本文介绍了字符串转换为一个JSON对象的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我想从出现从一个servlet在我的服务器的响应字符串创建JSON对象。
公共类JSONParser {静态InputStream为= NULL;
静态的JSONObject jObj = NULL;
静态JSON字符串=;//构造
公共JSONParser（）{} 公众的JSONObject getJSONFromUrl（URL字符串，字符串一）{ ArrayList的＆LT;＆的NameValuePair GT; postParameters =新的ArrayList＆LT;＆的NameValuePair GT;（）;
postParameters.add（新BasicNameValuePair（分支名称，一））; 尝试{
// defaultHttpClient
DefaultHttpClient的HttpClient =新DefaultHttpClient（）;
HttpPost httppost =新HttpPost（URL）;
httppost.setEntity（新UrlEn codedFormEntity（postParameters））; HTT presponse HTT presponse = httpClient.execute（httppost）;
HttpEntity httpEntity = HTT presponse.getEntity（）;
是= httpEntity.getContent（）; }赶上（UnsupportedEncodingException五）{
e.printStackTrace（）;
}赶上（ClientProtocolException E）{
e.printStackTrace（）;
}赶上（IOException异常五）{
e.printStackTrace（）;
} 尝试{
读者的BufferedReader =新的BufferedReader（新的InputStreamReader（
是，ISO-8859-1），8）;
StringBuilder的SB =新的StringBuilder（）;
串线= NULL;
而（（行= reader.readLine（））！= NULL）{
sb.append（行+\\ n）;
}
is.close（）;
JSON = sb.toString（）;
}赶上（例外五）{
Log.e（缓冲区错误，错误转换结果+ e.toString（））;
} //尝试分析字符串到一个JSON对象
尝试{
jObj =新的JSONObject（JSON）;
}赶上（JSONException E）{
Log.e（JSON解析器，错误分析数据+ e.toString（））;
}
返回jObj;
}}
这是我的servlet。
公共类AvailabilityResponse延伸的HttpServlet {@覆盖
保护无效的doPost（HttpServletRequest的请求，HttpServletResponse的响应）
抛出了ServletException，IOException异常{ PrintWriter的OUT = response.getWriter（）; 串br_id;
br_id =的request.getParameter（分支名称）; 尝试{
的Class.forName（com.mysql.jdbc.Driver）的newInstance（）。
连接CON =的DriverManager.getConnection（JDBC的：mysql：//本地主机：8888 / atmlivedetails，根，根）;
声明ST = con.createStatement（）;
结果集RS = st.executeQuery（'+ br_id +从哪里location_stat = act_brname选择atmbrno，atmbrname'）;
INT I = 0;
JSONArray jArray =新JSONArray（）;
而（rs.next（））{ 字符串s = rs.getString（atmbrno）;
串T = rs.getString（atmbrname）; JSONObject的arrayObj =新的JSONObject（）; arrayObj.put（atmbrno，S）;
arrayObj.put（atmbrname，T）; jArray.add（ⅰ，arrayObj）;
我++;
}
rs.close（）;
st.close（）;
的out.print（jArray）;
} 赶上（例外五）{
的out.print（E）;
}
}}
但是当我运行这个它说错误
分析数据org.json.JSONException：值[{atmbrname：阿努拉德普勒
[ATM 2]，atmbrno：ATM084}，{atmbrname：马纳尔分校
，atmbrno：类型org.json.JSONArray的ATM344}]无法转换
到的JSONObject
什么是错在这种情况下？
下面是我调用类
公共类ListAtmActivity扩展ListActivity {私有静态字符串URL =http://10.0.2.2:8080/hello/AvailabilityResponse;// TextView的错误;
//字符串brName;//包B = getIntent（）getExtras（）。
//字符串brName = b.getString（钥匙）;私有静态最后弦乐TAG_CONTACTS =接触;
私有静态最后弦乐TAG_ID =ID;
私有静态最后弦乐ATM_NO =atmbrno;
私有静态最后弦乐ATM_PLACE =atmbrname;//接触JSONArray
