Php JsonException:用户ID没有值

我遇到了一个问题，我的json对象数组没有传递值。我知道它们在那里，但我无法让它们进入我的android应用程序中的变量

The Java function to handle the JSONObject:/**
     * Function to get all userlogs
     **/
    public JSONObject getUserLogs(String userid){
     // userlogs JSONArray
       List<NameValuePair> params = new ArrayList<NameValuePair>();
       params.add(new BasicNameValuePair("tag", getUserLogs_tag ));
       params.add(new BasicNameValuePair("userid", userid));
       JSONObject json = jsonParser.getJSONFromUrl(getAllUserLogs,params);

       return json;
   }

PHP：

else if ($tag == 'getUserLogs') {
       $userid = $_POST['userid'];
       //Special case (normally conntections handled in functions/config or connect)
       $link = new mysqli($host, getDBuser(), getDBpassword(), getDBname());
        /* check connection */
        if (mysqli_connect_errno()) {
            $response["error"] = 1;
            $response["error_msg"] = mysqli_connect_error();
            echo json_encode($response);
        }
        $query = "SELECT * FROM userlog WHERE userid = '$userid'";
        $result = mysqli_query($link, $query);
        $row_cnt = mysqli_num_rows($result);
        //If results, then
        if( $row_cnt > 0){
            $response["userlogs"] = array();
            while($row = mysqli_fetch_array($result, MYSQLI_BOTH)){
                $userlog = array();
                $userlog["userlogs"]["userid"] = $row["userid"];
                $userlog["userlogs"]["medname"] = $row["medname"];
                $userlog["userlogs"]["medtaken"] = $row["medtaken"];
                $userlog["userlogs"]["tabsleft"] = $row["tabsleft"];
                array_push($response["userlogs"], $userlog);
            }
            $response["success"] = 1;
            echo json_encode($response);
            mysqli_close($link);
        }
        else {
            // userlogs not found
            // echo json with error = 1
            $response["error"] = 1;
            $response["error_msg"] = "No Userlogs found";
            echo json_encode($response);
        }
    }
    else {
        $response["error"] = 3;
        $response["error_msg"] = "JSON ERROR";
        echo json_encode($response);
    }
} 

有什么建议吗？

你有什么建议吗

"userlogs": [ // jsonarray userlogs
    {         // json object node
        "userlogs": {   // json object userlogs

然后

---

是的，很有道理。非常感谢：）

else if ($tag == 'getUserLogs') {
       $userid = $_POST['userid'];
       //Special case (normally conntections handled in functions/config or connect)
       $link = new mysqli($host, getDBuser(), getDBpassword(), getDBname());
        /* check connection */
        if (mysqli_connect_errno()) {
            $response["error"] = 1;
            $response["error_msg"] = mysqli_connect_error();
            echo json_encode($response);
        }
        $query = "SELECT * FROM userlog WHERE userid = '$userid'";
        $result = mysqli_query($link, $query);
        $row_cnt = mysqli_num_rows($result);
        //If results, then
        if( $row_cnt > 0){
            $response["userlogs"] = array();
            while($row = mysqli_fetch_array($result, MYSQLI_BOTH)){
                $userlog = array();
                $userlog["userlogs"]["userid"] = $row["userid"];
                $userlog["userlogs"]["medname"] = $row["medname"];
                $userlog["userlogs"]["medtaken"] = $row["medtaken"];
                $userlog["userlogs"]["tabsleft"] = $row["tabsleft"];
                array_push($response["userlogs"], $userlog);
            }
            $response["success"] = 1;
            echo json_encode($response);
            mysqli_close($link);
        }
        else {
            // userlogs not found
            // echo json with error = 1
            $response["error"] = 1;
            $response["error_msg"] = "No Userlogs found";
            echo json_encode($response);
        }
    }
    else {
        $response["error"] = 3;
        $response["error_msg"] = "JSON ERROR";
        echo json_encode($response);
    }
} 

"userlogs": [ // jsonarray userlogs
    {         // json object node
        "userlogs": {   // json object userlogs

userlogs = json.getJSONArray(KEY_USERLOGS); // fine 
for (int i = 0; i < userlogs.length(); i++) {
JSONObject c = userlogs.getJSONObject(i);

JSONObject userlogsobject = c.getJSONObject("userlogs");
String userID = userlogsobject.getString(KEY_USERID);