Question

我有一个JSON形式的联系人。现在我想将它们解码为String数组。有两个数组; names和phones。我使用此代码：

    String[] names;
    String[] phones;

    String test = "[{\"name\":\"A\",\"phone\":\"911\"},{\"name\":\"A1\",\"phone\":\"911\"},{\"name\":\"Abid\",\"phone\":\"371812\"}]";
    try {
        JSONArray jsonArray = new JSONArray(test);
        JSONObject jsonObject = new JSONObject(jsonArray.toString());
        Log.i("INFO", String.valueOf(jsonObject.length()));
    } catch (JSONException e) {
        e.printStackTrace();
    }

这会产生错误。如何在names阵列和phones阵列中的所有手机中添加所有名称。与names[0]一样，A分配了名字，phones[0]分配了911，这是与名字对应的第一个电话号码。我怎么能这样做，我是android的新手？

Answer 1

问题出在这一行

JSONObject jsonObject = new JSONObject(jsonArray.toString());

您正在尝试转换JSONArray中的JSONObject，但您无法转换JSONObject。如果你想访问JSONArray内的String[] names; String[] phones; String test = "[{\"name\":\"A\",\"phone\":\"911\"},{\"name\":\"A1\",\"phone\":\"911\"},{\"name\":\"Abid\",\"phone\":\"371812\"}]"; try { JSONArray jsonArray = new JSONArray(test); phones = names = new String[jsonArray.length()]; for(int i = 0 ; i < jsonArray.length(); i ++){ JSONObject jsonObject = jsonArray.getJSONObject(i); names[i] = jsonObject.getString("name"); phones[i] = jsonObject.getString("phone"); Log.i("INFO", "name : " + jsonObject.getString("name") + " , phone : " + jsonObject.getString("phone")); } } catch (JSONException e) { e.printStackTrace(); }，你必须循环遍历每个，或者你可以通过它的索引获得特定的对象。

public IEnumerable<voucher> index(string word, string DivisionCode, string yearlabel, int pageno)
        {
            using (IDbConnection dbConnection = Connection)
            {
                var data = dbConnection.Query<fydates>(@"SELECT REPLACE(CONVERT(NVARCHAR(10),fystart,6),' ','-') as fystart,
                REPLACE(CONVERT(NVARCHAR(10),fyend,6),' ','-') as fyend FROM fydates WHERE yearlabel = @yearlabel",
                new { yearlabel = yearlabel }).Single();

                List<voucher> DataQuery = dbConnection.Query<voucher, accounts, division, LockStatus, voucher>(@"SELECT V.id,V.voucherno,
                CASE WHEN V.vouchertype='P' THEN 'Paid' ELSE 'Receipt' END AS vouchertype,
                REPLACE(CONVERT(NVARCHAR(10),V.dated,6),' ','-') AS dated,V.subno,V.amount,COALESCE(V.syscreated,' ') AS syscreated,V.bankcode,
                COALESCE(V.cheqno,' ') AS Cheqno,V.Cheqdate,V.posted,V.currency,A.id as accountsid,A.achead,A.shortname,
                D.id as divisionid,D.divname,D.divncode,L.id as lockstatusid,L.name FROM voucher V
                LEFT JOIN Accounts A ON V.bankcode = A.mainac
                LEFT JOIN divisionmst D ON V.divisionid = D.divncode
                LEFT JOIN LockStatus L ON V.posted=L.Type
                WHERE (V.voucherno + V.Subno+V.cheqno + A.achead + LEFT(yearvoucherno,3) LIKE @word AND
                D.divncode LIKE @DivisionCode) AND (dated BETWEEN @fystart AND @fyend OR dated='31-Mar-49')
                Order By V.id DESC OFFSET @pageno *25 ROWS FETCH NEXT 25 ROWS only",
                (voucher, accounts, division, LockStatus) =>
                {
                    voucher.accounts = accounts;
                    voucher.division = division;
                    voucher.LockStatus = LockStatus;
                    return voucher;
                },
                new { word = word, fystart = data.fystart, fyend = data.fyend, DivisionCode = DivisionCode, pageno = pageno - 1 },
                splitOn: "accountsid, divisionid, lockstatusid").ToList();
                return DataQuery;
            }
        }

Answer 2

您无法将json arrya转换为json对象：

try {
        JSONArray jsonArray = new JSONArray(test);
        for(int i=0; i<jsonArray.length(); i++) {
           JSONObject jsonObject = jsonArray.getJSONObject(i);
           // Your code goes here..  
        }

    } catch (JSONException e) {
        e.printStackTrace();
    }

Answer 3

这是必需的代码。

 ArrayList<String> names = new ArrayList<>();
    ArrayList<String> phones = new ArrayList<>();

    String test = "[{\"name\":\"A\",\"phone\":\"911\"},{\"name\":\"A1\",\"phone\":\"911\"},{\"name\":\"Abid\",\"phone\":\"371812\"}]";
    try {
        JSONArray jsonArray = new JSONArray(test);
        for (int i = 0; i < jsonArray.length(); i++) {
            JSONObject jsonObject = jsonArray.getJSONObject(i);
            String name = jsonObject.getString("name");
            String phone = jsonObject.getString("phone");
            names.add(name);
            phones.add(phone);
            Log.i("INFO", name + ", " + phone);
        }
    } catch (JSONException e) {
        e.printStackTrace();
    }

上面你使用了简单的数组，它们只能存储固定大小的数据。但是arrayList和List可以存储可变大小的数据，或者你不需要修改它们的大小。

解码Android中的JSON字符串

3 个答案: