Question

单击ExpandableListView项时，我想用ActivityFragment替换WebView Fragmnet。

这是我的WebView片段：

 View mview;

 WebView view ;

 ProgressDialog mProgress ;

 @Override

  public View onCreateView(LayoutInflater inflater, ViewGroup container,
      Bundle savedInstanceState) {

     mview = inflater.inflate(R.layout.siteview,container, false);

     Bundle bundle = getArguments();

       String url = bundle.getString("url");
         view = (WebView) mview.findViewById(R.id.webView1);

      WebSettings settings = view.getSettings();
        settings.setJavaScriptEnabled(true);


       view.loadUrl(url);

       view.setWebViewClient(new WebViewClient() {
           @Override
           public boolean shouldOverrideUrlLoading(WebView view, String url) {
                view.loadUrl(url);
                return true;
           }


       });

    return mview;
  }


}

这是我的片段活动OnChildclick of Expendabels List

            FragmentTransaction transaction = getSupportFragmentManager().beginTransaction();

                Fragment siteview = new siteview();

               // Bundle bundle = new Bundle();


                //setTitle(menutitles[position]);
            // Click Jadval's Child
            if (groupPosition == 2 & childPosition == 0){
                mNav.toggleRightDrawer();  

               // bundle.putString("url", customHtml1.toString());

               // siteview.setArguments(bundle);

                transaction.replace(R.id.fram, siteview);

                // Commit the transaction
            }
                else if (groupPosition == 2 & childPosition == 1 ){
                mNav.toggleRightDrawer();  

            }
            transaction.commit()    ;           

            return false;

        }

本准则不起作用。问题出在哪儿？请给我提示

Answer 1

如果我没有错，你正在寻找，当你点击列表视图中的任何项目时，它必须打开特定的片段。为此你需要处理列表的位置。并在该特定位置设置一个特定的片段。像

这样的东西

switch(position)
{
   case 0:
    case 1:
 }


    and fragment class should be like this


 public class SearchFragment extends Fragment {
WebView wv;
public SearchFragment(){}

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
        Bundle savedInstanceState) {

    View rootView = inflater.inflate(R.layout.fragment_search, container, 
    false);

    wv=(WebView)rootView.findViewById(R.id.search_web);
    wv.getSettings().setJavaScriptEnabled(true);
    wv.loadUrl("http://www.google.com");

    wv.setWebViewClient(new WebViewClient() {
        @Override
        public boolean shouldOverrideUrlLoading(WebView view, String url) {
            view.loadUrl(url);
            return false;
   }
    });
    return rootView;
  }
}

如何在另一个片段中加载WebView的URL

1 个答案: