如何在成功回调函数上向WebSQL传递额外的参数

Question

tx.executeSql('SELECT * FROM bookmarks WHERE bookmarkID = ?', [newSync[i].id],
  function(tx, results) {
    console.log('results.rows.item(0).bookmarkID', results.rows.item(0).bookmarkID);

    tx.executeSql('UPDATE bookmarksSync SET thumbnail=?, ts_created=?, visits=?,
      visits_morning=?, visits_afternoon=?, visits_evening=?, visits_night=?, position=?,
      idgroup=? WHERE bookmarkID=?', 
      [
        results.rows.item(0).thumbnail, 
        results.rows.item(0).ts_created, 
        results.rows.item(0).visits, 
        results.rows.item(0).visits_morning, 
        results.rows.item(0).visits_afternoon, 
        results.rows.item(0).visits_evening, 
        results.rows.item(0).visits_night, 
        results.rows.item(0).position, 
        0, 
        newSync[i].id
      ], speeddial.storage.onError);
  }, speeddial.storage.onError);

newSync [i] .id似乎未定义，我很确定我必须将它传递给回调函数，但我不知道如何......有什么想法吗？ 我希望能够将newSync [i] .id和SQL选择的结果传递给另一个函数，这将更新WebSQL表