Dart 如何判断字符串是否可以是json.decode
我的缓存类Dart 如何判断字符串是否可以是json.decode,dart,flutter,Dart,Flutter,我的缓存类 import 'dart:async'; import 'dart:convert'; import 'package:shared_preferences/shared_preferences.dart'; class CacheUtil{ static set(String key, value) async{ if(value is Map || value is List){ value = json.encode(value); }
import 'dart:async';
import 'dart:convert';
import 'package:shared_preferences/shared_preferences.dart';
class CacheUtil{
static set(String key, value) async{
if(value is Map || value is List){
value = json.encode(value);
}
SharedPreferences preferences = await SharedPreferences.getInstance();
preferences.setString(key, json.encode(value));
}
static get(String key) async{
SharedPreferences preferences = await SharedPreferences.getInstance();
String data = preferences.getString(key);
return data;
}
}
在get方法中,我想看看值是否可以是json.decode
我该怎么办?只要试着解码它并捕获
FormatException
就可以知道它什么时候失败了:
void main() {
var jsonString = '{"abc';
var decodeSucceeded = false;
try {
var decodedJSON = json.decode(jsonString) as Map<String, dynamic>;
decodeSucceeded = true;
} on FormatException catch (e) {
print('The provided string is not valid JSON');
}
print('Decoding succeeded: $decodeSucceeded');
}
void main(){
var jsonString='{“abc';
var=false;
试一试{
var decodedJSON=json.decode(jsonString)作为映射;
解码成功=真;
}关于格式化异常捕获(e){
print('提供的字符串不是有效的JSON');
}
打印('Decoding successed:$decodecoccessed');
}