// JSONArray接触= NULL;@覆盖
公共无效的onCreate（捆绑savedInstanceState）{
super.onCreate（savedInstanceState）;
的setContentView（R.layout.list_item）; 字符串brName = getIntent（）getExtras（）的getString（密钥）。;
/ *包B = getIntent（）getExtras（）。
brName = b.getString（钥匙）;
* /
//为哈希映射的ListView
ArrayList的＆LT;＆HashMap的LT;字符串，字符串＆GT;＆GT; contactList =新的ArrayList＆LT;＆HashMap的LT;字符串，字符串＆GT;＆GT;（）; //创建JSON解析器实例
JSONParser jParser =新JSONParser（）; // URL从获取JSON字符串
JSONObject的JSON = jParser.getJSONFromUrl（URL，brName）;
尝试{
//字符串的结果;
JSONArray接触= json.getJSONArray（TAG_CONTACTS）; 的for（int i = 0; I＆LT; contacts.length（）;我++）{
JSONObject的json_data = contacts.getJSONObject（I） //存储在变量中的每个JSON项目
字符串atm_id = json_data.getString（ATM_NO）;
字符串atm_name = json_data.getString（ATM_PLACE）; HashMap的＆LT;字符串，字符串＆GT;地图=新的HashMap＆LT;字符串，字符串＆GT;（）; map.put（ATM_NO，atm_id）;
map.put（ATM_PLACE，atm_name）; contactList.add（地图）;
}
} 赶上（JSONException E）{
e.printStackTrace（）; } / **
*更新解析JSON数据到ListView控件
* * /
ListAdapter适配器=新SimpleAdapter（这一点，contactList，
R.layout.list_main，
新的String [] {ATM_NO，ATM_PLACE}，新的INT [] {
R.id.name，R.id.email}）; setListAdapter（适配器）;}}
解决方案
您可以JSON对象变成JSON数组，
公共类JSONParser {静态InputStream为= NULL;
静态JSONArray jObj = NULL;
静态JSON字符串=;//构造
公共JSONParser（）{} 公共JSONArray getJSONFromUrl（URL字符串，字符串一）{ ArrayList的＆LT;＆的NameValuePair GT; postParameters =新的ArrayList＆LT;＆的NameValuePair GT;（）;
postParameters.add（新BasicNameValuePair（分支名称，一））; 尝试{
// defaultHttpClient
DefaultHttpClient的HttpClient =新DefaultHttpClient（）;
HttpPost httppost =新HttpPost（URL）;
httppost.setEntity（新UrlEn codedFormEntity（postParameters））; HTT presponse HTT presponse = httpClient.execute（httppost）;
HttpEntity httpEntity = HTT presponse.getEntity（）;
是= httpEntity.getContent（）; }赶上（UnsupportedEncodingException五）{
e.printStackTrace（）;
}赶上（ClientProtocolException E）{
e.printStackTrace（）;
}赶上（IOException异常五）{
e.printStackTrace（）;
} 尝试{
读者的BufferedReader =新的BufferedReader（新的InputStreamReader（
是，ISO-8859-1），8）;
StringBuilder的SB =新的StringBuilder（）;
串线= NULL;
而（（行= reader.readLine（））！= NULL）{
sb.append（行+\\ n）;
}
is.close（）;
JSON = sb.toString（）;
}赶上（例外五）{
Log.e（缓冲区错误，错误转换结果+ e.toString（））;
} //尝试分析字符串到一个JSON数组
尝试{
jObj =新JSONArray（JSON）;
}赶上（JSONException E）{
Log.e（JSON解析器，错误分析数据+ e.toString（））;
}
返回jObj;
}}
您返回的数据是JSON数组，但你转换JSON对象，这是唯一的例外加薪。
静态JSONArray jObj = NULL; //尝试分析字符串到一个JSON数组
尝试{
jObj =新JSONArray（JSON）;
}赶上（JSONException E）{
Log.e（JSON解析器，错误分析数据+ e.toString（））;
}
I want to create json object from a string that comes as a response from a servlet in my server.
