如何用省道映射地图的返回

Question

我试图映射一个映射的返回，但它看起来像第一个映射，不返回迭代的结果。我在这里有点困惑。

我试试那个密码。它只需迭代“行”并返回迭代的数据：

void main() {
  final Map<String, dynamic> treeData = {
    "title": "test",
    "row": [
      [
        {
          "img":
              "https://cdn.pixabay.com/photo/2014/11/30/14/11/cat-551554_960_720.jpg"
        },
        {
          "img":
              "https://cdn.pixabay.com/photo/2015/03/27/13/16/maine-coon-694730_960_720.jpg"
        }
      ],
      [
        {
          "img":
              "https://cdn.pixabay.com/photo/2014/11/30/14/11/cat-551554_960_720.jpg"
        }
      ]
    ],
  };

var mappedData = treeData['row'].map((data) {
    print("");
    print("!!!!!!!!!!!!!!!!!!!");
    print(data);
    return data;
  }).toList().map((data) {
    print("");
    print("-------------------");
    print(data);
    return data;
  }).toList();
  
  print("");
  print("???????????");
  print(mappedData);
}

我得到的结果与我所期望的不一样。正如您所看到的，这两个迭代看起来是完全相同的。

!!!!!!!!!!!!!!!!!!!
[{img: https://cdn.pixabay.com/photo/2014/11/30/14/11/cat-551554_960_720.jpg}, {img: https://cdn.pixabay.com/photo/2015/03/27/13/16/maine-coon-694730_960_720.jpg}]

!!!!!!!!!!!!!!!!!!!
[{img: https://cdn.pixabay.com/photo/2014/11/30/14/11/cat-551554_960_720.jpg}]

-------------------
[{img: https://cdn.pixabay.com/photo/2014/11/30/14/11/cat-551554_960_720.jpg}, {img: https://cdn.pixabay.com/photo/2015/03/27/13/16/maine-coon-694730_960_720.jpg}]

-------------------
[{img: https://cdn.pixabay.com/photo/2014/11/30/14/11/cat-551554_960_720.jpg}]

???????????
[[{img: https://cdn.pixabay.com/photo/2014/11/30/14/11/cat-551554_960_720.jpg}, {img: https://cdn.pixabay.com/photo/2015/03/27/13/16/maine-coon-694730_960_720.jpg}], [{img: https://cdn.pixabay.com/photo/2014/11/30/14/11/cat-551554_960_720.jpg}]]

我希望有这样的东西，第二次迭代遍历第二个数组并返回对象值。

!!!!!!!!!!!!!!!!!!!
[{img: https://cdn.pixabay.com/photo/2014/11/30/14/11/cat-551554_960_720.jpg}, {img: https://cdn.pixabay.com/photo/2015/03/27/13/16/maine-coon-694730_960_720.jpg}]

!!!!!!!!!!!!!!!!!!!
[{img: https://cdn.pixabay.com/photo/2014/11/30/14/11/cat-551554_960_720.jpg}]

-------------------
{img: https://cdn.pixabay.com/photo/2014/11/30/14/11/cat-551554_960_720.jpg}
{img: https://cdn.pixabay.com/photo/2015/03/27/13/16/maine-coon-694730_960_720.jpg}

-------------------
{img: https://cdn.pixabay.com/photo/2014/11/30/14/11/cat-551554_960_720.jpg}

有办法让它这样运作吗？最好的选择是什么？

谢谢你的回答。

Answer 1

哦，我觉得很晚了，当我问这个问题的时候我很累。我发现了我对我的问题的解决办法。

                Row(children: [
              ...treeData['row']
                  .map((data) => Column(children: [
                        ...data.map((item) => Image.network(
                              item['img'],
                              fit: BoxFit.cover,
                              alignment: Alignment.topCenter,
                            ))
                      ]))
                  .toList()

很简单很容易..。似乎有时候我们需要睡觉，伙计们(我)。><

Answer 2

是的，有一种方法可以让它这样运作：

void main() {
  final Map<String, dynamic> treeData = {
    "title": "test",
    "row": [
      [
        {
          "img":
              "https://cdn.pixabay.com/photo/2014/11/30/14/11/cat-551554_960_720.jpg"
        },
        {
          "img":
              "https://cdn.pixabay.com/photo/2015/03/27/13/16/maine-coon-694730_960_720.jpg"
        }
      ],
      [
        {
          "img":
              "https://cdn.pixabay.com/photo/2014/11/30/14/11/cat-551554_960_720.jpg"
        }
      ]
    ],
  };

  var mappedData = treeData['row']
      .map((data) {
        print("");
        print("!!!!!!!!!!!!!!!!!!!");
        print(data);
        return data;
      })
      .toList()
      .map((data) {
        for (var item in data) {
          print("");
          print(item);
        }
        print("-------------------");

        return data;
      })
      .toList();

  print("");
  print("???????????");
  print(mappedData);
}

本质上，您希望打印要传递给第二个函数的列表中的每一项。您将获得第二次迭代的相同结果，因为您正在将第一次迭代的输出转换为函数正在接收的相同输入。