java gson在序列化时替换密码值 [英] java gson replace password value while serialization

问题描述

在使用Gson反序列化对象时，如何用 XXX 替换密码字段的值？我发现这篇文章： Gson：如何排除特定字段从序列化没有注释，基本上跳过这个领域。这将是一个选择，但我仍然希望用 XXX

来替换值。我也试过这个：

  GsonBuilder builder = new GsonBuilder（）。setPrettyPrinting（）; 
 builder.registerTypeAdapter（String.class，new JsonSerializer< String>（）{
 
 @Override public JsonElement serialize（String value，Type arg1，JsonSerializationContext arg2）{
 //找不到确定字段名称的方法
 return new JsonPrimitive（value）; 
} 
}）; 
  

不幸的是，我无法确定该字段的名称。那么还有其他选择吗？

我使用Gson以漂亮的方式记录一些对象，所以在阅读日志时我不需要格式化。

解决方案

发布此答案时，我感到非常跛脚。但是，这是你可以做到的，它基本上是在序列化之前复制和更改Java对象。

pre $ public class User {
private static final Gson gson = new Gson（）;
公共字符串名称;
公共字符串密码;
$ b $ public public User（String name，String pwd）{
this.name = name;
this.password = pwd;

$ b @Override
保护对象clone（）抛出CloneNotSupportedException {
返回新用户（this.name，this.password）;
}

public static void main（String [] aa）{
JsonSerializer< User> ser = new JsonSerializer< User>（）{
@Override
public JsonElement serialize（User u，Type t，JsonSerializationContext ctx）{
try {
User clone = u.clone（）;
clone.password = clone.password.replaceAll（。，x）;
return（gson.toJsonTree（clone，User.class））;
} catch（CloneNotSupportedException e）{
//如果你不克隆克隆，请做些什么。
}
返回gson.toJsonTree（u，User.class）;
}
};
Gson g = new GsonBuilder（）。registerTypeAdapter（User.class，ser）.create（）;
System.out.println（g.toJson（new User（naishe，S3cr37）））;

获得序列化为：

  {name：naishe，password：xxxxxx} 
  

How can I replace the value of a password field with XXX while de-serializing an object with Gson? I found this post: Gson: How to exclude specific fields from Serialization without annotations that basically skips the field. This would be an option, but I still would prefer to replace the value with XXX

I also tried this:

GsonBuilder builder = new GsonBuilder().setPrettyPrinting();
builder.registerTypeAdapter(String.class, new JsonSerializer<String>(){

  @Override public JsonElement serialize(String value, Type arg1, JsonSerializationContext arg2){
        // could not find a way to determine the field name     
        return new JsonPrimitive(value);
  }
});

Unfortunately, I wasn't able to determine the name of the field. So is there any other option?

I use Gson to log some objects the "pretty" way, so I don't need to bother with the formatting while reading the logs.

解决方案

I feel pretty lame while posting this answer. But, it's what you can, it essentially copies and changes the Java object, before serializing.

public class User {
    private static final Gson gson = new Gson();
    public String name;
    public String password;

    public User(String name, String pwd){
        this.name = name;
        this.password = pwd;
    }

    @Override
    protected Object clone() throws CloneNotSupportedException {
        return new User(this.name, this.password);
    }

    public static void main(String[] aa){
        JsonSerializer<User> ser = new JsonSerializer<User>() {
            @Override
            public JsonElement serialize(User u, Type t, JsonSerializationContext ctx) {
                try {
                    User clone = (User)u.clone();
                    clone.password = clone.password.replaceAll(".","x");
                    return (gson.toJsonTree(clone, User.class));
                } catch (CloneNotSupportedException e) {
                    //do something if you dont liek clone.
                }
                return gson.toJsonTree(u, User.class);
            }
        };
        Gson g = new GsonBuilder().registerTypeAdapter(User.class, ser).create();
        System.out.println(g.toJson(new User("naishe", "S3cr37")));
    }
}

Gets serialized to:

{"name":"naishe","password":"xxxxxx"}

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