如何URL转换为URI？ [英] How to convert URL to a URI?

问题描述

我想缓存使用此code这些图片...

但我一直在这里得到一个语法错误？

 乌里imageUri =新的URI（aURL）;
 

下面是code IM使用。

  URL aURL =新的URL（myRemoteImages [位置]）;                                乌里imageUri =新的URI（aURL）;
                                如果（新文件（新文件（myContext.getCacheDir（），缩略图），+ imageUri.hash code（））。存在（））
                                {
                                    字符串cachFile =+ imageUri.hash code（）;
                                    FIS的FileInputStream;                                    尝试{
                                        FIS =新的FileInputStream（cachFile）;
                                        位图BM = BitmapFactory.de codeStream（FIS）;
                                         i.setImageBitmap（BM）;
                                            i.setScaleType（ImageView.ScaleType.FIT_CENTER）;
                                            / *设置的ImageView的宽度/高度。 * /
                                            如果（Build.VERSION.SDK_INT＆GT; = 11）{
                                                i.setLayoutParams（新Gallery.LayoutParams（450，300））;
                                            }
                                            其他{
                                                i.setLayoutParams（新Gallery.LayoutParams（125，125））;
                                            }
                                    }赶上（FileNotFoundException异常五）{                                           Log.e（DEBUGTAG，Remtoe图像异常，E）;                                            / *为宽/高设置图片应缩放。 * /
                                            i.setScaleType（ImageView.ScaleType.FIT_CENTER）;
                                            / *设置的ImageView的宽度/高度。 * /
                                            如果（Build.VERSION.SDK_INT＆GT; = 11）{
                                                i.setLayoutParams（新Gallery.LayoutParams（450，300））;                                            返回我;
                                            }
                                                i.setLayoutParams（新Gallery.LayoutParams（125，125））;
                                                返回我;
                                            }
 

解决方案

阅读的文档。你不能与URL创建一个URI。使用类似

  URI URI =新的URI（aURL.toString（））;
 

捕获任何必要的例外ofcourse。

I am trying to cache these images using this code...

But i keep getting a syntax error here?

 Uri imageUri = new Uri(aURL);

Here is the code im using.

                                URL aURL = new URL(myRemoteImages[position]);



                                Uri imageUri = new Uri(aURL);
                                if (new File(new File(myContext.getCacheDir(), "thumbnails"), "" + imageUri.hashCode()).exists())
                                {
                                    String cachFile = ""+imageUri.hashCode();
                                    FileInputStream fis;

                                    try {
                                        fis = new FileInputStream(cachFile);
                                        Bitmap bm = BitmapFactory.decodeStream(fis); 
                                         i.setImageBitmap(bm);
                                            i.setScaleType(ImageView.ScaleType.FIT_CENTER);
                                            /* Set the Width/Height of the ImageView. */
                                            if(Build.VERSION.SDK_INT >= 11){
                                                i.setLayoutParams(new Gallery.LayoutParams(450, 300));
                                            }
                                            else{
                                                i.setLayoutParams(new Gallery.LayoutParams(125, 125));
                                            }


                                    } catch (FileNotFoundException e) {

                                           Log.e("DEBUGTAG", "Remtoe Image Exception", e);





                                            /* Image should be scaled as width/height are set. */
                                            i.setScaleType(ImageView.ScaleType.FIT_CENTER);
                                            /* Set the Width/Height of the ImageView. */
                                            if(Build.VERSION.SDK_INT >= 11){
                                                i.setLayoutParams(new Gallery.LayoutParams(450, 300));

                                            return i;
                                            }
                                                i.setLayoutParams(new Gallery.LayoutParams(125, 125));
                                                return i;
                                            }

解决方案

Read the docs. You can't create an URI with an URL. Use something like

URI uri = new URI(aURL.toString());

Catching any necessary exceptions ofcourse.

这篇关于如何URL转换为URI？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IT屋！