cURL不會顯示PHP的整個GET參數

Question

這是我用於向我的應用發送通知的代碼。

include("connection.php"); 
    include("utils.php"); 

    $Info = mysql_fetch_array(mysql_query("SELECT * FROM data WHERE iddata = ".$_GET["dataId"])); 
    $ownerInfo = mysql_fetch_array(mysql_query("SELECT * FROM dataUsers WHERE id = ".$dataInfo['id'])); 
    $data = mysql_fetch_array(mysql_query("SELECT dataUsers.id, dataUsers.token, data.idData, data.id FROM data INNER JOIN data ON data.iddata = '".$_GET["data"]."' AND data.id = dataUsers.id"))["token"]; 

    $datagOwnerDeviceToken = getDeviceTokenFromUserToken($dataOwnerToken); 
    //call server where simplePush.php is located -ports opened- 
    $curl = curl_init(); 
    curl_setopt_array($curl, array(
     CURLOPT_RETURNTRANSFER => 1, 
     CURLOPT_URL => "http://www.myserver.com/sendNotif.php?msg=hello guys&deviceToken=xxxxxxxxxxxx", 
     CURLOPT_USERAGENT => 'whatever' 
     )); 

     $resp = curl_exec($curl); 
     curl_close($curl); 

的問題是，我收到通知，而不是「好傢伙」，我已收到，「你好」，那是說，串偷竊時，它開創一個空格。我認爲這可能是由於編碼，但我真的不知道如何使其工作。

如果我將msg參數更改爲msg = helloguys，我會收到整個字符串。所以空白處有問題。

非常感謝。

這是現在的實際代碼：

$curl = curl_init(); 

curl_setopt_array（$捲曲，陣列（ CURLOPT_RETURNTRANSFER => 1，

$url = "http://server.com/notifications.php?".http_build_query(array(
    'msg' => "hello this is ".$dataInfo["wastringText"], 
    'deviceToken' => $dataOwnerDeviceToken 
)); 

回聲 「這是我的URL：」。$網址; curl_setopt_array（$捲曲，陣列（ CURLOPT_RETURNTRANSFER => 1， CURLOPT_URL => $網址， CURLOPT_USERAGENT => '無論' ））;
$ resp = curl_exec（$ curl）; curl_close（$ curl）;

Answer 1

您需要將空間編碼爲%20或+。但最好使用http_build_query來編碼整個查詢字符串。

$url = "http://www.myserver.com/sendNotif.php?" . 
    http_build_query(array(
     'msg' => 'hello guys', 
     'deviceToken' => 'xxxxxxxxxxxx' 
    )); 
curl_setopt_array($curl, array(
    CURLOPT_RETURNTRANSFER => 1, 
    CURLOPT_URL => $url, 
    CURLOPT_USERAGENT => 'whatever' 
    ));