Question

所以我尝试在一个Activity中实现两个API，并在同一个Activity中显示我的第二个API的listview和webview。我设法让列表视图失效了。这是MainActivity的一部分，我试图通过onClick方法同时调用两个API

    txtSearch = (EditText)webView.findViewById(R.id.txtSearch);
    searchbtn = (Button) webView.findViewById(R.id.searchbtn);
    webView = (WebView)webView.findViewById(R.id.webView);

    searchbtn.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            String item = txtSearch.getText().toString();
            new JsonSearchTask(item).execute();
        }
    });
}

    private class JsonSearchTask extends AsyncTask<Void, Void, Void> {

        String searchResult = "";
        String search_url =   "http://ajax.googleapis.com/ajax/services/search/web?v=1.0&q=";
        String search_query;

        JsonSearchTask(String item){

            try {
                search_item = URLEncoder.encode(item, "utf-8");
            } catch (UnsupportedEncodingException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

            search_query = search_url + search_item;
        }

        @Override
        protected Void doInBackground(Void... arg0) {

            try {
                searchResult = ParseResult(sendQuery(search_query));
            } catch (JSONException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

            return null;
        }

        @Override
        protected void onPreExecute() {

            searchbtn.setEnabled(false);
            searchbtn.setText("Wait...");
            super.onPreExecute();
        }

        @Override
        protected void onPostExecute(Void result) {

            webView.loadData(searchResult,
                    "text/html",
                    "UTF-8");

            searchbtn.setEnabled(true);
            searchbtn.setText("Search");

            super.onPostExecute(result);
        }

    }

private String sendQuery(String query) throws IOException{
    String result = "";

    URL searchURL = new URL(query);

    HttpURLConnection httpURLConnection = (HttpURLConnection) searchURL.openConnection();

    if(httpURLConnection.getResponseCode() == HttpURLConnection.HTTP_OK){
        InputStreamReader inputStreamReader = new InputStreamReader(httpURLConnection.getInputStream());
        BufferedReader bufferedReader = new BufferedReader(
                inputStreamReader,
                8192);

        String line = null;
        while((line = bufferedReader.readLine()) != null){
            result += line;
        }

        bufferedReader.close();
    }

    return result;
}

private String ParseResult(String json) throws JSONException{
    String parsedResult = "";

    JSONObject jsonObject = new JSONObject(json);
    JSONObject jsonObject_responseData = jsonObject.getJSONObject("responseData");
    JSONArray jsonArray_results = jsonObject_responseData.getJSONArray("results");

    //parsedResult += "Google Search APIs (JSON) for : <b>" + search_item + "</b><br/>";
    //parsedResult += "Number of results returned = <b>" + jsonArray_results.length() + "</b><br/><br/>";

    for(int i = 0; i < jsonArray_results.length(); i++){

        JSONObject jsonObject_i = jsonArray_results.getJSONObject(i);

        String iTitle = jsonObject_i.getString("title");
        String iContent = jsonObject_i.getString("content");
        String iUrl = jsonObject_i.getString("url");

        parsedResult += "<a href='" + iUrl + "'>" + iTitle + "</a><br/>";
        parsedResult += iContent + "<br/><br/>";
    }

    return parsedResult;
}

这是GoogleSearch Api public void onCreate（Bundle savedInstanceState）{ super.onCreate（savedInstanceState）; 的setContentView（R.layout.twit_list）;

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:orientation="vertical" android:layout_width="match_parent"
android:layout_height="wrap_content"
android:id = "@+id/activitymain"
>




<RelativeLayout
    android:layout_width="match_parent"
    android:layout_height="match_parent"
    android:orientation="horizontal">
    <EditText
        android:layout_width="match_parent"
        android:layout_height="wrap_content"
        android:id = "@+id/txtSearch"/>
    <Button
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:layout_below="@+id/txtSearch"
        android:text="Search"
        android:id="@+id/searchbtn" />
    <Button
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:layout_below="@+id/txtSearch"
        android:layout_toRightOf="@+id/searchbtn"
        android:layout_toEndOf="@+id/searchbtn"
        android:id="@+id/save"
        android:text="     Save     " />

    <Button
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:text="Saved Searches"
        android:id="@+id/savedSearches"
        android:layout_alignTop="@+id/save"
        android:layout_toRightOf="@+id/save"
        android:layout_toEndOf="@+id/save" />
</RelativeLayout>



<ListView
    android:layout_width="match_parent"
    android:layout_height="174dp"
    android:id = "@android:id/list"
    android:background="#FF498CDE">

</ListView>

<WebView
    android:layout_width="match_parent"
    android:layout_height="202dp"
    android:id="@+id/webView"
    android:layout_gravity="center_horizontal" />

XML

if( ConstructionType == "New Residential" ) 
app.alert("You must enter the Residential Finished Floor Area field before continuing");
else if ( ConstructionType == "Accessory Suite")
    app.alert("You must fill in the Residential Finished Floor Area field before continuing");

基本上我想在点击搜索按钮时同时运行两个api。请有人指出我正确的方向，我对此完全陌生。感谢

Answer 1

问题是您在JsonSearchTask AsyncTask中没有得到任何响应。

问题是您用于Google网络搜索的API目前无法使用。您应该使用Google自定义搜索API（https://developers.google.com/custom-search/）。

如何在一个Activity中显示listview和webview？

1 个答案: