傳遞變量到curl_setopt

Question

我希望能夠做到以下幾點：

$search_terms[0]='frank'; 
    $search_terms[1]='sinatra'; 
    $search_terms[2]='beyonce'; 

    foreach($search_terms as $term){ 
    $ch = curl_init(); 
    $url ='http://search.twitter.com/search.json?q=' + $term +'&rpp=100'; 
    curl_setopt($ch, CURLOPT_URL,$url); 
    curl_setopt($ch,CURLOPT_RETURNTRANSFER,1); 

    $var = curl_exec($ch); 


    curl_close($ch); 

    $obj = json_decode($var, true); 

echo $term; 
    var_dump($obj); 
    } 

但我得到一個空對象，當我傾倒$ OBJ，即使$長期版畫確定。

Answer 1

您應該.而不是+ Concat的網址：

$url ='http://search.twitter.com/search.json?q=' . $term . '&rpp=100';