在Android的视频上传 [英] video upload in Android
本文介绍了在Android的视频上传的处理方法，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
我试图将视频上传到服务器，在这里我有存储在资产文件夹中的视频，并试图upload.i使用了下列code：
私人无效uploadVideo（）{
字符串upLoadServerUri =HTTP：//xxxxx/video_upload.php
//的String [] =串sourceFileUri;
字符串文件名=文件：/// android_asset / clip0003
HttpURLConnection的康恩= NULL;
DataOutputStream类DOS = NULL;
DataInputStream以inStream中= NULL;
字符串lineEnd =\\ r \\ n;
串twoHyphens = - ;
字符串边界=*****;
INT读取动作，方bytesAvailable，缓冲区大小;
INT serverResponse code = 1;
字节[]缓冲区;
INT MAXBUFFERSIZE = 1 * 1024 * 1024;
串responseFromServer =; 文件的资源文件=新的文件（文件名）;
如果（！sourceFile.isFile（））{
Log.e（Huzza，源文件不存在）; } 尝试{//打开一个URL连接到这个Servlet 的FileInputStream的FileInputStream =新的FileInputStream（新文件（文件名））;
网址URL =新的URL（upLoadServerUri）;
康恩=（HttpURLConnection类）url.openConnection（）; //打开HTTP连接的URL
conn.setDoInput（真）; //允许输入
conn.setDoOutput（真）; //允许输出
conn.setUseCaches（假）; //不要使用缓存副本
conn.setRequestMethod（POST）;
conn.setRequestProperty（连接，保持活动）;
conn.setRequestProperty（ENCTYPE，多部分/表单数据）;
conn.setRequestProperty（内容类型，的multipart / form-data的;边界=+边界）;
conn.setRequestProperty（视频文件名）;
DOS =新的DataOutputStream类（conn.getOutputStream（））; dos.writeBytes（twoHyphens +边界+ lineEnd）;
dos.writeBytes（内容处置：表格数据;名称= \\uploaded_file \\;文件名= \\+文件名+\\+ lineEnd）;
dos.writeBytes（lineEnd）; 参考bytesAvailable = fileInputStream.available（）; //创建最大大小的缓冲区
Log.i（Huzza，初始.available：+信息bytesAvailable）; BUFFERSIZE = Math.min（方bytesAvailable，MAXBUFFERSIZE）;
缓冲区=新的字节[缓冲区大小] //读取文件，并将其写入形式...
读取动作= fileInputStream.read（缓冲液，0，缓冲区大小）; 而（读取动作大于0）{
dos.write（缓冲液，0，缓冲区大小）;
参考bytesAvailable = fileInputStream.available（）;
BUFFERSIZE = Math.min（方bytesAvailable，MAXBUFFERSIZE）;
读取动作= fileInputStream.read（缓冲液，0，缓冲区大小）;
} //发送文件数据后necesssary多部分的表单数据...
