But i need to knw one thing, if i use Array adapter before assigning the
values of str[i] its displaying listview also if i set adapter after
assigning values to Str[i] also dispalying listview . Why so ?
For an example this line :-
str1[i]=jsonArray.getJSONObject(i).getJSONObject("FORM").
getJSONArray("FPROP")
.getJSONObject(0).getString("FNAME");
}
ArrayAdapter<String>adapter=new ArrayAdapter<String>
(this,android.R.layout.simple_list_item_1,android.R.id.text1,str1);
lv.setAdapter(adapter);
}
And my code is here ,
JSONArray jsonArray=jsonObj.getJSONArray("formlist");
int length = jsonArray.length();
String[] str1 =new String[length];
// ArrayAdapter<String>adapter=new ArrayAdapter<String>
(this,android.R.layout.simple_list_item_1,android.R.id.text1,str1);
// lv.setAdapter(adapter);
// ArrayList<String> items = new ArrayList<String>();
for (int i = 0; i < length; i++) {
str1[i]=jsonArray.getJSONObject(i).getJSONObject("FORM").
getJSONArray("FPROP")
.getJSONObject(0).getString("FNAME");
}
ArrayAdapter<String>adapter=new ArrayAdapter<String>
(this,android.R.layout.simple_list_item_1,android.R.id.text1,str1);
lv.setAdapter(adapter);
}
catch (JSONException e) {
On Sun, Aug 5, 2012 at 12:55 AM, Meena Rengarajan <[email protected]>wrote:
> Yeah , exactly what you have said is right . Now works on well ! I havent
> parsed JSON. I am new to this topic. Anywys thankyou :)
>
>
> On Sat, Aug 4, 2012 at 1:21 PM, HideCheck <[email protected]> wrote:
>
>> I guess
>> Show of ListView rather than wrong, You have failed to parse JSON.
>>
>> > ArrayAdapter<String> adapter = new
>> > ArrayAdapter<String>(this,android.R.layout.simple_list_item_1,
>> > android.R.id.text1,str1);
>> > lv.setAdapter(adapter);
>> this code is correct.
>>
>> Structure of the JSON is?
>>
>> 2012/8/4 Meena Rengarajan <[email protected]>:
>> > How should i use Array adapter here to get listview values .. I am very
>> new
>> > to Android can anyone help me please..
>> >
>> > I used this in my code,
>> >
>> > ArrayAdapter<String> adapter = new
>> > ArrayAdapter<String>(this,android.R.layout.simple_list_item_1,
>> > android.R.id.text1,str1);
>> > lv.setAdapter(adapter);
>> >
>> > On Sat, Aug 4, 2012 at 12:47 PM, Meena Rengarajan <
>> [email protected]>
>> > wrote:
>> >>
>> >> How do i wanna split a string into two parts here by omitting first
>> >> character ?
>> >>
>> >>
>> >> On Saturday, August 4, 2012 10:36:18 AM UTC+5:30, Meena Rengarajan
>> wrote:
>> >>>
>> >>> This is my code. Listview using Json in Android . But my lists is not
>> >>> displaying when i click a button.. Only displaying first activity
>> alone.
>> >>> Whats wrong here ? can anyone help me ..
>> >>> try {
>> >>> JSONObject json1=new JSONObject(resultJSON);
>> >>> JSONArray jsonArray=json1.getJSONArray("");
>> >>> int length = jsonArray.length();
>> >>> System.out.println("json1:"+json1);
>> >>> System.out.println("jsonarray:"+jsonArray);
>> >>> String[] str1 = new String[length];
>> >>> for (int i = 0; i < str1.length; i++) {
>> >>>
>> >>>
>> str1[i]=jsonArray.getJSONObject(i).getJSONObject("FORM").getJSONArray("FPROP".getJSONObject(0).getString("FNAME");
>> >>>
>> >>> }
>> >>> } catch (Exception e) {
>> >>> // TODO: handle exception
>> >>> Toast.makeText(NewActivity.this,"sorry", Toast.LENGTH_LONG).show();
>> >>> }
>> >>> ArrayAdapter<String> adapter = new
>> >>> ArrayAdapter<String>(this,android.R.layout.simple_list_item_1,
>> >>> android.R.id.text1,str1);
>> >>>
>> >> --
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