【发布时间】:2016-09-27 15:50:59
【问题描述】:
我在从 mysql 解析 android json 时遇到问题,当我尝试执行我的项目时,它显示“org.json.jsonexception end of input at character 0 of” 这是我的解析代码:
protected String doInBackground(String... arg0) {
for(String url: arg0){
try{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);
HttpResponse response = httpclient.execute(httppost);
inp = response.getEntity().getContent();
}
catch (ClientProtocolException e){
error = "ClientProtocolException: " + e.getMessage();
}catch (IOException e){
error = "ClientProtocolException: " + e.getMessage();
}
}
BufferedReader reader;
try{
reader = new BufferedReader(new InputStreamReader(inp, "iso-8859-1"),8);
String line=null;
while ((line = reader.readLine()) != null){
txt += line+"\n";
}
inp.close();
}catch (UnsupportedEncodingException e){
error = "Unsupport Encoding: "+e.getMessage();
}catch (IOException e){
error = "Error IO: "+e.getMessage();
}
list1 = new ArrayList<Pharm>();
try{
JSONArray jArray = new JSONArray(txt);
for(int i=0;i<jArray.length();i++){
JSONObject jsonData = jArray.getJSONObject(i);
Pharm pharm = new Pharm();
pharm.setNomPharm(jsonData.getString("nomPharm"));
pharm.setAdressePharm(jsonData.getString("adressePharm"));
list1.add(pharm);
}
}catch (JSONException e){
error = "Error convert to JSON or Error JSON format: "+ e.getMessage();
}
return error;
}
我需要你的帮助。
【问题讨论】:
-
你能告诉我们返回的
JSON吗? -
这将是我们解决问题的必须...
-
{ 药店:[ { idPharm:“1”,nomPharm:“pharm1”,adressePharm:“hay saada”,lat:“ssssssss”,lon:“ddddddddd”}]}
-
我用 127.0.0.1 测试它
-
我必须显示在 listView 中返回的 JSON 的一些键
标签: android json listview parsing