Question

在网上搜索答案如何实现搜索选项并将其链接到WebView时，我找到了一个例子，但是执行搜索时出现问题，然后启动第二个Activity，我该如何搜索没有初始化新活动的网页？

示例代码：

\ AndroidManifest.xml中

<intent-filter>
                <action android:name="android.intent.action.SEARCH"/>
            </intent-filter>

            <meta-data android:name="android.app.searchable"
                android:resource="@xml/searchable"/>

\ main.xml中

<?xml version="1.0" encoding="utf-8"?>
<menu xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:appcompat="http://schemas.android.com/apk/res-auto">

    <item
        android:id="@+id/menu_search"
        android:title="@string/menu_search"
        appcompat:actionViewClass="android.support.v7.widget.SearchView"
        appcompat:showAsAction="always"/>

</menu>

\ searchable.xml

<?xml version="1.0" encoding="utf-8"?>
<searchable android:label="@string/search"
android:hint="@string/search_hint"
android:searchSuggestAuthority=""
android:searchSuggestIntentAction="android.intent.action.VIEW"
android:searchSuggestIntentData=""
xmlns:android="http://schemas.android.com/apk/res/android" />

\ MainActivity.java

@Override
public boolean onCreateOptionsMenu(Menu menu) {
    MenuInflater inflater = getMenuInflater();

    inflater.inflate(R.menu.main, menu);

    SearchView searchView = (SearchView) menu.findItem(R.id.menu_search).getActionView();
    SearchManager searchManager = (SearchManager) getSystemService(SEARCH_SERVICE);
    searchView.setSearchableInfo(searchManager.getSearchableInfo(getComponentName()));
    Intent searchIntent = getIntent();
    String query = searchIntent.getStringExtra(SearchManager.QUERY);
    if(Intent.ACTION_SEARCH.equals(searchIntent.getAction())) {

        loadPage("https://sayt.com//index.php?do="+query);

    }


    return super.onCreateOptionsMenu(menu);

}

使用搜索栏和webview

0 个答案: