Question

我有以下JSON响应，希望转换为Java，然后将数据保存到数据库。

我查看了各种工具，但无法找到合适的解决方案。

我做错了什么但却无法理解差距在哪里。

以下是我的JSON：

{
    "release-1.0": [{
        "id": 55,
        "resourceId": "126",
        "allGraphs": null,
        "isChecked": true
    }, {
        "id": 56,
        "resourceId": "125",
        "allGraphs": null,
        "isChecked": true
    }, {
        "id": 58,
        "resourceId": "140",
        "allGraphs": null,
        "isChecked": true
    }]
}

这是我的Java类映射到JSON之上。

@DatabaseTable(tableName = "test_group")
public class TestGroup {
    private List<TestGroup> testGroup;

    public TestGroup() {
        // ORMLite needs a no-arg constructor
    }

    @DatabaseField
    private List<String> test_group_id;

    @DatabaseField
    private String id;

    @DatabaseField
    private String test_details;

    @DatabaseField
    private String graph_id;

    public void setTestGroupID(List<String> testGroupId) {
        this.test_group_id = testGroupId;
    }

    public void setId(String id) {
        this.id = id;
    }

    public void testDetails(String testDetails) {
        this.test_details = testDetails;
    }

    public void setGraphId(String allGraphs) {
        this.graph_id = allGraphs;
    }

    public List<TestGroup> getAllGraphs() {
        return testGroup;
    }
}

我使用了杰克逊，但收到如下错误：

com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "release-1.0" (class org.example.model.TestGroup), not marked as ignorable (4 known properties: "testGroupID", "graphId", "id", "allGraphs"])
 at [Source: {"release-1.0":[{"id":55,"resourceId":"126","allGraphs":null,"isChecked":true},{"id":56,"resourceId":"125","allGraphs":null,"isChecked":true},{"id":58,"resourceId":"140","allGraphs":null,"isChecked":true}]}; line: 1, column: 17]

请帮忙。

提前致谢。

Answer 1

您需要一个周围的容器类来将字段“release-1.0”映射为List。因为json表达式："testcases": [引用了一个列表。

 // will map to the new field release by name
 private String release;

 // Or mapped by named property
 @JsonProperty("testcases")
 private List<TestGroup> release10 = new ArrayList<TestGroup>();

创建一个包含该字段的类，jackson将一个TestGroups列表绑定到它。

Answer 2

如错误所示，"release-1.0"中的JSON字段为Unrecognized - 这意味着您TestGroup中没有包含该名称的字段类。

JSON个文件必须与Class个数据成员匹配：

[
    {
        "id": 55,
        "resourceId": "126",
        "allGraphs": null,
        "isChecked": true
    }, {
        "id": 56,
        "resourceId": "125",
        "allGraphs": null,
        "isChecked": true
    }, {
        "id": 58,
        "resourceId": "140",
        "allGraphs": null,
        "isChecked": true
    }]
]

如果List<TestGroup>是TestGroup，则匹配为@DatabaseTable(tableName = "test_group") public class TestGroup { public TestGroup() { // ORMLite needs a no-arg constructor } @DatabaseField private String id; @DatabaseField private String resourceId; @DatabaseField private String allGraphs; @DatabaseField private Bollean isChecked; // Getters and setters - preferably auto-generated since NAMES MUST MATCH. }：

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使用DAO进行JSON到Java的转换和映射

2 个答案: