0

如何从该页面遍历并显示from_usertext属性:

http://search.twitter.com/search.json?q=stackexchange

使用此代码?:

$json = json_decode(file_get_contents('http://search.twitter.com/search.json?q=stackexchange'));        

        foreach($json as $obj){
            var_dump($obj);
        }

这是此页面的 var_dump():

float 0.021
float 3.2906566512425E+17
string '329065665124249601' (length=18)
string '?page=2&max_id=329065665124249601&q=stackexchange' (length=49)
int 1
string 'stackexchange' (length=13)
string '?since_id=329065665124249601&q=stackexchange' (length=44)
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int 15
int 0
string '0' (length=1)
4

2 回答 2

1
foreach($json->results as $result) {
    echo $result->from_user;
}
于 2013-04-30T03:12:39.323 回答
1

有一个对象集合,每个对象都包含from_usertext属性存储在一个名为 的数组中results。您可以循环此数组并获取如下属性:

    $json = json_decode(file_get_contents('http://search.twitter.com/search.json?q=stackexchange'));        
    foreach($json->results as $result){
       $from_user = $result->from_user;
       $text = $result->text;
       echo $from_user . ' ' .$text;
    }
于 2013-04-30T03:17:38.163 回答