java Jackson 解析器 json setter 将值作为字符串数组
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Hymanson parser json setter to take value as string array
提问by javaMan
I have below json:
我有以下json：
"[{\"movieName\":\"A\",\"Leadactor\":\"\",\"leadActress\":\"\",\"movieTitle\":\"\",\"hero\":\"\",\"heroine\":\"\",\"source\":\"IMDB\"}," +
"{\"movieName\":\"\",\"Leadactor\":\"\",\"leadActress\":\"\",\"movieTitle\":\"B\",\"hero\":\"B1\",\"heroine\":\"B2\",\"source\":\"Netflix\"}," +
"{\"movieName\":\"C\",\"Leadactor\":\"C1\",\"leadActress\":\"C2\",\"movieTitle\":\"\",\"hero\":\"\",\"heroine\":\"\",\"source\":\"IMDB\"}," +
"{\"movieName\":\"D\",\"Leadactor\":\"D1\",\"leadActress\":\"D2\",\"movieTitle\":\"\",\"hero\":\"\",\"heroine\":\"\",\"source\":\"IMDB\"}," +
"{\"movieName\":\"\",\"Leadactor\":\"\",\"leadActress\":\"\",\"movieTitle\":\"E\",\"hero\":\"E1\",\"heroine\":\"E2\",\"source\":\"Netflix\"}]";
I am using Hymanson parser to map it to a class:
我正在使用 Hymanson 解析器将它映射到一个类：
I want movieName and movieTitle to map into Name property in the java class. So i wrote the below class:
我希望 movieName 和 movieTitle 映射到 java 类中的 Name 属性。所以我写了下面的类：
public static class MovieData {
@JsonProperty("Name")
private String name;
@JsonSetter({"movieName"})
private void setMovieName(final String name) {
if((name != null) && (! name.equals(""))) {
setNameInternal(name);
}
}
@JsonSetter("movieTitle")
private void setMovieTitle(final String name) {
if((name != null) && (! name.equals(""))) {
setNameInternal(name);
}
}
private void setNameInternal(final String name) {
this.name = name;
}
}
}
In my real json there are so many fields like movieName, movieTitle which i want to normalize into a common name.
在我真正的 json 中有很多字段，如 movieName、movieTitle，我想将它们规范化为一个通用名称。
Is there any simple syntax like the below which can reduce code duplication:
有没有像下面这样的简单语法可以减少代码重复：
public static class MovieData {
@JsonProperty("Name")
private String name;
@JsonSetter(value = { "movieName", "movieTitle" })
private void setName(final String name) {
if((name != null) && (! name.equals(""))) {
this.name=name;
}
}
}
The above code gave me error on jsonSetter:
上面的代码在 jsonSetter 上给了我错误：
Type mismatch: cannot convert from String[] to String.
EDIT
编辑
If Hymanson doesn't support it, can GSON Support this operation.
如果 Hymanson 不支持，GSON 可以支持这个操作。
Thanks
谢谢
采纳答案by Micha? Ziober
You can use @JsonAnySetter, what it is mean you can find on Hymanson Core (Data-Binding) Annotationspage.
您可以使用@JsonAnySetter，这意味着您可以在Hymanson Core (Data-Binding) Annotations页面上找到。
I have created simple bean which is related to your example:
我创建了与您的示例相关的简单 bean：
class MovieData {
private static List<String> NAME_PROPERTIES = Arrays.asList("movieName", "movieTitle");
private String name;
public void setName(String name) {
this.name = name;
}
@JsonAnySetter
private void parseUnknownProperties(String propertyName, String propertyValue) {
if (NAME_PROPERTIES.contains(propertyName) && !propertyValue.isEmpty()) {
this.name = propertyValue;
}
}
@Override
public String toString() {
return name;
}
}
Now when I deserialize your JSON in this way:
现在，当我以这种方式反序列化您的 JSON 时：
ObjectMapper objectMapper = new ObjectMapper();
System.out.println(Arrays.toString(objectMapper.readValue(json, MovieData[].class)));
As result I can see:
结果我可以看到：
[A, B, C, D, E]
回答by Ingreatway
Dont do this much. it is very simple with Gson
不要做这么多。使用 Gson 非常简单
create class for your single set record like
为您的单组记录创建类
class Movie{
private String movieName;
private String Leadactor;
private String leadActress;
//put getter and setter for your fields
}
in main file
Type type = new TypeToken<List<Movie>>(){}.getType();
List<Movie> data = new Gson().fromJson(json_string,type);
回答by Eugene Petrenko
You declare the field like @JsonProperty String[] data
你像这样声明这个领域 @JsonProperty String[] data
and use
并使用
new ObjectMapper()
.enable(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY)
.readValue(json, YOUT_TYPE.class)
So this option make Hymanson part both "foo":"bar" and "foo": ["a", "b", "c"].
所以这个选项让Hyman逊同时成为“foo”：“bar”和“foo”：[“a”，“b”，“c”]。
All you need to do is to merge string array back into the string with right line separators. I do it via Guava data == null ? null : Jointer.on("\n").join(data)
您需要做的就是使用正确的行分隔符将字符串数组合并回字符串。我是通过番石榴做的data == null ? null : Jointer.on("\n").join(data)
