Android的JSON结果处理 [英] Android Json result processing

问题描述

我有JSON结果是这样的：array.getJSONObject（J） -

I have Json result like this: array.getJSONObject(j) --

{执行code：WAT2，FreeIssue preFIX：结果
  \",\"DisPaySchedule$p$pfix\":\"\",\"NextFreeIssueNo\":\"1\",\"NextReturnNo\":\"20\",\"UploadedType\":\"1\",\"DisNextFreeIssueNo\":\"1\",\"DisNextFOCNo\":\"1\",\"NextVisitNo\":\"15\",\"DisNextOrderNo\":\"1\",\"UploadedOn\":\"Jun
  17 2011
  下午6点33分，NextReturnAcceptNo：1，BusinessUnit：HEMA，TXNReference preFIX：20110708
  ，订单preFIX：OR 4，UploadedMethod：3，FOC preFIX：搜索
  ，返回preFIX：RT4
  \",\"Retailer$p$pfix\":\"TEM4\",\"NextRetailerNo\":\"10\",\"NextInvoiceNo\":\"1\",\"NextGRNNo\":\"1\",\"Invoice$p$pfix\":\"IN4
  ，NextTXNReference：2，NextOrderNo：37，ReturnAccept preFIX：搜索
  \",\"PaySchedule$p$pfix\":\"PS4\",\"NextReceiptNo\":\"1\",\"NextFOCNo\":\"20\",\"NextPayScheduleNo\":\"41\",\"NextGRONo\":\"1\",\"DisReturn$p$pfix\":\"
  ，DisReceipt preFIX：，DisNextReturnNo：1，病症preFIX：搜索
  ，DisNextReceiptNo：1，DisNextPayScheduleNo：1，NextActivityNo：1，DisInvoice preFIX：
  ，DisNextInvoiceNo：1，UploadedBy：WAT2，DisFreeIssue preFIX：搜索
  ，DisFOC preFIX：，收据preFIX：RP4}

{"ExecutiveCode":"WAT2 ","FreeIssuePrefix":"
","DisPaySchedulePrefix":"","NextFreeIssueNo":"1","NextReturnNo":"20","UploadedType":"1","DisNextFreeIssueNo":"1","DisNextFOCNo":"1","NextVisitNo":"15","DisNextOrderNo":"1","UploadedOn":"Jun 17 2011 6:33PM","NextReturnAcceptNo":"1","BusinessUnit":"HEMA","TXNReferencePrefix":"20110708 ","OrderPrefix":"OR4 ","UploadedMethod":"3","FOCPrefix":"
","ReturnPrefix":"RT4 ","RetailerPrefix":"TEM4","NextRetailerNo":"10","NextInvoiceNo":"1","NextGRNNo":"1","InvoicePrefix":"IN4 ","NextTXNReference":"2","NextOrderNo":"37","ReturnAcceptPrefix":"
","PaySchedulePrefix":"PS4","NextReceiptNo":"1","NextFOCNo":"20","NextPayScheduleNo":"41","NextGRONo":"1","DisReturnPrefix":" ","DisReceiptPrefix":" ","DisNextReturnNo":"1","DisOrderPrefix":"
","DisNextReceiptNo":"1","DisNextPayScheduleNo":"1","NextActivityNo":"1","DisInvoicePrefix":" ","DisNextInvoiceNo":"1","UploadedBy":"WAT2","DisFreeIssuePrefix":"
","DisFOCPrefix":" ","ReceiptPrefix":"RP4 "}

和我有拿vaue＆安培;命名列表中的

And I have take vaue & name in the list:

我已经这样写的：

try {
  // getSoapResponseTableDataJson(responsePrimitiveData,
  // null,tablesName.get(i));
  String result = responsePrimitiveData.toString();
  JSONObject jsonobject = new JSONObject(result);
  ArrayList<String> toFieldList = new ArrayList<String>();
  ArrayList<String> toFieldValList = new ArrayList<String>();
  JSONArray array = jsonobject.getJSONArray("Table1");
  int max = array.length();

  HashMap<String, String> applicationSettings = new HashMap<String, String>();
   for (int j = 0; j < max; j++) {
     System.out.println(" -- array.getJSONObject(j) -- "
        + array.getJSONObject(j));
     String value = array.getJSONObject(j).getString("value");
     String name = array.getJSONObject(j).getString("name");
     applicationSettings.put(name, value);
   }
  System.out.println(" ---- json --- "+ applicationSettings);
 } catch (JSONException e) {
  e.printStackTrace();
}

这是说：
  org.json.JSONException：对于价值不价值

什么名字＆安培;值？

what is name & value?

其实我想拿到钥匙单独的列表和放大器;值对在单独的列表....
我想利用这样的
       名称，如： {执行code，FreeIssue preFIX，DisPaySchedule preFIX，......}
       像值： {WAT2，，测试，...}

Actually I want to get the Key separate list & value pair in separate list.... I want to take like this name like : {ExecutiveCode,FreeIssuePrefix,DisPaySchedulePrefix,.....} value like : {WAT2, "","TEst",.....}

请帮我。

在此先感谢....

推荐答案

只是觉得......没有 {名：东西} 在你的JSON ...

just think ... there is no {"name": "something"} in your JSON ...

修改 ...我念错的JSONObject文档... ...有没有JSONObject.getNames（）

EDIT ... i read wrong JSONObject documentation ... there is no JSONObject.getNames()

使用obj.names（）

use obj.names()

JSONObject obj = array.getJSONObject(j);
JSONArray[] names = obj.names();

迭代名

String name = names.getString(i);
String value = obj.getString(name);
applicationSettings.put(name, value)

旧的答案是

使用：

JSONObject obj = array.getJSONObject(j);
String[] names = JSONObject.getNames(obj);

迭代名

value = obj.getString(names[i])

这篇关于Android的JSON结果处理的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！