Question

我创建了一个WebView，其中一些单词显示为标签，这些标签必须是可点击的，以便在点击标签后，向用户显示新页面中的相关内容列表。为此，我将这些单词放在WebView的超链接中。我需要在每个标签中找到文本＆＃34; a＆＃34;单击超链接后。但是为webviewclient定义WebView工作。这是我的主要活动：

 public class MainActivity extends AppCompatActivity {

    String[] wordCloud = new String[]{"Donut", "Eclair", "Froyo", "Gingerbread", "Honeycomb",
            "Ice Cream Sandwich", "Jelly Bean", "KitKat", "Lollipop", "Marshmallow"};

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        final WebView d3 = (WebView) findViewById(R.id.d3);



        WebSettings ws = d3.getSettings();
        ws.setJavaScriptEnabled(true);
        String span = "";
        String html = "<HTML><head>\n" +
                "    <meta charset=\"utf-8\">\n" +
                "    <meta name=\"viewport\" content=\"width=device-width, user-scalable=no\">\n" +
                "</head>\n" +
                "<body>\n";
        for (int i = 0; i < wordCloud.length; i++) {
            span = span + "<span  data-weight=\"" + 5 + "\" >" + wordCloud[i] + "</span><a href=\"example.com/tags/"+wordCloud[i] +"\">"+ wordCloud[i] +"</a><br/>\n";
        }
        html= html +span +
                "<span  data-weight=\"6\" > some example </span><br/>\n" +
                "</body></HTML>";

        d3.loadData(html, "text/html; charset=utf-8", "UTF-8");


        d3.setWebViewClient(new WebViewClient() {
            @Override
            public boolean shouldOverrideUrlLoading(WebView view, String url) {
                String[] parts = url.split("/");
                String tag=parts[2];
                Toast.makeText(getApplicationContext(),tag, Toast.LENGTH_LONG).show();
                return true;
            }


        });


    }

}

如何在不打开链接的情况下获取android webview中超链接的url

0 个答案: