uitableview - parse.com 中多节 tableView 中的 cellForRowAtIndexPath 问题

Question

我在 Parse.com 的 tableView 中使用 cellForRowAtIndexPath 时遇到问题我正在运行的代码是:

override func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {
    let cell = tableView.dequeueReusableCellWithIdentifier("ListadoInventarioCell", forIndexPath: indexPath) as UITableViewCell

        var sectionTitle:String = timelineData1.objectAtIndex(indexPath.section) as String
        var sectionAnimals:Array = porBodega[sectionTitle]!

        // This is the line with problems
        var animal:String = sectionAnimals.objectAtIndex(indexPath.row) as String
        cell.textLabel?.text = animal
}

问题是变量 porBodega 是 String: ArrayOfPFObject 类型的字典，因此sectionAnimals 是 PFObjects 的数组。如果我执行 println(porBodega)，我会得到:

[Oficina: [
<InventarioObjetos:wObhxQ98oz:(null)> {
    Bodega = Oficina;
    Categoria = "Equipo de monta\U00f1a";
    Descripcion = Chalecos;
    FechaDeEntrada = "2012-04-01 19:44:00 +0000";
    Marca = "O'rageous";
    Modelo = "VM-001";
    NumeroDeInventario = CSOV002;
    NumeroDeSerie = NA;
    Status = Usado;
    Tamano = M;
}, 
<InventarioObjetos:u3Yn80lXU8:(null)> {
    Bodega = Oficina;
    Categoria = "Equipo de monta\U00f1a";
    Descripcion = Chalecos;
    FechaDeEntrada = "2012-04-01 19:44:00 +0000";
    Marca = "O'rageous";
    Modelo = "VM-001";
    NumeroDeInventario = CSOV003;
    NumeroDeSerie = NA;
    Status = Usado;
    Tamano = M;
}
]]

所以我不知道如何从我的对象之一获取特定数据，例如变量 Modelo，然后将其设置在单元格文本标签中。有什么想法吗？？

Answer 1

编辑:

override func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {
    let cell = tableView.dequeueReusableCellWithIdentifier("ListadoInventarioCell", forIndexPath: indexPath) as UITableViewCell

        var sectionTitle:String = timelineData1.objectAtIndex(indexPath.section) as String
        var sectionAnimals:Array = porBodega[sectionTitle]!

        let animal:PFObject = sectionAnimals[indexPath.row] as PFObject
        cell.textLabel?.text = animal.objectForKey("modelo") as String
}