飞镖：不带运行功能的返回变量

由一起疯了发布于 2020-05-24 03:29:07

我有一系列看起来像这样的物品：

  final List<Map<String, String>> dateIdeas = [
    {'Description': 'Drink Cocktails','Image': 'assets/images/cocktailsfull.jpg'},
    {'Description': 'Go for dinner!', 'Image': "assets/images/dinner.jpg"},
];

当用户点击按钮时，第一项将从数组中删除。

removeItem() {
    setState(() {
      dateIdeas.removeAt(index);
    });
  }

这可以正常工作。但是，我希望用户能够撤消此操作。我对此的解决方案是执行以下操作：

Have the removeItem function return what the array looked like before the removal:

  Map<String, String> removeItemLiked() {
    final removed = dateIdeas[index];
    setState(() {
      _controller.reset();
      likes.add(dateIdeas[index]['Description']);
      dateIdeas.removeAt(index);
    });
    return removed;
  }

Have an undo function that adds the item back into the array at index 0:

  void undo() {
    setState(() {
      Map<String, String> result = removeItemLiked();
      dateIdeas.insert(0, result);
    });
  }

The issue here is that when triggering the undo function it also retriggers removeItem. How can I have the old array stored to a variable so I can always use it later on?

小萝莉m

2020-05-24 03:29:07

The removeItem is getting called by you within the undo function. To prevent this, separate the function call from the removed variable.

// within stateful widget, declare 'removed'...
Map<String, String> removed;

您的其他函数将像这样被修改；

  void removeItemLiked() {
    removed = dateIdeas[index];
    setState(() {
      _controller.reset();
      likes.add(dateIdeas[index]['Description']);
      dateIdeas.removeAt(index);
    });
 }

  void undo() {
    if(removed != null) {
        setState(() {
          dateIdeas.insert(0, removed);
        });
    }
 }

Jim

2020-05-24 03:29:07

You can store the removed item in a variable like lastRemoved and store it's index and value both. Here's a simple example:

class MyNums {
  List<int> numbers = [123, 34, 5, 436, 56, 677];
  Map<String, dynamic> lastRemoved;

  removeItem(int index) {
    lastRemoved = {
      'index': index,
      'item': numbers[index],
    };
    numbers.removeAt(index);
  }

  undoRemove() {
    if (lastRemoved != null) {
      numbers.insert(lastRemoved['index'], lastRemoved['item']);
    }
    lastRemoved = null;
  }
}

测试：

main() {
  MyNums nums = MyNums();
  print('initial:\t\t ${nums.numbers}');

  nums.removeItem(2);
  print('remove from index 2:\t ${nums.numbers}');

  nums.undoRemove();
  print('undo index 2:\t\t ${nums.numbers}');

  nums.removeItem(2);
  print('remove from index 2:\t ${nums.numbers}');

  nums.removeItem(2);
  print('remove from index 2:\t ${nums.numbers}');

  nums.undoRemove();
  print('undo index 2:\t\t ${nums.numbers}');

  nums.undoRemove();
  print('nothing to undo:\t ${nums.numbers}');
}

输出：

initial:                 [123, 34, 5, 436, 56, 677]
remove from index 2:     [123, 34, 436, 56, 677]
undo index 2:            [123, 34, 5, 436, 56, 677]
remove from index 2:     [123, 34, 436, 56, 677]
remove from index 2:     [123, 34, 56, 677]
undo index 2:            [123, 34, 436, 56, 677]
nothing to undo:         [123, 34, 436, 56, 677]

一起疯了

这家伙很懒，什么都没留下

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