Question

我想将包含来自后台服务的数据的自定义矢量对象传递给活动，以便更新该活动的UI。我无法通过意图传递我的数据。请帮助我，我需要做什么..

以下是我正在使用的代码段。在下面的代码中，我想将文章列表传递给一个活动。

Vector<RowData> getArticleLog() throws JSONException{

    Vector<RowData> articleList = new Vector<RowData>();

    RowData rd;
    InputStream is = null;
    String result = null;
    JSONArray jarray = new JSONArray();

    //Add data to be send.
    ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
    nameValuePairs.add(new BasicNameValuePair("article_title", "Flying Horse"));

    try {

        HttpClient httpclient = new DefaultHttpClient();

        // for local server xampp
        HttpPost httppost = new HttpPost("http://10.0.2.2/groupbook/return_article_log.php");

        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        HttpResponse response = httpclient.execute(httppost);

        HttpEntity entity = response.getEntity();
        is = entity.getContent();
        Log.i("postData", response.getStatusLine().toString());

    } catch (Exception e) {
        Log.e(TAG, "json error : " + e.toString());
    }
    //convert response to string
    try {
         BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
         StringBuilder sb = new StringBuilder();
         String line = null;
         while ((line = reader.readLine()) != null) {
                 sb.append(line + "\n");
         }
         is.close();

         result=sb.toString();
    } catch (Exception e) {
         Log.e(TAG, "Error converting result "+e.toString());
    }

    try {
        jarray = new JSONArray(result);
    } catch (JSONException e) {
        e.printStackTrace();
    }
    //return jArray in the form of Array list;
    JSONObject rowElement = null;
    int viewers = 0;
    int votes = 0;

    for (int i = 0; i < jarray.length(); i++) {
        rowElement = jarray.getJSONObject(i);

        viewers = rowElement.getInt("viewers");
        votes = rowElement.getInt("votes");

        rd = new RowData(i, 
                rowElement.getString("article_title"), 
                Integer.toString(i), 
                rowElement.getString("last_post_by"),
                "2 hours", 
                rowElement.getString("catagory_tag"), 
                Integer.toString(viewers), 
                rowElement.getString("privacy_tag"), 
                Integer.toString(votes), 
                rowElement.getString("catagory_title"));
        articleList.add(rd);
    }

    return articleList;
}

RowData类如下：

public class RowData {

protected int mId;
public String articleTitle;
public String articleCounter;
public String lastPostBy;
public String updatedTime;
public String catagoryTag;
public String viewingNow;
public String privacyTag;
public String votes;
public String catagoryTitle;

public RowData(int mId, String articleTitle, String articleCounter,
        String lastPostBy, String updatedTime, String catagoryTag,
        String viewingNow, String privacyTag, String votes,
        String catagoryTitle) {
    this.mId = mId;
    this.articleTitle = articleTitle;
    this.articleCounter = articleCounter;
    this.lastPostBy = lastPostBy;
    this.updatedTime = updatedTime;
    this.catagoryTag = catagoryTag;
    this.viewingNow = viewingNow;
    this.privacyTag = privacyTag;
    this.votes = votes;
    this.catagoryTitle = catagoryTitle;
}
}

以下方式是正确的吗？

private void DisplayingInfo() throws JSONException{
    Log.d(TAG, "entered DisplayLoggingInfo");
    intent.putExtra("articleLog", getArticleLog());
    sendBroadcast(intent);
}

Answer 1

您可以做的一件事是制作自定义类工具Serializable（或Parcelable），将带有putExtra()方法的可序列化对象发送到活动并使用getSerializableExtra()在你的活动上得到它。

编辑1： 一个简单的例子：

在您的自定义类中：

import java.io.Serializable;

@SuppressWarnings("serial")
public class YourCustomVectorClass implements Serializable {
    // ...
}

在您要开始活动的服务中：

Intent intent = new Intent(yourContext, yourActivity.class);
intent.putExtra("theNameOfTheObject", yourObject);
startActivity(intent);

在您的活动中：

YourCustomVectorClass yourVector = (YourCustomVectorClass) getIntent().getSerializableExtra("theNameOfTheObject");

编辑2 ：再次阅读该问题后，我意识到您正在将Vector个RowData个对象传递给您的活动。

由于Java Vector class实现Serializable，我认为您不应该做任何事情，只需使用putExtra()将该向量传递给活动，并在活动上使用getSerializableExtra()获取它。

将对象从后台服务传递到活动

1 个答案: