java – Gson Json解析器数组数组

发布时间：2020-12-14 05:38:16 所属栏目：Java 来源：网络整理

导读：希望解析一些Json并解析数组数组.不幸的是我无法弄清楚如何处理json中的嵌套数组. JSON { "type": "MultiPolygon","coordinates": [ [ [ [ -71.25,42.33 ],[ -71.25,42.33 ] ] ],[ [ [ -71.23,[ -71.23,42.33 ] ] ] ]} 当我只是一个阵列时,我实现了什么. pub

希望解析一些Json并解析数组数组.不幸的是我无法弄清楚如何处理json中的嵌套数组.

JSON

{
    "type": "MultiPolygon","coordinates": [
        [
            [
                [
                    -71.25,42.33
                ],[
                    -71.25,42.33
                ]
            ]
        ],[
            [
                [
                    -71.23,[
                    -71.23,42.33
                ]
            ]
        ]
    ]
}

当我只是一个阵列时,我实现了什么.

public class JsonObjectBreakDown {
    public String type; 
    public List<List<String[]>> coordinates = new ArrayList<>();
    public void setCoordinates(List<List<String[]>> coordinates) {
        this.coordinates = coordinates;
    }




}

解析呼叫

JsonObjectBreakDown p = gson.fromJson(withDup,JsonObjectBreakDown.class);

解决方法

你有一组数组的字符串数组数组.你需要

public List<List<List<String[]>>> coordinates = new ArrayList<>();

下列

public static void main(String args[]) {
    Gson gson = new Gson();
    String jsonstr ="{  "type": "MultiPolygon","coordinates": [        [            [                [                    -71.25,42.33                ],[                    -71.25,42.33                ]            ]        ],[            [                [                    -71.23,[                    -71.23,42.33                ]            ]        ]    ]}";
    JsonObjectBreakDown obj = gson.fromJson(jsonstr,JsonObjectBreakDown.class);

    System.out.println(Arrays.toString(obj.coordinates.get(0).get(0).get(0)));
}

public static class JsonObjectBreakDown {
    public String type; 
    public List<List<List<String[]>>> coordinates = new ArrayList<>();
    public void setCoordinates(List<List<List<String[]>>> coordinates) {
        this.coordinates = coordinates;
    }
}

版画

[-71.25,42.33]

（编辑：李大同）

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