Question

制作拖放后我无法从列表框中获取新的有序模型列表放下操作。在视图中我可以看到它是有序的，但当我尝试获取新的项目列表时，它返回原始列表。有没有办法在屏幕上按顺序获取listmodel？

Answer 1

我解决了它在模型列表中进行手动排序。我意识到当使用拖放时它只能直观地改变列表项目的顺序。这是我的工作代码

<listbox id="listBox_Action"
                                        fixedLayout="true" width="300px"
                                        model="@{mainCtrl.comboModelAction}"
                                        selectedItem="@{mainCtrl.selectedAction}" mold="paging"
                                        pageSize="25" forward="onSelect=onSelectedPanelChange()"
                                        onSelect="bd.value=self.selectedItem.label;bd.close();">


                                        <listhead>
                                            <listheader
                                                label="${c:l('SYSADM.SelectAction.combo.ActionName')}" />

                                        </listhead>
                                        <listitem
                                            self="@{each=ACTION_LIST}">
                                            <listcell
                                                label="@{ACTION_LIST.action_func}"
                                                value="@{ACTION_LIST.action_func}" />

                                        </listitem>
                                    </listbox>

这里是java类

public void onChangeList(ForwardEvent event) {
        DropEvent dropEvent = (DropEvent) event.getOrigin();

        // Listitem li = (Listitem)event.getTarget();
        // Listbox lb =(Listbox)event.getOrigin().getTarget();

        int indexOfdraggedListitem =0;
        int indexOfdroppedListitem =0;

        Listitem draggedListitem = (Listitem) dropEvent.getDragged();

        if (dropEvent.getTarget() instanceof Listitem) {

            droppedListitem = (Listitem) dropEvent.getTarget();

            indexOfdraggedListitem = left.getIndexOfItem(draggedListitem);

            indexOfdroppedListitem = left.getIndexOfItem(droppedListitem);

            if (indexOfdraggedListitem > indexOfdroppedListitem) {
                droppedListitem.getParent().insertBefore(draggedListitem, droppedListitem);

                draggableListLeft.add(indexOfdroppedListitem, draggableListLeft.get(indexOfdraggedListitem));

                // Collections.swap(draggableListLeft, indexOfdraggedListitem,
                // indexOfdroppedListitem);

                draggableListLeft.remove(indexOfdraggedListitem + 1);
                // draggableListLeft.add(indexOfdroppedListitem + 1,
                // temp_Fie_Dif_Request_Set);

            } else {

                droppedListitem.getParent().insertBefore(draggedListitem, droppedListitem);

                draggableListLeft.add(indexOfdroppedListitem, draggableListLeft.get(indexOfdraggedListitem));

                draggableListLeft.remove(indexOfdraggedListitem);

            }

        } else   {
            indexOfdraggedListitem = left.getIndexOfItem(draggedListitem);
            left.appendChild(draggedListitem);
            draggableListLeft.add(draggableListLeft.get(indexOfdraggedListitem));
            draggableListLeft.remove(indexOfdraggedListitem);
        }


    }

Answer 2

在您的情况下，您最好维护模型中的顺序，而不是组件集。换句话说，当拖放受影响时，您应该更改模型中的顺序。

ZK拖放问题

2 个答案: