我会从我的api json响应中创建一些listview,但是我在代码中停留了此LinkedTreeMap错误。谁能帮我解决这个问题?

public class KategoriListAdapter extends BaseAdapter {

    Context context;
    ArrayList<Barang> barang;

    public KategoriListAdapter(Context context, ArrayList<Barang> barang) {
        this.context = context;
        this.barang = barang;
    }

    @Override
    public int getCount() {
        return barang.size();
    }

    @Override
    public Barang getItem(int i) {
        return this.barang.get(i);
    }

    @Override
    public long getItemId(int i) {
        return i;
    }

    @TargetApi(Build.VERSION_CODES.KITKAT)
    @Override
    public View getView(final int i, View view, ViewGroup viewGroup) {
        if (view == null) {
            view = LayoutInflater.from(context).inflate(R.layout.custom_list_view_kategori, viewGroup, false);
        }

        TextView tvNama = (TextView) view.findViewById(R.id.tv_nama);
        TextView tvHarga = (TextView) view.findViewById(R.id.tv_harga);
        TextView tvUsername = (TextView) view.findViewById(R.id.tv_username);


        Object getrow = this.barang.get(i);
        LinkedTreeMap<Object, Object> rowmap = (LinkedTreeMap) getrow;
        String nama = rowmap.get("nama").toString();
        String harga = rowmap.get("harga").toString();
        String username = rowmap.get("username").toString();

        tvNama.setText(nama);
        tvHarga.setText(harga);
        tvUsername.setText(username);

        return view;
    }

}

public class Barang {

    @SerializedName("id")
    @Expose
    private Integer id;
    @SerializedName("username")
    @Expose
    private String username;
    @SerializedName("nama")
    @Expose
    private String nama;
    @SerializedName("harga")
    @Expose
    private String harga;
    @SerializedName("gambar")
    @Expose
    private String gambar;

    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    public String getNama() {
        return nama;
    }

    public void setNama(String nama) {
        this.nama = nama;
    }

    public String getHarga() {
        return harga;
    }

    public void setHarga(String harga) {
        this.harga = harga;
    }

    public String getGambar() {
        return gambar;
    }

    public void setGambar(String gambar) {
        this.gambar = gambar;
    }

}


我运行活动时的日志结果是

java.lang.ClassCastException: com.example.barangkoz.model.Barang cannot be
cast to com.google.gson.internal.LinkedTreeMap
    at     com.example.barangkoz.activities.KategoriListAdapter.getView(KategoriListAdapter.java:57)`

最佳答案

您正在适配器构造函数中传递“ Barang”对象的ArrayList,这意味着您的适配器中已经有了对象Barang的列表,您可以直接使用它而无需转换为TreeMap。

在适配器的getView方法中更改此

Object getrow = this.barang.get(i);




Barang barang = barang.get(i);


它将使Barang对象位于Barang列表中i的位置。
这样您就可以使用在对象Barang内部定义的getters方法从该对象获取数据。

String harga = barang.getHarga();
String nama = barang.getNama();
String userName = barang.getUsername();


并将其设置为您的TextView或

您可以将数据直接从Barang对象设置为TextView(在设置为TextView之前无需执行其他步骤即可将其设置为变量),就像这样

tvHarga.setText(barang.getHarga());
tvNama.setText(barang.getNama());
tvUsername.setText(barang.getUsername());

08-03 18:32