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是否有一种简单的方法可以为 Android TextView 组件创建自定义链接并处理它们的“触摸”行为?

我在互联网上没有找到任何解决方案,但我自己提供了。

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1 回答 1

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有一种简单的方法可以为 textview 创建自定义链接并根据触摸事件处理它们的行为。为了避免编写和编写自己的模式包装器,使用了 Html 包装器。

    TextView tView = ((TextView)v.findViewById(R.id.otp_activation_notification));
    Spanned ssBuilder =  Html.fromHtml("Not a link <a href=\"foo://haha/arg1/arg2?q1=1&q2=2\">The first link</a> bla bla "
    + " <a href=\"foo://haha2?q3=3\">The second link</a>");
    tView.setText(ssBuilder);
    tView.setMovementMethod(new LinkMovementMethod(){

        @Override
        public boolean onTouchEvent(TextView widget, Spannable buffer, MotionEvent event) {
            //TODO: In order to override the links actions
            int x = (int) event.getX();
            int y = (int) event.getY();

            Layout layout = widget.getLayout();
            int line = layout.getLineForVertical(y);
            int off = layout.getOffsetForHorizontal(line, x);

            try {
                URLSpan[] urlSpans = buffer.getSpans(off, off, URLSpan.class);
                if (urlSpans != null && urlSpans.length > 0) {
                    Uri uri = Uri.parse(urlSpans[0].getURL());
                    String scheme = uri.getScheme();
                    if ("foo".equals(scheme)) {
                        String command = uri.getAuthority();
                        if ("haha".equals(command)) {
                            List<String> arguments = uri.getPathSegments();
                            String q1 = uri.getQueryParameter("q1");
                            String q2 = uri.getQueryParameter("q2");
                            //TODO: Execute command (pay attention for MotionEvent)
                            return true;
                        } else if ("haha2".equals(command)) {
                            String q3 = uri.getQueryParameter("q2");
                            //TODO: Execute command2 (pay attention for MotionEvent)
                            return true;
                        }
                        return false;
                    }
                }
            } catch (Exception e) {
                //Log: unable to parse link;
            }
            //return false in case you don't want to use default behavior.
            return super.onTouchEvent(widget, buffer, event);

        }
    });
于 2017-09-03T14:35:03.370 回答