dos.writeBytes（lineEnd）;
dos.writeBytes（twoHyphens +边界+ twoHyphens + lineEnd）; //服务器的响应（code和消息）
serverResponse code = conn.getResponse code（）;
串serverResponseMessage = conn.getResponseMessage（）; Log.i（上传文件到服务器，HTTP响应是：+ serverResponseMessage +：+ serverResponse code）;
//关闭流
Log.i（上传文件到服务器，文件名+文件写入）;
fileInputStream.close（）;
dos.flush（）;
dos.close（）;
}赶上（MalformedURLException的前）{
ex.printStackTrace（）;
Log.e（上传文件到服务器，错误：+ ex.getMessage（），除息）;
}赶上（例外五）{
e.printStackTrace（）;
}
//这个块会给上传链接的响应
尝试{
RD的BufferedReader =新的BufferedReader（新的InputStreamReader（conn.getInputStream（）））;
串线;
而（（行= rd.readLine（））！= NULL）{
Log.i（Huzza，RES消息：+线）;
}
rd.close（）;
}赶上（IOException异常ioex）{
Log.e（Huzza，错误：+ ioex.getMessage（），ioex）;
}
} //结束upLoad2Server
但我不能能够上传由于以下错误！可能有人帮助我.Thanks！
错误：
03-20 18：09：07.033：E / AndroidRuntime（1529）：致命异常：主要
03-20 18：09：07.033：E / AndroidRuntime（1529）：显示java.lang.NullPointerException
03-20 18：09：07.033：E / AndroidRuntime（1529）：在com.example.galleryframe.videoupload.uploadVideo（videoupload.java:233）
03-20 18：09：07.033：E / AndroidRuntime（1529）：在com.example.galleryframe.videoupload.access $ 0（videoupload.java:150）
03-20 18：09：07.033：E / AndroidRuntime（1529）：在com.example.galleryframe.videoupload $ 3.onClick（videoupload.java:114）
03-20 18：09：07.033：E / AndroidRuntime（1529）：在android.view.View.performClick（View.java:2485）
03-20 18：09：07.033：E / AndroidRuntime（1529）：在android.view.View $ PerformClick.run（View.java:9080）
03-20 18：09：07.033：E / AndroidRuntime（1529）：在android.os.Handler.handleCallback（Handler.java:587）
03-20 18：09：07.033：E / AndroidRuntime（1529）：在android.os.Handler.dispatchMessage（Handler.java:92）
03-20 18：09：07.033：E / AndroidRuntime（1529）：在android.os.Looper.loop（Looper.java:123）
03-20 18：09：07.033：E / AndroidRuntime（1529）：在android.app.ActivityThread.main（ActivityThread.java:3683）
03-20 18：09：07.033：E / AndroidRuntime（1529）：在java.lang.reflect.Method.invokeNative（本机方法）
03-20 18：09：07.033：E / AndroidRuntime（1529）：在java.lang.reflect.Method.invoke（Method.java:507）
03-20 18：09：07.033：E / AndroidRuntime（1529）：在com.android.internal.os.ZygoteInit $ MethodAndArgsCaller.run（ZygoteInit.java:839）
03-20 18：09：07.033：E / AndroidRuntime（1529）：在com.android.internal.os.ZygoteInit.main（ZygoteInit.java:597）
03-20 18：09：07.033：E / AndroidRuntime（1529）：在dalvik.system.NativeStart.main（本机方法）
解决方案
不知道这线是233，我将做一个猜测，说这个问题行
的FileInputStream的FileInputStream =新的FileInputStream（新文件（文件名））;
您文件名被定义为一个URI，但你想创建一个输入流，它作为一个文件。我怀疑这个输入流是空。通过这次与调试，以确认步骤。
编辑：如果您想阅读您的文件资产文件夹，使用 getAssets（）方法：
字符串文件名=clip0003;
InputSteam fileInputSteam =新的InputStreamReader（getAssets（）打开（文件名）。）;
请注意，这code假设你是一个活动的内部扩展或其它类上下文 。如果没有，你需要访问上下文和呼叫 context.getAssets（）。
I am trying to upload video to server ,here i have store the video in assets folder,and trying to upload.i have used the following code:
private void uploadVideo() {
String upLoadServerUri = "http://xxxxx/video_upload.php";
// String [] string = sourceFileUri;
String fileName = "file:///android_asset/clip0003";
HttpURLConnection conn = null;
DataOutputStream dos = null;
DataInputStream inStream = null;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
int serverResponseCode=1;
byte[] buffer;
int maxBufferSize = 1 * 1024 * 1024;
String responseFromServer = "";
File sourceFile = new File(fileName);
if (!sourceFile.isFile()) {
Log.e("Huzza", "Source File Does not exist");
}
try { // open a URL connection to the Servlet
FileInputStream fileInputStream = new FileInputStream(new File(fileName));
URL url = new URL(upLoadServerUri);
conn = (HttpURLConnection) url.openConnection(); // Open a HTTP connection to the URL
conn.setDoInput(true); // Allow Inputs
conn.setDoOutput(true); // Allow Outputs
conn.setUseCaches(false); // Don't use a Cached Copy
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("ENCTYPE", "multipart/form-data");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
conn.setRequestProperty("video", fileName);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"uploaded_file\";filename=\""+ fileName + "\"" + lineEnd);
dos.writeBytes(lineEnd);
bytesAvailable = fileInputStream.available(); // create a buffer of maximum size
Log.i("Huzza", "Initial .available : " + bytesAvailable);
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// read file and write it into form...
