如何用php类解析json文件的值
我尝试用如下格式解析json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\/",
{
"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'];
}