2013-12-14 81 views
0

我有這個php代碼,我不知道我怎麼能把所有這些數組發送到代碼的末尾,只有一個數組,以便我可以發送這個數組到android?我在哪裏接收我在android中發送的對象?

$username = $_POST["username"]; 
$locations_string = ""; 
$names = array(); 
$response1= array(); 
$response2= array(); 
$user_record = mysql_query("SELECT * FROM users WHERE username='$username'"); 
$value = mysql_fetch_array($user_record); 
$userlocation = $value['location']; 

$query = mysql_query("SELECT * FROM AllFriendsExceptBlocked WHERE username ='$username'  "); 

while($row = mysql_fetch_array($query)){ 
array_push($names , $value["friendname"]); 

echo $row['friendname']. " - ". $row['location']; 
echo "<br />"; 
} 
for($i=0; $i < count($names) ; $i++){ 
$friend = $names[$i]; 
$user_record = mysql_query("SELECT * FROM AllFriendsExceptBlocked WHERE  friendname='$friend' "); 
$value = mysql_fetch_array($user_record); 
$locations_string = $locations_string . "," . $value['location']; 

} 
$response["message"] = "send locations of friends in AllFriendsExceptBlocked"; 
$response1["panicedUserLocation"] = $userlocation; 
$response2["locations"] = $locations_string; 

echo json_encode($response); 
json_encode($response); 
echo json_encode($response2); 
json_encode($response2); 
echo json_encode($response2); 
json_encode($response2); 

現在我怎麼能接收到我從Android中發送的PHP響應?

回答

0

從http調用中獲得響應的位置嘗試將響應轉換爲json對象。例。

InputStream JsonStream=responce.getEntity().getContent(); 
BufferedReader reader= new BufferedReader(new InputStreamReader(JsonStream)); 
StringBuilder builder = new StringBuilder(); 
String line; 
while((line=reader.readLine())!=null) 
{ 

    builder.append(line); 
} 
String JSONdata = builder.toString(); 
Log.i("JsonData",JSONdata); 
JSONObject json = new JSONObject(JSONdata); 
+0

好吧我做錯了我應該只發送一個數組我不能發送所有這些數組!所以有關如何將所有這些響應綁定在一起的想法? – user3101219

+0

你可以在一個數組中嵌套數組,例如:{{},{}} –

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