遍历对象数组

Question

{
"Catalog": {
    "shirts": [
        {
            "id": "93453951-8427394-234723908",
            "name": "Demin Shirt",
            "price": 100.0,
            "category": "Random Category",
            "available": true,
        },
        {
            "id": "93453951-8427394-40325978",
            "name": "Random Shirt",
            "price": 500.0,
            "category": "Random Category",
            "available": true,
        }
    ],
    "Jeans": [
        {
            "id": "4802345-348579-5983452-23423",
            "name": "Bare Denim Jeans",
            "price": 2000.0,
            "category": "Some Category",
            "available": true,
        },
        {
            "id": "143682137-3481293-239842",
            "name": "Levis jeans",
            "price": 1000.0,
            "category": "Some Category",
            "available": true,
       }
    ]
}

}

如何遍历这个对象数组，以便能够在衬衫类别下显示所有不同类型的衬衫，在牛仔裤标题下显示所有牛仔裤。

类似这样的：

• 衬衫
  • 牛仔衬衫。
  • 随机衬衫
• 牛仔裤
  • 裸色牛仔裤
  • 李维斯牛仔裤

Answer 1

通用解决方案

这是一个对象，而不是一个列表，所以你不能把它插入一个循环中。如果你想遍历所有的值和子值，你将不得不递归地进行。

void recurseObject(dynamic value) {
  if (value is List) {
    // value is a list, iterate over its children
    for (var child in value) {
      recurseObject(object);
    }
  } else if (value is Map) {
    // value is an object, iterate over its keys and recurse the corresponding values
    for (var key in object.keys) {
      print(key);
      recurseObject(object[key]);
    }
  } else {
    // value is a primitive, so just print it
    print(value);
  }
}

---

特定于您的对象

如果你的对象有一个固定的结构，那么你可以抓住所有你想要的位并设置一些嵌套循环。

final catalog = rootObj['Catalog'] as Map<String, Dynamic>;

for (var category in catalog.keys) {
  final categoryList = catalog[category] as List<dynamic>;
  print(category);
  for (var value in categoryList) {
    final name = value['name'];
    print(name);
  }
}

Answer 2

你所拥有的是纯 JSON。您需要先将其转换为 Dart 对象。

有几种方法可以进行转换。我发现最简单的方法是使用在线转换器，例如这个网站：https://app.quicktype.io/

将 JSON 转换为对象后，您可以像这样遍历它：

printCatalog() {
 var catalog =
    Catalog.fromRawJson("Here's where you need to pass in your JSON");

 print("Shirts: ");

 catalog.shirts.forEach((shirt) {
   print(shirt.name);
 });

 print("Jeans: ");

 catalog.jeans.forEach((jean) {
   print(jean.name);
 });
}

Answer 3

只需使用 dart:convert 库和 json.decode。然后在你的 json 中访问你想要的任何东西。 我修复了您的 JSON，因为它没有正确设置，这是完整的解决方案。

import 'dart:convert';

void main() {
  var jsonstring = '''{
    "Catalog": {
        "shirts": [{
                "id": "93453951-8427394-234723908",
                "name": "Demin Shirt",
                "price": 100.0,
                "category": "Random Category",
                "available": "true"
            },
            {
                "id": "93453951-8427394-40325978",
                "name": "Random Shirt",
                "price": 500.0,
                "category": "Random Category",
                "available": true
            }
        ],
        "Jeans": [{
                "id": "4802345-348579-5983452-23423",
                "name": "Bare Denim Jeans",
                "price": 2000.0,
                "category": "Some Category",
                "available": true
            },
            {
                "id": "143682137-3481293-239842",
                "name": "Levis jeans",
                "price": 1000.0,
                "category": "Some Category",
                "available": true
            }
        ]
    }
}''';

  Map<String, dynamic> data = json.decode(jsonstring);

  print("--COOL SHIRTS--");
  for (var shirt in data["Catalog"]["shirts"]) {
    print(shirt["name"]);
  }
  print("\n--COOL JEANS--");
  for (var jeans in data["Catalog"]["Jeans"]) {
    print(jeans["name"]);
  }
}

输出：

--COOL SHIRTS--
Demin Shirt
Random Shirt

--COOL JEANS--
Bare Denim Jeans
Levis jeans