问题描述
当我尝试从URL加载图像,使用以下code(去掉实际图像路径):
When I attempt to load an image from a URL, using the following code (real image path removed):
Bitmap bitmap = BitmapFactory.decodeStream((InputStream)new URL("http://some-path/img.png").getContent());
我收到以下错误:
I receive the following error:
Error reading from ./org/apache/harmony/awt/www/content/image/png.class
这是什么可能导致错误有什么想法?
Any thoughts on what might be causing the error?
我使用的是GoogleTV的AVD,如果该事项。
I am using a GoogleTV AVD, if that matters.
推荐答案
我希望这将是足够的。
如果您使用的是PHP;
If you are using php;
echo base64_encode($imgBinary); // You can get the imagebinary by using the fread and fopen methods provided by php
在Android:
on android:
HttpClient client = new DefaultHttpClient();
HttpResponse response = client.execute(new HttpGet(url));
HttpEntity entity = httpResponse.getEntity();
if(entity != null) {
InputStream is = entity.getContent();
byte[] decodedString = Base64.decode(convertStreamToString(is), Base64.DEFAULT);
Bitmap decodedByte = BitmapFactory.decodeByteArray(decodedString, 0, decodedString.length);
}
这可能不是最有效的方式,但它应该做的工作。
从那里,你可以建立:)
This is probably not the most efficient way, but it should do the job.From there on you can build :)
可以COM preSS位图成后PNG和安全的。例如:
You can compress the bitmap into a PNG after, and safe it. example:
decodedByte.compress(compressformat, quality, stream);//suported compress formats can be used like so: Bitmap.CompressFormat.PNG etc
convertStreamToString是很容易找到的方法。只是做一个快速谷歌搜索,或写你自己的。
convertStreamToString are easily found methods. Just do a quick google search, or write your own.
这篇关于从错误读取png.class的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!