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HomeBackend DevelopmentPHP Tutorial 这一个数据怎样解析,该如何处理

这一个数据怎样解析
我用了json
但是输不出来动西,
当然也无法操作了。
我主要是想得到data[]里面每一个数据。比如
我想得到。og_id 我应该怎么操作呢????


public function getJson($url){
$str = file_get_contents($url);
$data = json_decode($str,true);
echo "

";print_r($data);exit();<br>		return $data;<br>	}<br><br>在浏览器里找打开连接是这样的。<br><br>
Java code
<!--

Code highlighting produced by Actipro CodeHighlighter (freeware)
http://www.CodeHighlighter.com/

-->SINAFINANCE132436359014192800(({data:[{og_id:"1061971",contest_id:"8",sid:"2096647540",StockCode:"sz002230",StockName:"科大讯飞",SellBuy:"0",OrderPrice:"34.700",DealAmount:"4200",OrderAmount:"4200",IfDealt:"1",OrderTime:"2011-10-25 09:58:44",mtime:"2011-10-25 09:58:44",remark:""}],count:"100"}))


------解决方案--------------------
你这个不是json来的
------解决方案--------------------
这是jsonp
------解决方案--------------------
http://api.jquery.com/jQuery.getJSON/
------解决方案--------------------
1、不知道你的这个串是从哪里来的,因为你没给出出处
2、不知道什么是 jsonp,但此串显然不符合那个连接中的描述
3、但我知道这是一个js指令,他通过先前已加载过的 SINAFINANCE132436359014192800 函数进行某个操作。
4、我还知道,传递的 json 串是微软系列语言使用的标准格式:键名没有双引号
这种格式的 json 不能被目前 php 的内置函数所接受,不能不说是个败笔

js 分析
JScript code
<script>
function SINAFINANCE132436359014192800(v) {
  for(t in v) {
    if(typeof(v[t]) == 'object') SINAFINANCE132436359014192800(v[t])
    document.write(t + ':'+v[t]+'<br>');
  }
}

SINAFINANCE132436359014192800(({data:[{og_id:"1061971",contest_id:"8",sid:"2096647540",StockCode:"sz002230",StockName:"科大讯飞",SellBuy:"0",OrderPrice:"34.700",DealAmount:"4200",OrderAmount:"4200",IfDealt:"1",OrderTime:"2011-10-25 09:58:44",mtime:"2011-10-25 09:58:44",remark:""}],count:"100"}))
</script>
<br><font color="#e78608">------解决方案--------------------</font><br>
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这是jsonp
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