onBackPressed() - 无法解析符号' webView'

时间:2015-07-04 20:38:30

标签: java android xml webview

我试图在我的WebView应用程序中取回按钮功能,但是onBackPressed()函数正在给出错误'无法解析符号webView'

这是我的代码:

@SuppressLint("SetJavaScriptEnabled")
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    AdView mAdView;
    mAdView = (AdView) findViewById(R.id.adView);
    AdRequest adRequest = new AdRequest.Builder().build();
    mAdView.loadAd(adRequest);

    WebView webView = (WebView) findViewById(R.id.webview);

    WebSettings webSettings = webView.getSettings();
    webSettings.setJavaScriptEnabled(true);

    // To keep browser in width of phone
    webView.getSettings().setUseWideViewPort(true);
    webView.getSettings().setLoadWithOverviewMode(true);

    // Sets default text size
    webSettings.setTextZoom(175);

    MyWebViewClient webViewClient = new MyWebViewClient();
    webView.setWebViewClient(webViewClient);

    webView.loadUrl("http://www.google.com");
}

@Override
public void onBackPressed() {
    if (webView.canGoBack()) {
        webView.goBack();
    } else {
        super.onBackPressed();
    }
}

1 个答案:

答案 0 :(得分:2)

你可以在onBackPressed中执行此操作:

total_amount = models.DecimalField(max_digits=10,decimal_places=2,blank = True,null = True,default = Decimal('0.00'))

编辑: 你可以这样做(带字段):

    @Override
public void onBackPressed() {
    WebView webView = (WebView) findViewById(R.id.webview);
    if (webView.canGoBack()) {
        webView.goBack();
    } else {
        super.onBackPressed();
    }
}