Question

我使用StaticLayout将HTML中的文本绘制到画布上。我需要弄清楚的是我如何在文本中点击链接。我认为最简单的方法是获取文本中链接的坐标，并制作视图上的自定义点击事件。

到目前为止，这就是我的上帝：

Spanned mySpannedHTML = Html.fromHtml(html_text, null, null);

StaticLayout sl = new StaticLayout(mySpannedHTML, mWhite, mWidth, Layout.Alignment.ALIGN_NORMAL, 1.0f, 0.0f, false);

sl.draw(canvas);

//FINE SO FAR, but how to get the coorinates of the links?


//GETTING THE LINKS

URLSpan[] getUrls =((Spanned) sl.getText()).getSpans(0, sl.getText().length(), URLSpan.class);

for (URLSpan urlSpan : getUrls) {

                int start =((Spanned) sl.getText()).getSpanStart(urlSpan);
                int end =((Spanned) sl.getText()).getSpanEnd(urlSpan);

//LOG LINKS URL AND LINK TEXT

                Log.i("spans", "url: "+urlSpan.getURL() + " Link text: "+ ((Spanned) sl.getText()).subSequence(start, end) + " ");

            }

所以，我真的卡在这里......之所以我需要文本中链接的坐标，是因为我将画布加载到openGL纹理上进行渲染。说得通？

Answer 1

我创建了一个函数来从调用者传递MotionEvent和View。然后我得到urlSpan的路径，获取边界（参见rectF）并将偏移量添加到矩形（因为我的staticlayout没有覆盖整个屏幕，而rectF显示静态布局内部的位置）生成的rectF给出了URLSpan的边界。

 public boolean onTextTouched(View widget,MotionEvent event)
{
    //If the text contains an url, we check its location and if it is touched: fire the click event.
    Spanned spanned = (Spanned)text;
    URLSpan[] urls = spanned.getSpans(0,spanned.length(),URLSpan.class);
    for(int i=0;i<urls.length;i++)
    {
        //get the start and end points of url span
        int start=spanned.getSpanStart(urls[i]);
        int end=spanned.getSpanEnd(urls[i]);

        Path dest = new Path();
        mStaticLayout.getSelectionPath(start, end, dest);

        RectF rectF = new RectF();
        dest.computeBounds(rectF, true);

        //Add the left and top margins of your staticLayout here.
        rectF.offset(getContentLeft(),getContentTop());

        if(rectF.contains(event.getX(),event.getY()))
        {
            urls[i].onClick(widget);
            return true;
        }
    }
    return false;
}

Answer 2

使用paint并使用measureText确定x坐标，然后使用行号和上升和下降来确定y。

获取staticlayout中的链接位置

2 个答案: