WHOIS API JSON 数组内容

Question

我需要一些帮助。正如您所想象的那样，由于我已经在这方面搞砸了将近五个小时，现在我有点沮丧。

我的挫败感主要是由于我对 JSON 数组缺乏经验，尤其是更复杂的数组以及如何在 PHP 中处理它们。到目前为止，无论我尝试修改下面的示例代码，都会导致大量错误或没有任何脚本执行......

我正在使用以下服务 -> WHOIS API 来查找与某个域/IP 有关的数据。

服务返回一个 JSON 数组，例如：`

{
  "WhoisRecord": {
    "createdDate": "1997-09-15T00:00:00-0700",
    "updatedDate": "2015-06-12T10:38:52-0700",
    "expiresDate": "2020-09-13T21:00:00-0700",
    "registrant": {
      "name": "Dns Admin",
      "organization": "Google Inc.",
      "street1": "Please contact contact-admin@google.com, 1600 Amphitheatre Parkway",
      "city": "Mountain View",
      "state": "CA",
      "postalCode": "94043",
      "country": "UNITED STATES",
      "email": "dns-admin@google.com",
      "telephone": "16502530000",
      "fax": "16506188571",
      "rawText": "Registrant Name: Dns Admin\nRegistrant Organization: Google Inc.\nRegistrant Street: Please contact contact-admin@google.com, 1600 Amphitheatre Parkway\nRegistrant City: Mountain View\nRegistrant State/Province: CA\nRegistrant Postal Code: 94043\nRegistrant Country: US\nRegistrant Phone: +1.6502530000\nRegistrant Fax: +1.6506188571\nRegistrant Email: dns-admin@google.com"
    },
    "administrativeContact": {
      "name": "DNS Admin",
      "organization": "Google Inc.",
      "street1": "1600 Amphitheatre Parkway",
      "city": "Mountain View",
      "state": "CA",
      "postalCode": "94043",
      "country": "UNITED STATES",
      "email": "dns-admin@google.com",
      "telephone": "16506234000",
      "fax": "16506188571",
      "rawText": "Admin Name: DNS Admin\nAdmin Organization: Google Inc.\nAdmin Street: 1600 Amphitheatre Parkway\nAdmin City: Mountain View\nAdmin State/Province: CA\nAdmin Postal Code: 94043\nAdmin Country: US\nAdmin Phone: +1.6506234000\nAdmin Fax: +1.6506188571\nAdmin Email: dns-admin@google.com"
    }
  }
}`

我感兴趣的只是 WhoisRecord -> 注册人部分（姓名、组织、街道 1、城市、州、邮政编码、国家/地区、电子邮件等）

到目前为止，一切都很好。

但是，当我运行 PHP 代码示例时，他们提供的 API 内容开始让我有点困惑。代码如下所示：

  <?php
  $username="YOUR_USERNAME";
  $password="YOUR_PASSWORD";    
  $contents = file_get_contents("http://www.whoisxmlapi.com//whoisserver/WhoisService?domainName=google.com&username=$username&password=$password&outputFormat=JSON");
  //echo $contents;
  $res=json_decode($contents);
  if($res){
    if($res->ErrorMessage){
        echo $res->ErrorMessage->msg;
    }   
    else{
        $whoisRecord = $res->WhoisRecord;
        if($whoisRecord){
            echo "Domain name: " . print_r($whoisRecord->domainName,1) ."<br/>";
            echo "Created date: " .print_r($whoisRecord->createdDate,1) ."<br/>";
            echo "Updated date: " .print_r($whoisRecord->updatedDate,1) ."<br/>";
            if($whoisRecord->registrant)echo "Registrant: <br/><pre>" . print_r($whoisRecord->registrant->rawText, 1) ."</pre>";
            //print_r($whoisRecord);
        }
    }
  }

?>

当我执行它时，我立即被以下错误猛烈抨击，当某些数据丢失（例如注册人的姓名）时，错误数量会增加。

注意：未定义属性：stdClass::$ErrorMessage in /home/users/pcsnlftp/india.pcs-nl.com/includes/scripts/test/test-processor.php 上线47
域名：google.com
创建日期： 1997-09-15T00:00:00-0700
更新日期： 2015-06-12T10:38:52-0700
注册人：
注册人姓名：Dns 管理员注册人组织：Google Inc. 注册人街道：请 联系contact-admin@google.com，1600 Amphitheatre Parkway Registrant 城市：山景城 注册人州/省：CA Registrant Postal 代码：94043 注册人国家：美国注册人电话：+1.6502530000 注册人传真：+1.6506188571 注册人电子邮件：dns-admin@google.com

我的问题是两方面的；

1. 我如何摆脱基本上没有用（对我而言）的错误？
2. 如何将所需的数据放入我可以插入到 MySQL 数据库中的变量中？

任何帮助将不胜感激！

Answer 1

应该这样做：

<?php
  $username="YOUR_USERNAME";
  $password="YOUR_PASSWORD";    
  $contents = file_get_contents("http://www.whoisxmlapi.com//whoisserver/WhoisService?domainName=google.com&username=$username&password=$password&outputFormat=JSON");
  $res=json_decode($contents, true);
  if($res){
    if(isset($res['ErrorMessage'])){
        echo $res['ErrorMessage'];
    } else {
        if(isset($res['WhoisRecord'])){
            echo "Domain name: " . $res['WhoisRecord']['domainName']."<br/>";
            echo "Created date: " .$res['WhoisRecord']['createdDate']."<br/>";
            echo "Updated date: " .$res['WhoisRecord']['updatedDate']."<br/>";
            if(isset($res['WhoisRecord']['registrant']))
                echo "Registrant: <br/><pre>" . $res['WhoisRecord']['registrant']['rawText'] ."</pre>";
        }
    }
  }

?>

Answer 2

使用它从 json 数组中获取数据。

echo "Name".$res['WhoisRecord']['registrant']['name'];
echo "Organization".$res['WhoisRecord']['registrant']['organization'];

如果不行，先$res=json_decode($contents);到$res=json_decode($contents, true);再用这个

echo "Name".$res['WhoisRecord']['registrant']['name'];
echo "Organization".$res['WhoisRecord']['registrant']['organization'];