将图像从服务器检索到 android 应用程序

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时间:2020-08-20 04:08:41  来源:igfitidea点击:

Retriving image from server to android app

androidandroid-layoutandroid-serviceandroid-image

提问by user1389233

package com.sample.downloadImage;
import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.net.HttpURLConnection;
import java.net.URL;
import java.net.URLConnection;
import org.apache.http.util.ByteArrayBuffer;
import android.app.Activity;
import android.graphics.Bitmap;
import android.graphics.BitmapFactory;
import android.os.Bundle;
import android.widget.ImageView;

public class downloadImage extends Activity {

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

  Bitmap bitmap = DownloadImage("http://www.allindiaflorist.com/imgs/arrangemen4.jpg");


        ImageView img = (ImageView) findViewById(R.id.img);
        img.setImageBitmap(bitmap);
    }

    private InputStream OpenHttpConnection(String urlString) 
    throws IOException
    {
        InputStream in = null;
        int response = -1;

        URL url = new URL(urlString); 
        URLConnection conn = url.openConnection();

        if (!(conn instanceof HttpURLConnection))                     
            throw new IOException("Not an HTTP connection");

        try{
            HttpURLConnection httpConn = (HttpURLConnection) conn;
            httpConn.setAllowUserInteraction(false);
            httpConn.setInstanceFollowRedirects(true);
            httpConn.setRequestMethod("GET");
            httpConn.connect();
            response = httpConn.getResponseCode();                 
            if (response == HttpURLConnection.HTTP_OK) {
                in = httpConn.getInputStream();                                 
            }                     
        }
        catch (Exception ex)
        {
            throw new IOException("Error connecting");            
        }
        return in;     
    }
    private Bitmap DownloadImage(String URL)
    {        
        Bitmap bitmap = null;
        InputStream in = null;  



        try {
            in = OpenHttpConnection(URL);
            BufferedInputStream bis = new BufferedInputStream(in, 8190);

            ByteArrayBuffer baf = new ByteArrayBuffer(50);
            int current = 0;
            while ((current = bis.read()) != -1) 
            {
                baf.append((byte)current);
            }
            byte[] imageData = baf.toByteArray();
            bitmap =BitmapFactory.decodeByteArray(imageData, 0, imageData.length);
            in.close();
        } 
       catch (IOException e1) 
       {

            e1.printStackTrace();
        }
        return bitmap;                
    }
}

wanna retrive images from server , so i tried to post a image in server and retrive through url but it works good for small images and when it comes for big image more than 60kb , could some one give a idea to solve the problem

想从服务器检索图像,所以我尝试在服务器中发布图像并通过 url 检索,但它适用于小图像,当涉及超过 60kb 的大图像时,有人可以提出解决问题的想法

package com.sample.downloadImage;
import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.net.HttpURLConnection;
import java.net.URL;
import java.net.URLConnection;
import org.apache.http.util.ByteArrayBuffer;
import android.app.Activity;
import android.graphics.Bitmap;
import android.graphics.BitmapFactory;
import android.os.Bundle;
import android.widget.ImageView;

public class downloadImage extends Activity {

    HttpURLConnection httpConn;
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

  Bitmap bitmap = DownloadImage("http://www.allindiaflorist.com/imgs/arrangemen4.jpg");


        ImageView img = (ImageView) findViewById(R.id.img);
        img.setImageBitmap(bitmap);
    }

    private InputStream OpenHttpConnection(String urlString) 
    throws IOException
    {
        InputStream in = null;
        int response = -1;

        URL url = new URL(urlString); 
        URLConnection conn = url.openConnection();

        if (!(conn instanceof HttpURLConnection))                     
            throw new IOException("Not an HTTP connection");

        try{
             httpConn = (HttpURLConnection) conn;
            httpConn.setAllowUserInteraction(false);
            httpConn.setInstanceFollowRedirects(true);
            httpConn.setRequestMethod("GET");
            httpConn.connect();
            response = httpConn.getResponseCode();                 
            if (response == HttpURLConnection.HTTP_OK) {
                in = httpConn.getInputStream();  

                DownloadImage(urlString);
            }                     
        }
        catch (Exception ex)
        {
            throw new IOException("Error connecting");            
        }
        return in;     
    }

    private Bitmap DownloadImage(String URL)
    {        
        Bitmap bitmap = null;
        //InputStream is = null;  
        InputStream in;
        try
        {
            in = httpConn.getInputStream();
            BufferedInputStream bis = new BufferedInputStream(in, 3 *1024);
            ByteArrayBuffer baf = new ByteArrayBuffer(50);
            int current = 0;
            while ((current = bis.read()) != -1)
            {
                baf.append((byte)current);
                byte[] imageData = baf.toByteArray();
                bitmap =BitmapFactory.decodeByteArray(imageData, 0, imageData.length);
                return bitmap;     
            }
        }
        catch (IOException e) 
        {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        return bitmap;  
          }
}

回答by Dheeresh Singh

form : Load Large Image from server on Android

形式:从 Android 上的服务器加载大图像

it is not uncommon for BitmapFactory.decodeFromStream() to give up and just return null when you connect it directly to the InputStream of a remote connection. Internally, if you did not provide a BufferedInputStream to the method, it will wrap the supplied stream in one with a buffer size of 16384. One option that sometimes works is to pass a BufferedInputStream with a larger buffer size like:

当您将 Bi​​tmapFactory.decodeFromStream() 直接连接到远程连接的 InputStream 时,它放弃并返回 null 的情况并不少见。在内部,如果您没有为该方法提供 BufferedInputStream,它会将提供的流包装在一个缓冲区大小为 16384 的流中。有时可行的一种选择是传递一个具有更大缓冲区大小的 BufferedInputStream,例如:

BufferedInputStream bis = new BufferedInputStream(is, 32 * 1024); A more universally effective method is to download the file completely first, and then decode the data like this:

BufferedInputStream bis = new BufferedInputStream(is, 32 * 1024); 一种更普遍有效的方法是先完全下载文件,然后像这样解码数据:

InputStream is = connection.getInputStream();
BufferedInputStream bis = new BufferedInputStream(is, 8190);

ByteArrayBuffer baf = new ByteArrayBuffer(50);
int current = 0;
while ((current = bis.read()) != -1) {
    baf.append((byte)current);
}
byte[] imageData = baf.toByteArray();
BitmapFactory.decodeByteArray(imageData, 0, imageData.length);

FYI, the buffer sizes in this example are somewhat arbitrary. As has been said in other answers, it's a fantastic idea not to keep an image that size in memory longer than you have to. You might consider writing it directly to a file and displaying a downsampled version.

仅供参考,此示例中的缓冲区大小有些随意。正如在其他答案中所说的那样,不要在内存中保留该大小的图像的时间超过您需要的时间是一个绝妙的主意。您可能会考虑将其直接写入文件并显示降采样版本。

Hope that helps!

希望有帮助!

回答by i leaf

Look at this page and download the sample code. it will solve your problem

查看此页面并下载示例代码。它会解决你的问题

http://developer.android.com/training/displaying-bitmaps/index.html

http://developer.android.com/training/displaying-bitmaps/index.html