Problem in the organization of the data of a Listview - Basedapter

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Problem in the organization of the data of a Listview - Basedapter

Asked 9 years, 7 months ago

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I’m having a little problem organizing the personal data of an external Mysql database. Like, I created a Infopeoplesdapter class inheriting from Basedapter where I pick up name, address, neighborhood, zip code, phone and remarks. I played on listview the following way:

List<InfoPessoais> dadosPessoais = new ArrayList<InfoPessoais>();
    InfoPessoais ip = new InfoPessoais();

    for (String k : dados) {
        ip = new InfoPessoais();
        ip.setNome(k);
        ip.setEndereco(k);
        ip.setBairro(k);
        ip.setCep(k);
        ip.setTelefone(k);
        ip.setObs(k);
        dadosPessoais.add(ip);
    }
    lvInformacoesPessoais.setAdapter(new InfoPessoaisAdapter(this, dadosPessoais));

The problem is that at the time of showing on the screen of my device, it shows the name repeatedly for all fields, hence the next field repeats again and so on... it gets all wrong as in the image below:

inserir a descrição da imagem aqui

dice:

try{
        respostaRetornada = ConexaoHttpClient.executaHttpPost(url, parametrosPost);
        String resposta = respostaRetornada.toString();
        resposta = resposta.replaceAll("\\s+", "");
        Log.i("Informações", "Informações: "+resposta);

        char separador = '#';
        int contadados = 0;

        for (int i=0; i < resposta.length(); i++){
            if (separador == resposta.charAt(i)){
                contadados++;
                dados = new String[contadados];
            }
        }

        char caracterLido = resposta.charAt(0);
        String nome = "";

        for (int i=0; caracterLido != '^'; i++){
            caracterLido = resposta.charAt(i);
            Log.i("Chars", "Chars do Paciente"+caracterLido);

            if (caracterLido != '#'){
                if (caracterLido == '*'){
                    nome = nome + " ";
                }else
                nome+= (char) caracterLido;

            }else{
                Log.i("Nome", "Nome: "+nome);
                dados[posicao] =""+ nome;
                Log.i("Nome posição ["+posicao+"]", ""+dados[posicao]);
                posicao = posicao + 1;
                nome = "";
            }
        }
        Log.i("Fim", "Fim do for");

    }catch(Exception erro){
        Toast.makeText(getBaseContext(), "Erro: "+erro, Toast.LENGTH_LONG).show();
    }

in your class InfoPessoais was made the @Override of the method ToString()?

– Enzo Tiezzi

2014/11/12 at 15:55
To tell you the truth I don’t even have this Tostring method()

– Emerson Moraes

2014/11/12 at 15:56
String k : data in for tu ta setando all fields with the value of k... so the fields will exit repeated

– Pedro Rangel

2014/11/12 at 15:57
what exactly you want to show?

– Enzo Tiezzi

2014/11/12 at 15:57
in your case, always has the value of k

– Enzo Tiezzi

2014/11/12 at 15:58
you have to do something like this... ip.setName(data.getitem(i).getName()); this getitem is just an example to be able to item in position i.

– Pedro Rangel

2014/11/12 at 16:00
show how you did the "data" class in String k : data.

– Pedro Rangel

2014/11/12 at 16:02
better explaining which class belongs to the data variable? show how you implemented this class in your question would be a good one.

– Pedro Rangel

2014/11/12 at 16:08
@Pedrorangel the data is all I’m searching for from my php file, which contains all the fields I need...

– Emerson Moraes

2014/11/12 at 16:34
@Enzotiezzi how would I do not always set k? I can not let repeat everything...

– Emerson Moraes

2014/11/12 at 16:35
@Pedrorangel said there, but now that you’ve posted all the code, let’s take a look

– Enzo Tiezzi

2014/11/12 at 16:36
Yes, but by the variable data, I can’t pull any of that... it’s just a String[] where I stored all the php data

– Emerson Moraes

2014/11/12 at 16:38
the data are coming in what order? are all sequential within data[] ?? need to know how is the vector given after obtaining the values.

– Enzo Tiezzi

2014/11/12 at 16:39
Yes, name, address and so on...

– Emerson Moraes

2014/11/12 at 16:40

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by Enzo Tiezzi • **1,194** points · Answer 1 · 2014-11-12T16:47:14+00:00

Instead of making the following code :

for (String k : dados) {
    ip = new InfoPessoais();
    ip.setNome(k);
    ip.setEndereco(k);
    ip.setBairro(k);
    ip.setCep(k);
    ip.setTelefone(k);
    ip.setObs(k);
    dadosPessoais.add(ip);
}

Do the following:

int i = 0;
while(i < dados.length)
{
    ip = new InfoPessoais();
    ip.setNome(dados[i++]);
    ip.setEndereco(dados[i++]);
    ip.setBairro(dados[i++]);
    ip.setCep(dados[i++]);
    ip.setTelefone(dados[i++]);
    ip.setObs(dados[i++]);
    dadosPessoais.add(ip);
}

This is very impractical to do, but for now to solve your problem, this is enough.