public class JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
// constructor
public JSONParser() {
}
public JSONObject getJSONFromUrl(String url,String a) {
ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>();
postParameters.add(new BasicNameValuePair("branchname", a));
try {
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);
httppost.setEntity(new UrlEncodedFormEntity(postParameters));
HttpResponse httpResponse = httpClient.execute(httppost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
return jObj;
}}
This is my servlet..
public class AvailabilityResponse extends HttpServlet {
@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
PrintWriter out=response.getWriter();
String br_id;
br_id=request.getParameter("branchname");
try{
Class.forName("com.mysql.jdbc.Driver").newInstance();
Connection con=DriverManager.getConnection("jdbc:mysql://localhost:8888 /atmlivedetails","root","root");
Statement st=con.createStatement();
ResultSet rs=st.executeQuery("select atmbrno, atmbrname from location_stat where act_brname='"+br_id+"'");
int i=0;
JSONArray jArray = new JSONArray();
while(rs.next()){
String s = rs.getString("atmbrno");
String t = rs.getString("atmbrname");
JSONObject arrayObj = new JSONObject();
arrayObj.put("atmbrno",s);
arrayObj.put("atmbrname",t);
jArray.add(i,arrayObj);
i++;
}
rs.close ();
st.close ();
out.print(jArray);
}
catch(Exception e){
out.print(e);
}
}}
but when I run this it says Error
"parsing data org.json.JSONException: Value [{"atmbrname":"ANURADAPURA
[ATM 2]","atmbrno":"ATM084"},{"atmbrname":"MANNAR BRANCH
","atmbrno":"ATM344"}] of type org.json.JSONArray cannot be converted
to JSONObject"
what is the wrong in this case?
Here is my calling class
public class ListAtmActivity extends ListActivity{
private static String url ="http://10.0.2.2:8080/hello/AvailabilityResponse";
//TextView error;
//String brName;
//Bundle b = getIntent().getExtras();
//String brName = b.getString("key");
private static final String TAG_CONTACTS = "contacts";
private static final String TAG_ID = "id";
private static final String ATM_NO = "atmbrno";
private static final String ATM_PLACE = "atmbrname";
// contacts JSONArray
//JSONArray contacts = null;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.list_item);
String brName=getIntent().getExtras().getString("key");
/*Bundle b = getIntent().getExtras();
brName = b.getString("key");
*/
// Hashmap for ListView
ArrayList<HashMap<String, String>> contactList = new ArrayList<HashMap<String, String>>();
// Creating JSON Parser instance
JSONParser jParser = new JSONParser();
// getting JSON string from URL
JSONObject json = jParser.getJSONFromUrl(url,brName);
try{
//String results;
JSONArray contacts = json.getJSONArray(TAG_CONTACTS);
for(int i = 0; i < contacts.length(); i++){
JSONObject json_data = contacts.getJSONObject(i);
// Storing each json item in variable
String atm_id = json_data.getString(ATM_NO);
String atm_name = json_data.getString(ATM_PLACE);
HashMap<String, String> map = new HashMap<String, String>();
map.put(ATM_NO, atm_id);
map.put(ATM_PLACE, atm_name);
contactList.add(map);
}
}
catch(JSONException e) {
e.printStackTrace();
}
/**
* Updating parsed JSON data into ListView
* */
ListAdapter adapter = new SimpleAdapter(this, contactList,
R.layout.list_main,
new String[] { ATM_NO, ATM_PLACE }, new int[] {
R.id.name , R.id.email });
setListAdapter(adapter);
}}
解决方案
you can change json object to json array,
public class JSONParser {
static InputStream is = null;
static JSONArray jObj = null;
static String json = "";
// constructor
public JSONParser() {
}
public JSONArray getJSONFromUrl(String url,String a) {
ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>();
postParameters.add(new BasicNameValuePair("branchname", a));
try {
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);
httppost.setEntity(new UrlEncodedFormEntity(postParameters));
HttpResponse httpResponse = httpClient.execute(httppost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON Array
try {
jObj = new JSONArray(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
return jObj;
}}
Your return data is json array, but you convert json object ,that's only Exception raise.
static JSONArray jObj = null;
// try parse the string to a JSON Array
try {
jObj = new JSONArray(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
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