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
dos.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// Responses from the server (code and message)
serverResponseCode = conn.getResponseCode();
String serverResponseMessage = conn.getResponseMessage();
Log.i("Upload file to server", "HTTP Response is : " + serverResponseMessage + ": " + serverResponseCode);
// close streams
Log.i("Upload file to server", fileName + " File is written");
fileInputStream.close();
dos.flush();
dos.close();
} catch (MalformedURLException ex) {
ex.printStackTrace();
Log.e("Upload file to server", "error: " + ex.getMessage(), ex);
} catch (Exception e) {
e.printStackTrace();
}
//this block will give the response of upload link
try {
BufferedReader rd = new BufferedReader(new InputStreamReader(conn.getInputStream()));
String line;
while ((line = rd.readLine()) != null) {
Log.i("Huzza", "RES Message: " + line);
}
rd.close();
} catch (IOException ioex) {
Log.e("Huzza", "error: " + ioex.getMessage(), ioex);
}
} // end upLoad2Server
But i cannot able to upload due to following error!! could anybody help me .Thanks !!!
Error:
03-20 18:09:07.033: E/AndroidRuntime(1529): FATAL EXCEPTION: main
03-20 18:09:07.033: E/AndroidRuntime(1529): java.lang.NullPointerException
03-20 18:09:07.033: E/AndroidRuntime(1529): at com.example.galleryframe.videoupload.uploadVideo(videoupload.java:233)
03-20 18:09:07.033: E/AndroidRuntime(1529): at com.example.galleryframe.videoupload.access$0(videoupload.java:150)
03-20 18:09:07.033: E/AndroidRuntime(1529): at com.example.galleryframe.videoupload$3.onClick(videoupload.java:114)
03-20 18:09:07.033: E/AndroidRuntime(1529): at android.view.View.performClick(View.java:2485)
03-20 18:09:07.033: E/AndroidRuntime(1529): at android.view.View$PerformClick.run(View.java:9080)
03-20 18:09:07.033: E/AndroidRuntime(1529): at android.os.Handler.handleCallback(Handler.java:587)
03-20 18:09:07.033: E/AndroidRuntime(1529): at android.os.Handler.dispatchMessage(Handler.java:92)
03-20 18:09:07.033: E/AndroidRuntime(1529): at android.os.Looper.loop(Looper.java:123)
03-20 18:09:07.033: E/AndroidRuntime(1529): at android.app.ActivityThread.main(ActivityThread.java:3683)
03-20 18:09:07.033: E/AndroidRuntime(1529): at java.lang.reflect.Method.invokeNative(Native Method)
03-20 18:09:07.033: E/AndroidRuntime(1529): at java.lang.reflect.Method.invoke(Method.java:507)
03-20 18:09:07.033: E/AndroidRuntime(1529): at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:839)
03-20 18:09:07.033: E/AndroidRuntime(1529): at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:597)
03-20 18:09:07.033: E/AndroidRuntime(1529): at dalvik.system.NativeStart.main(Native Method)
解决方案
Without knowing which line is 233, I'll make a guess and say that the problem line is
FileInputStream fileInputStream = new FileInputStream(new File(fileName));
Your filename is defined as a URI, but you're trying to create an input stream to it as a File. I suspect this input stream is null. Step through this with a debugger to confirm.
EDIT: If you want to read a file from you assets folder, use getAssets() method:
String fileName = "clip0003";
InputSteam fileInputSteam = new InputStreamReader(getAssets().open(filename));
Note that this code assumes that you're inside an Activity or other class that extends Context. If not, you'll need access to a Context and call context.getAssets().
这篇关于在Android的视频上传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！
查看全文
