如何用php类解析json文件的值

如何用php类解析json文件的值,php,json,class,Php,Json,Class,我尝试用如下格式解析json { "bossresponse":{ "responsecode":"200", "web":{ "start":"0", "count":"50", "totalresults":"88300000", "results":[ { "date":"", "clickurl":"http:\/\/naruto.viz.com\/",

我尝试用如下格式解析json

{
  "bossresponse":{
    "responsecode":"200",
    "web":{
      "start":"0",
      "count":"50",
      "totalresults":"88300000",
      "results":[
        {
          "date":"",
          "clickurl":"http:\/\/naruto.viz.com\/",
          "url":"http:\/\/naruto.viz.com\/",
          "dispurl":"naruto<\/b>.viz.com",
          "title":"NARUTO<\/b> Shippuden - OFFICIAL U.S. Site - Watch the Anime ...",
          "abstract":"Check out the latest discussions! TOPICS. AUTHOR"
        },
        {
          "date":"",
          "clickurl":"http:\/\/naruto.com\/",
          "url":"http:\/\/naruto.com\/",
          "dispurl":"naruto<\/b>.com",
          "title":"NARUTO<\/b>",
          "abstract":""
        }
      ]
    }
  }
}
这是成功的,但我认为我的代码是疯狂的代码,你可以看看foreach和foreach非常不好维护。我想在OOP中像类一样实现。。。
你能帮我怎么做吗?有什么建议吗?

有什么原因这行不通吗

$json = json_decode($json, TRUE);
$results = $json['bossresponse']['web']['results'];
foreach($results as $result) {
    echo $result['title'];
}

这是工作中提琴…与此代码是简单的,但强大的…非常感谢。。。
$json = json_decode($json, TRUE);
$results = $json['bossresponse']['web']['results'];
foreach($results as $result) {
    echo $result['title'];
}