属性文件转换为json,Java代码
//我的属性文件:属性文件转换为json,Java代码,java,json,properties-file,Java,Json,Properties File,//我的属性文件: key1 = value1 key2 = value2,value3 key3 = value4, value5, value6 Requirement in json "Arguments": { "key1":"value1", "key2":["value2","value3"] "key3":["value4","value5","value6"] } //请提供java代码将您
key1 = value1
key2 = value2,value3
key3 = value4, value5, value6
Requirement in json
"Arguments":
{
"key1":"value1",
"key2":["value2","value3"]
"key3":["value4","value5","value6"]
}
//请提供java代码将您的“参数”放入一个类中,我将其称为参数
。然后使用将该类序列化为JSON
例如:
Gson gson = new GsonBuilder().setPrettyPrinting().create();
String response = gson.toJson(argumentInstance);
Files.write(path, Arrays.asList(response.split("\n")));
然后,您可以像这样将其读回,提供类:
Gson gson = new Gson();
argumentInstance = gson.fromJson(Files.newBufferedReader(path), Argument);
当然。你还想和代码一起喝咖啡吗?首先你需要提到你试图解决的问题我该如何问一个好问题?问题是如何将一个键的多个值放入json对象。如上面示例中的键2和键3。我们可以做单值。谢谢。有帮助