如何编码Map<int, List<int>>类型映射？解决json.encode编码失败问题

Hey there! I see you're hitting a snag when trying to encode your Map<int, List<int>> with json.encode()—that "Converting object to an encodable object failed: _LinkedHashMap" error makes total sense once you know the root cause.

Why This Happens #

JSON’s core specification only allows strings as object keys. Dart’s default json.encode() doesn’t automatically convert integer keys to strings, so it can’t process your Map<int, ...> directly, hence the encoding failure.

Solutions to Encode Your blocksMap #

1. Convert Integer Keys to Strings (Simplest Approach)

Create a new map where all integer keys are converted to strings before encoding—this plays nice with JSON’s requirements:

import 'dart:convert';

void main() {
  Map<int, List<int>> blocksMap = {
    0: [18, 117, 26, 200, 4, 30, 43, 110],
    1: [18, 117, 26, 200, 4, 30, 43, 110],
    // ... your remaining entries
  };

  // Convert int keys to strings
  Map<String, List<int>> stringKeyMap = blocksMap.map((key, value) => MapEntry(key.toString(), value));
  
  // Now encode successfully
  String jsonString = json.encode(stringKeyMap);
  print(jsonString);
}

2. Restore Integer Keys When Decoding

If you need to get back the original Map<int, List<int>> after decoding, parse the string keys back to integers:

// Decode the JSON string first
Map<String, dynamic> decodedMap = json.decode(jsonString);

// Convert string keys back to int and cast lists to List<int>
Map<int, List<int>> restoredBlocksMap = decodedMap.map(
  (key, value) => MapEntry(int.parse(key), List<int>.from(value))
);

3. Custom Encoder (Advanced, Optional)

If you want to skip creating an intermediate map, you can write a tiny custom encoder—though this is overkill for most use cases:

String customEncodeMap(Map<int, List<int>> map) {
  return json.encode({
    for (var entry in map.entries) entry.key.toString(): entry.value
  });
}

// Usage
String jsonString = customEncodeMap(blocksMap);

Quick Tips #

• Double-check that your integer keys can safely convert to/from strings (no invalid numeric values when decoding).
• Using List<int>.from(value) during decoding ensures you get a strongly-typed integer list instead of a dynamic one.

内容的提问来源于stack exchange，提问作者Titas Černiauskas