Android的广场毕加索不会加载波斯人物形象网址 [英] Android Square Picasso not load persian character image url
本文介绍了Android的广场毕加索不会加载波斯人物形象网址的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我要创建由方毕加索的Web应用程序,但如果图像的URL中包含波斯字符(ا,ب,ج,ی...)毕加索不会加载图像。
此网址不工作:
Picasso.with(mContext).load(\"http://www.shutterstock.ir/thumbs/10006/74601661-گربه-چشم-ابی-ولاغر-سیامی-در-یک-پس-زمینه-،-وکتور-سفید.jpg\")
.placeholder(R.drawable.ic_launcher)
.error(R.drawable.face_top_image).noFade()。调整(100,100)
.into(ImageView的);
此网址工作
Picasso.with(mContext).load(\"http://www.shutterstock.ir/thumbs/10006/74601661-%DA%AF%D8%B1%D8%A8%D9%87-%DA%86%D8%B4%D9%85-%D8%A7%D8%A8%DB%8C-%D9%88%D9%84%D8%A7%D8%BA%D8%B1-%D8%B3%DB%8C%D8%A7%D9%85%DB%8C-%D8%AF%D8%B1-%DB%8C%DA%A9-%D9%BE%D8%B3-%D8%B2%D9%85%DB%8C%D9%86%D9%87-%D8%8C-%D9%88%DA%A9%D8%AA%D9%88%D8%B1-%D8%B3%D9%81%DB%8C%D8%AF.jpg\")
.placeholder(R.drawable.ic_launcher)
.error(R.drawable.face_top_image).noFade()。调整(100,100)
.into(ImageView的);
解决方案
您需要URI EN code中的URL。
查看文档
Uri.en code(URL);
或者,如果指定某些允许的字符了以下工作:
私有静态最后弦乐ALLOWED_URI_CHARS =@#&安培; = * + -_,:!?()/〜'%;
字符串urlEn codeD = Uri.en code(路径,ALLOWED_URI_CHARS);
I want to create web application by square picasso, but if image url contains persian characters (ا،ب،ج،ی، ...) Picasso not load image.
This url not working:
Picasso.with(mContext).load("http://www.shutterstock.ir/thumbs/10006/74601661-گربه-چشم-ابی-ولاغر-سیامی-در-یک-پس-زمینه-،-وکتور-سفید.jpg")
.placeholder(R.drawable.ic_launcher)
.error(R.drawable.face_top_image).noFade().resize(100, 100)
.into(imageView);
This url work
Picasso.with(mContext).load("http://www.shutterstock.ir/thumbs/10006/74601661-%DA%AF%D8%B1%D8%A8%D9%87-%DA%86%D8%B4%D9%85-%D8%A7%D8%A8%DB%8C-%D9%88%D9%84%D8%A7%D8%BA%D8%B1-%D8%B3%DB%8C%D8%A7%D9%85%DB%8C-%D8%AF%D8%B1-%DB%8C%DA%A9-%D9%BE%D8%B3-%D8%B2%D9%85%DB%8C%D9%86%D9%87-%D8%8C-%D9%88%DA%A9%D8%AA%D9%88%D8%B1-%D8%B3%D9%81%DB%8C%D8%AF.jpg")
.placeholder(R.drawable.ic_launcher)
.error(R.drawable.face_top_image).noFade().resize(100, 100)
.into(imageView);
解决方案
You need to URI encode the URL.
See the docs
Uri.encode(url);
Or, if specifying certain allowed characters the following works:
private static final String ALLOWED_URI_CHARS = "@#&=*+-_.,:!?()/~'%";
String urlEncoded = Uri.encode(path, ALLOWED_URI_CHARS);
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