Question

我有json到可序列化的claas然后我用它来填充listview，一切正常但我想在另一个xml页面上用一个可序列化的对象填充textview并且我读到最好的方法是捆绑意图，但我做错了编码。

public class FeedItem implements Serializable {

    private String title;
    private String date;
    private String attachmentUrl;
    private String id;
    private String content;
    private String url;

    public String getUrl() {
        return url;
    }

    public void setUrl(String url) {
        this.url = url;
    }

    public String getContent() {
        return content;
    }

    public void setContent(String content) {
        this.content = content;
    }

    public String getId() {
        return id;
    }

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

    public String getTitle() {
        return title;
    }

    public void setTitle(String title) {
        this.title = title;
    }

    public String getDate() {
        return date;
    }

    public void setDate(String date) {
        this.date = date;
    }

    public String getAttachmentUrl() {
        return attachmentUrl;
    }

    public void setAttachmentUrl(String attachmentUrl) {
        this.attachmentUrl = attachmentUrl;
    }




    @Override
    public String toString() {
        return "[ title=" + title + ", date=" + date + "]";
    }

    ArrayList<FeedItem> all_thumbs = new ArrayList<FeedItem>();
    all_thumbs.add(new FeedItem(title);
    Intent intent = new Intent(title);

    Bundle extras = new Bundle();

    extras.putSerializable("title",title);
    intent.putExtras(extras);

}

并且在我想要使用它的类中

public void updateList() {

    TextView infoz = (TextView) getView().findViewById(R.id.infoz);
    Bundle args;
    getArguments().getSerializable(title);

Answer 1

改为使用：

public void updateList() {

    TextView infoz = (TextView) getView().findViewById(R.id.infoz);
    Intent intent = getIntent();
    Bundle extras = intent.getExtras();
    String title = (String) extras.getSerializable("title");

Answer 2

如果要传递任何原始数据类型的值（String，int，long，float等），则不需要使用Bundle.putExtra（KEY，Serialized Object）和Bundle.getSerializable（KEY）。您可以使用Bundle.putExtra（KEY，原始数据）;

如果要将类对象传递给intent / bundle，则需要在该类中实现Serializable / Parsable，如：

public class Test implements Serializable
{
   private static final long serialVersionUID = 1L;
   int test1;
}

然后您可以将该对象实例传递给intent：

myIntent.putExtra( KEY, new Test() );

有关更多说明，请检查pass seriable object in intent example

Android意图捆绑传递字符串

2 个答案: