从 java 中的 POST 读取 json 文件

Question

How can I unmarhsaller json file like this?

{
    "packageId": "11",
    "jsScript": "var divideFn = function(a,b) { return a/b} ",
    "functionName": "divideFn",
    "tests": [
        {
            "testName": "test1",
            "expectedResult": "2.0",
            "params": [
                2,
                1
            ]
        }
    ]
}

I have a class that works well with packageId, jsScrript, functionName, but not with the tests

public class Data {
    private final int packageId;
    private final String jsScript;
    private final String functionName;
    private final List<Tests> tests;

    @JsonCreator
    public Data(@JsonProperty("packageId") String packageId,
               @JsonProperty("jsScript") String jsScript,
               @JsonProperty("functionName") String functionName,
               @JsonProperty("tests") List<Tests> tests) {
        this.packageId = Integer.parseInt(packageId);
        this.jsScript= jsScript;
        this.functionName = functionName;
        this.tests = tests;
    }
    }

    public class Tests{
    public final String testName;
    public final int expectedResult;

    @JsonCreator
    public Tests(@JsonProperty("testName") String testName,
                 @JsonProperty("expectedResult") String expectedResult){
        this.testName= testName;
        this.expectedResult = Integer.parseInt(expectedResult);
    }
}

What should I change in classes to make it work well? I also tried to read tests like String, but it didn't help

Answer 1

It seems that you have encountered some problems for deserializing the JSON array tests to objects. There are several methods to solve this.

Method 1
Add @JsonIgnoreProperties(ignoreUnknown = true) on your class Tests to prevent your code from the exception of Unrecognized field "params" .

Method 2
Deserialize JSON array tests to List<JsonNode> if it is not important and you won't further parse it in the future.

Method 3
Use follwing class for mapping JSON array tests to List<Test> .

class Test {
    private String testName;
    private Float expectedResult;
    private List<Integer> params;

    //general getters ans setters
}

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BTW, I don't think you need to use @JsonCreator for deserialization.