使用 PagerAdapter 的可点击链接

Question

我用 FragmentPagerAdapter 显示了一些文本。此文本包含一些我需要可点击的链接。有一个将 OnTouchEvent 与 View 一起使用的解决方案，但是我该如何获取坐标呢？或者PagerAdapter有更好的解决方案吗？

编辑：代码是：

private void setupPager() {
    if (pagerAdapter == null) {
        pagerAdapter = new TextFragmentPagerAdapter(getSupportFragmentManager());
    }
    pager.setAdapter(pagerAdapter);
    findViewById(R.id.progressBar1).setVisibility(View.GONE);
    pager.setCurrentItem(current_page);
    pager.setVisibility(View.VISIBLE);
}

private class TextFragmentPagerAdapter extends FragmentStatePagerAdapter {

    public TextFragmentPagerAdapter(FragmentManager fragmentManager) {
        super(fragmentManager);
    }

    @Override
    public Fragment getItem(int position) {
        spannable_page = null;
        Fragment fragment = new Fragment();
        new PageForm(position, line_array, spannable_main).start();
        while (true) {
            if (spannable_page != null) {
                fragment = PageFragment.newInstance(spannable_page.get());
                break;
            }
        }
        return fragment;
    }

    @Override
    public int getCount() {
        return total_pages;
    }
}

Inner class PageForm 像这样构建每个页面：

...
  Object[] spans_test = spannable.getSpans(start_range, end_range - 1, Object.class);
            CharSequence sequence_test = spannable.subSequence(start_range, end_range);
            SpannableStringBuilder builder_test = new SpannableStringBuilder();
            builder_test.insert(0, sequence_test);
            int sp_start, sp_end, flags, sp_s_fin, sp_e_fin;

            for (int i = 0; i < spans_test.length; i++) {
                sp_start = spannable.getSpanStart(spans_test[i]);
                sp_end = spannable.getSpanEnd(spans_test[i]);
                flags = spannable.getSpanFlags(spans_test[i]);
                sp_s_fin = sp_start - start_range;
                sp_e_fin = sp_end - start_range;
                if (sp_s_fin < 0) sp_s_fin = 0;
                if (sp_e_fin > builder_test.length() - 1) sp_e_fin = builder_test.length() - 1;
                builder_test.setSpan(spans_test[i], sp_s_fin, sp_e_fin, flags);
            }

            synchronized (this) {
                spannable_page = new WeakReference<Spannable>(builder_test);
            }

所以，我需要在每个页面中跟踪链接。

Answer 1

我找到了解决方案，而且比我想象的要容易。使用 PagerAdapter 使链接可点击的关键是将侦听器置于页面的 Fragment class.

public class PageFragment extends Fragment {
...
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
...


URLSpan[] links = spannable.getSpans(0, spannable.length(), URLSpan.class);
        for (URLSpan span : links) {
            textLinks(spannable, span);
        }
        textView.setText(spannable);
            textView.setMovementMethod(LinkMovementMethod.getInstance());
        return view;
    }


private void textLinks(Spannable text, final URLSpan span) {
        int start = text.getSpanStart(span);
        int end = text.getSpanEnd(span);
        int flags = text.getSpanFlags(span);
        ClickableSpan clickable = new ClickableSpan() {
            @Override
            public void onClick(View view) {
                System.out.println("LINK CHECK: " + span.getURL());
            }
        };
        text.setSpan(clickable, start, end, flags);
        text.removeSpan(span);
    }

使用 PagerAdapter 的可点击链接

Clickable links with PagerAdapter

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fragmentpageradapter

android-pageradapter