MVC3和Twitter Bootstrap TypeAhead無插件

Question

我已經獲得了TypeAhead與靜態數據正常工作，並且能夠正確調用我的控制器功能並獲取數據，但它是A：不正確解析數據或B：TypeAhead不是正確設置。

的JavaScript：

<input type="text" id="itemSearch" data-provide="typeahead" value="@ViewBag.Item" name="itemSearch"/> 




    $('#itemSearch').typeahead({ 
    source: function (query, process) { 
     parts = []; 
     map = {}; 

     $.ajax({ 
       url: '@Url.Action("MakePartsArray")', 
       dataType: "json", 
       type: "POST", 
       data: {query: query}, 
       success: function (data) { 
           $.each(data, function (i, part) { 
           map[part.description] = part; 
           parts.push(part.description); 
          }); 

          typeahead.process(parts); 
       } 
      }); 
    }, 
    updater: function (item) { 
     selectedPart = map[item].itemNumber; 
     return item; 
    }, 
    matcher: function (item) { 
     if (item.toLowerCase().indexOf(this.query.trim().toLowerCase()) != -1) { 
      return true; 
     } 
    }, 
    sorter: function (items) { 
     return items.sort(); 
    }, 
    highlighter: function (item) { 
     var regex = new RegExp('(' + this.query + ')', 'gi'); 
     return item.replace(regex, "<strong>$1</strong>"); 
    } 
}); 

控制器：

public ActionResult MakePartsArray(string query) 
    { 
     var possibleItem = query.ToLower(); 
     var allItems = Db.PartInventorys.Where(l => l.ItemNumber.Contains(possibleItem)).Select(x => new { itemNumber = x.ItemNumber, description = x.Description }); 
     return new JsonResult 
     { 
      ContentType = "text/plain", 
      Data = new { query, total = allItems.Count(), suggestions = allItems.Select(p => p.itemNumber).ToArray(), matches = allItems, }, 
      JsonRequestBehavior = JsonRequestBehavior.AllowGet 
     }; 
    } 

在我控制我看到被正確檢索到的數據，它似乎正確地解析，但沒有被顯示出來了我的預輸入。

任何想法如何驗證故障發生的確切位置或有人看到我的代碼中的直接錯誤？

Answer 1

問題是在阿賈克斯呼叫 -

$.ajax({ 
      url: '@Url.Action("MakePartsArray")', 
      dataType: "json", 
      type: "POST", 
      data: {query: query}, 
      success: function (data) { 
       $.each(data.matches, function (i, part) { 
        var composedInfo = part.description + ' (' + part.itemNumber + ')'; 
        map[composedInfo] = part; 
        parts.push(composedInfo); 
       }); 

       process(parts); 
      } 
     }); 

和返回類型

return new JsonResult 
     { 
      ContentType = "application/json", 
      Data = new { query, total = allItems.Count(), suggestions = allItems.Select(p => p.itemNumber).ToArray(), matches = allItems }, 
      JsonRequestBehavior = JsonRequestBehavior.AllowGet 
     };