Vin*_*oth 41 java android json
我对作为服务器响应的JSONObject进行了一些修复.
jsonObj = new JSONObject(resultString);
JSONObject sync_reponse = jsonObj.getJSONObject("syncresponse");
String synckey_string = sync_reponse.getString("synckey");
JSONArray createdtrs_array = sync_reponse.getJSONArray("createdtrs");
JSONArray modtrs_array = sync_reponse.getJSONArray("modtrs");
JSONArray deletedtrs_array = sync_reponse.getJSONArray("deletedtrs");
String deleted_string = deletedtrs_array.toString();
{"syncresponse":{"synckey":"2011-09-30 14:52:00","createdtrs":[],"modtrs":[],"deletedtrs":[{"companyid":"UTB17","username":"DA","date":"2011-09-26","reportid":"31341"}]
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正如你在我得到的响应中看到的那样,我正在解析JSONObject并创建syncresponse,synckey作为JSON对象 createdtrs,modtrs,deletedtrs作为JSONArray.我想从deletedtrs访问JSONObject,以便我可以拆分它们并使用这些值.即我想提取companyid,用户名,日期等.
我怎么能这样做?
感谢您的输入.
Lar*_*ars 118
JSONArray对象有一个函数getJSONObject(int index),你可以通过编写一个简单的for循环遍历所有的JSONObjects:
JSONArray array;
for(int n = 0; n < array.length(); n++)
{
JSONObject object = array.getJSONObject(n);
// do some stuff....
}
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Vin*_*kla 20
这是你的json:
{
"syncresponse": {
"synckey": "2011-09-30 14:52:00",
"createdtrs": [
],
"modtrs": [
],
"deletedtrs": [
{
"companyid": "UTB17",
"username": "DA",
"date": "2011-09-26",
"reportid": "31341"
}
]
}
}
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它正在解析:
JSONObject object = new JSONObject(result);
String syncresponse = object.getString("syncresponse");
JSONObject object2 = new JSONObject(syncresponse);
String synckey = object2.getString("synckey");
JSONArray jArray1 = object2.getJSONArray("createdtrs");
JSONArray jArray2 = object2.getJSONArray("modtrs");
JSONArray jArray3 = object2.getJSONArray("deletedtrs");
for(int i = 0; i < jArray3 .length(); i++)
{
JSONObject object3 = jArray3.getJSONObject(i);
String comp_id = object3.getString("companyid");
String username = object3.getString("username");
String date = object3.getString("date");
String report_id = object3.getString("reportid");
}
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