试试这个：

return json.decode(prefs.getString(key) ?? '');
// or
return prefs.getString(key) != null ? json.decode(prefs.getString(key)) : '';

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行prefs.getString(key)返回一个String?。这意味着可以返回一个String或null值。必须确保传递给json.decode()时该值不会为空。
查看documentation以了解更多信息。

更新代码

read(String key) async {
    final prefs = await SharedPreferences.getInstance();
    String? str = prefs.getString(key) ?? '';
    return json.decode(str); 
}

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这是因为getString方法的返回类型为String?

/// Reads a value from persistent storage, throwing an exception if it's not a
  /// String.
  String? getString(String key) => _preferenceCache[key] as String?;

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但json.decode以String为参数

/// Parses the string and returns the resulting Json object.
  ///
  /// The optional [reviver] function is called once for each object or list
  /// property that has been parsed during decoding. The `key` argument is either
  /// the integer list index for a list property, the string map key for object
  /// properties, or `null` for the final result.
  ///
  /// The default [reviver] (when not provided) is the identity function.
  dynamic decode(String source,
      {Object? reviver(Object? key, Object? value)?}) {
    reviver ??= _reviver;
    if (reviver == null) return decoder.convert(source);
    return JsonDecoder(reviver).convert(source);
  }

型
因此，必须将String传入json.decode方法。有很多方法可以做到这一点

// prefs.getString(key) return null, use '' as parameter
return json.decode(prefs.getString(key) ?? ''); 
// or
// throw exception when prefs.getString(key) return null. 
// Wrap your method with `try ... catch ... `
return json.decode(prefs.getString(key)!);

型