Question

我正在尝试解析一个齐射json响应，其中第一个数组Sub1在尝试解析第二个Sub2时工作正常，Sub2返回异常而没有Sub2的值。

[{
"Id": 1,
"Title1": "Title1",
"Title2": "title2",
"Title3": "title3",
"Link1": "link",
"Link2": "link",
"Link3": "link",
"Ord": 1,
"Sub1": [{
    "Id": 58,
    "Title1": "Title1",
    "Title2": "Title2",
    "Title3": "Title3",
    "Link1": "link",
    "Link2": "link",
    "Link3": "link",
    "Ord": 1,
    "Sub2": [{
        "Id": 69,
        "Title1": "Title1",
        "Title2": "Title2",
        "Title3": "Title3",
        "Link1": "link",
        "Link2": "link",
        "Link3": "link",
        "Ord": 1
    }]
}]
}]

这是代码的解析部分：

for (int i = 0; i < response.length(); i++) {
                JSONObject person = (JSONObject) response
                        .get(i);
                id = person.getString("Id");
                titleel = person.getString("Title1");
                linkel = person.getString("Link1");
                ordinal = person.getString("Ord");
                JSONArray sub = person.getJSONArray("Sub1");
                for(int j =0; j<sub.length(); j++){
                    JSONObject subobj = sub.getJSONObject(j);
                    subtitle = subobj.getString("Title1");
                    subtitleen = subobj.getString("Title2");
                    subtitlede = subobj.getString("Title3");
                    subid = subobj.getString("Id");
                    sublinkel = subobj.getString("Link1");
                    sublinken = subobj.getString("Link2");
                    sublinkde = subobj.getString("Link3");
                    subordinal =subobj.getString("Ord");

                    JSONArray suba = subobj.getJSONArray("Sub2");
                    for(int k =0; k<suba.length(); k++) {
                        JSONObject subobja = suba.getJSONObject(k);
                        subtitlea = subobja.getString("Title1");
                        subtitleena = subobja.getString("Title2");
                        subtitledea = subobja.getString("Title3");
                        subida = subobja.getString("Id");
                        sublinkela = subobja.getString("Link1");
                        sublinkena = subobja.getString("Link2");
                        sublinkdea = subobja.getString("Link3");
                        subordinala = subobja.getString("Ord");
                    }
                    }
            }

Answer 1

好了，经过两天的搜索，我终于找到了问题所在。就像Raghunandan说我替换了每个get方法的选择。

 for (int i = 0; i < response.length(); i++) {
            JSONObject person = (JSONObject) response
                    .get(i);
            id = person.optString("Id");
            titleel = person.optString("Title1");
            linkel = person.optString("Link1");
            ordinal = person.optString("Ord");
            JSONArray sub = person.optJSONArray("Sub1");
            for(int j =0; j<sub.length(); j++){
                JSONObject subobj = sub.optJSONObject(j);
                subtitle = subobj.optString("Title1");
                subtitleen = subobj.optString("Title2");
                subtitlede = subobj.optString("Title3");
                subid = subobj.optString("Id");
                sublinkel = subobj.optString("Link1");
                sublinken = subobj.optString("Link2");
                sublinkde = subobj.optString("Link3");
                subordinal =subobj.optString("Ord");

                JSONArray suba = subobj.optJSONArray("Sub2");
                for(int k =0; k<suba.length(); k++) {
                    JSONObject subobja = suba.optJSONObject(k);
                    subtitlea = subobja.optString("Title1");
                    subtitleena = subobja.optString("Title2");
                    subtitledea = subobja.optString("Title3");
                    subida = subobja.optString("Id");
                    sublinkela = subobja.optString("Link1");
                    sublinkena = subobja.optString("Link2");
                    sublinkdea = subobja.optString("Link3");
                    subordinala = subobja.optString("Ord");
                }
                }
        }

用嵌套的jsonarrays解析json响应

1 个答案: