )),
    )),
    1 => 
    stdClass::__set_state(array(
       'id' => 'c201',
       'name' => 'Johnny Depp',
       'email' => 'johnny_depp@gmail.com',
       'address' => 'xx-xx-xxxx,x - street, x - country',
       'gender' => 'male',
       'phone' => 
      stdClass::__set_state(array(
         'mobile' => '+91 0000000000',
         'home' => '00 000000',
         'office' => '00 000000',
      )),
    )),
  ),
));

你仍然无法创建对象，因为stdClass没有方法__set_state，你可以在这里阅读更多相关内容：http://blog.thoughtlabs.com/blog/2008/02/02/phps-mystical-__set_state-method

注意：还有额外的＆＃34;，＆＃34;在你应该删除的第二个conctact元素之后。

如何为此JSON创建PHP代码

4 个答案: