Question

以下是我从API获得的回复。

[{
        "claim1": [{
            "claim_no": "1",
            "claim_date": "14\/06\/2017 12:00:00 AM",
            "month_code": "1",
            "claim_amount": "8074"
        }]
    }, {
        "billing1": [{
            "claim_no": "141",
            "month_code": "1",
            "miti": "6",
            "amount": "7374",
            "paid": "7374",
            "fee1": "0"
        }]
    }, {
        "claim2": [{
            "claim_no": "26",
            "claim_date": "20\/06\/2017 12:00:00 AM",
            "month_code": "2",
            "claim_amount": "2424"
        }]
    }, {
        "billing2": []
    }, {
        "claim3": [{
            "claim_no": "40",
            "claim_date": "13\/07\/2017 12:00:00 AM",
            "month_code": "3",
            "claim_amount": "2924"
        }]
    }, {
        "billing3": [{
            "claim_no": "724",
            "month_code": "3",
            "miti": "13",
            "amount": "2924",
            "paid": "2924",
            "fee1": "0"
        }]
    }, {
        "claim4": [{
            "claim_no": "43",
            "claim_date": "09\/08\/2017 12:00:00 AM",
            "month_code": "4",
            "claim_amount": "2424"
        }]
    }, {
        "billing4": [{
            "claim_no": "402",
            "month_code": "4",
            "miti": "20",
            "amount": "5348",
            "paid": "3124",
            "fee1": "0"
        }]
    }]

我正在使用aQuery jsonarray类我试图在JSON对象中获取它，但是出现了类似

的错误

jsonarray无法转换为jsonobject

下面是我的代码此代码显示此错误

“org.json.JSONException：claim_no没有值”

我想获得claim_no

的值

aq.ajax("myapiurlhere", JSONArray.class, new AjaxCallback<JSONArray>(){
            @Override
            public void callback(String url, JSONArray array, AjaxStatus status) {
                super.callback(url, array, status);
                Log.i("response:", "response:" + url + array);

                for (int i = 0; i < array.length(); i++) {

                    try {

                        JSONObject object = array.getJSONObject(i);

                        bill_no = object.getString("claim_no");


                        Log.i("response:","claim: "+claim_no+" bill: "+object);
                    } catch (JSONException e) {
                        e.printStackTrace();
                    }


                }
            }
        });

我也试过下面的代码，但是我收到了错误

“org.json.JSONArray类型的claim1无法转换为JSONObject”

 aq.ajax("myapiurl", JSONArray.class, new AjaxCallback<JSONArray>(){
            @Override
            public void callback(String url, JSONArray array, AjaxStatus status) {
                super.callback(url, array, status);
                Log.i("response:", "response:" + url + array);

                for (int i = 0; i < array.length(); i++) {

                    try {
int mycount = i+1;
                        JSONObject object = array.getJSONObject(i);
                        JSONObject claimobject = object.getJSONObject("claim"+mycount);
                        claim_no = claimobject.getString("claim_no");


                        Log.i("response:","claim: "+claim_no);
                    } catch (JSONException e) {
                        e.printStackTrace();
                    }


                }
            }
        });

Answer 1

这是一个数组

"claim1": [

您正在获取对象

JSONObject claimobject = object.getJSONObject("claim"+mycount);

你需要先获得一个数组

 JSONObject claimobject = object.getJSONArray("claim"+mycount).getJSONObject(0);

此外，您的API结构不合理，int mycount = i+1;可能会因为您同时计算结算和声明元素而失败

试试这个

for (int i = 0; i < array.length(); i++) {
    int mycount = (i/2)+1;
    try {

获取编号的json键值

1 个答案: