我完全迷失了，我正试图通过Android应用程序将照片发送到php网页。
理论上一切都应该是正确的，但目的地数据已损坏或我不知道......
我可以获得帖子数据，我尝试使用一个简单的字符串，它运行正常，但是文件很重，数据似乎已损坏。
public class EncodingAndSending extends Thread{
ShareOnMyWebSiteActivity mycontext;
ContentResolver cr;
Uri uri;
public Handler mainHandler,sHandler;
public EncodingAndSending(ShareOnMyWebSiteActivity myctxt,Handler mainone,Uri URI,ContentResolver crr){
mycontext=myctxt;
cr=crr;
uri=URI;
mainHandler=mainone;
this.start();
}
return buffer.toString().toUpperCase();
}
public void run(){
InputStream is=null;
byte[] data=null;
try {
is = cr.openInputStream(uri);
// Get binary bytes for encode
data = getFileBytes(is);
} catch (Exception e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
String data_string=Base64.encodeToString(data,Base64.URL_SAFE);
if(data_string!=""){
SendRequest(data_string);
}
else{
}
}
public byte[] getFileBytes(InputStream ios) throws IOException {
ByteArrayOutputStream ous = null;
//InputStream ios = null;
try {
byte[] buffer = new byte[4096];
ous = new ByteArrayOutputStream();
//ios = new FileInputStream(file);
int read = 0;
while ((read = ios.read(buffer)) != -1)
ous.write(buffer, 0, read);
} finally {
try {
if (ous != null)
ous.close();
} catch (IOException e) {
// swallow, since not that important
}
try {
if (ios != null)
ios.close();
} catch (IOException e) {
// swallow, since not that important
}
}
return ous.toByteArray();
}
private void SendRequest(String data_string){
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("xxxxx.php");
try {
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("image", data_string));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
// Execute HTTP Post Request
HttpResponse response = httpclient.execute(httppost);
ResponseHandler<String> responseHandler=new BasicResponseHandler();
String responseBody = httpclient.execute(httppost, responseHandler);
} catch (Exception e) {
// TODO Auto-generated catch block
}
}
}
编辑：
这很有效。我可以编码和解码并预览图像。我没有使用getBytes（）函数，我不知道问题是否来自那里。
我会告诉你的。
super.onCreate(savedInstanceState);
setContentView(R.layout.test_image);
ImageView image = (ImageView) findViewById(R.id.imageView1);
FileInputStream in;
BufferedInputStream buf;
Intent intent = getIntent();
Bundle extras = intent.getExtras();
Uri uri = (Uri) extras.getParcelable(Intent.EXTRA_STREAM);
ContentResolver cr = getContentResolver();
Bitmap bMap=null;
try {
InputStream is = cr.openInputStream(uri);
bMap = BitmapFactory.decodeStream(is);
if (is != null) {
is.close();
}
} catch (Exception e) {
Log.e("Error reading file", e.toString());
}
ByteArrayOutputStream baos = new ByteArrayOutputStream();
bMap.compress(Bitmap.CompressFormat.JPEG, 100, baos);
byte[] b = baos.toByteArray();
String data_string=Base64.encodeToString(b,Base64.DEFAULT);
b=null;bMap=null;
b=Base64.decode(data_string, Base64.DEFAULT);
bMap=BitmapFactory.decodeByteArray(b, 0, b.length);
image.setImageBitmap(bMap);
答案 0 :(得分：2)
我写了一个小应用程序，允许将手机摄像头拍摄的图像发送到数据库。这就是我解决问题的方法......
public void writeCommentForRestaurant(int id, String comment, String author,
Bitmap image) {
if (image != null) {
/* Get the image as string */
// Normal
ByteArrayOutputStream full_stream = new ByteArrayOutputStream();
image.compress(Bitmap.CompressFormat.PNG, 100, full_stream);
byte[] full_bytes = full_stream.toByteArray();
String img_full = Base64.encodeToString(full_bytes, Base64.DEFAULT);
// Thumbnail
ByteArrayOutputStream thumb_stream = new ByteArrayOutputStream();
// The getScaledBitmap method only minimizes the Bitmap to a small icon!
getScaledBitmap(image, 72).compress(Bitmap.CompressFormat.JPEG, 75,
thumb_stream);
byte[] thumb_bytes = thumb_stream.toByteArray();
String img_thumbnail = Base64.encodeToString(thumb_bytes, Base64.DEFAULT);
// new HTTPWorker(ctx, mHandler, HTTPWorker.WRITE_COMMENT, true).execute(
// Integer.toString(id), comment, author, img_thumbnail, img_full);
} else {
// new HTTPWorker(ctx, mHandler, HTTPWorker.WRITE_COMMENT, true).execute(
// Integer.toString(id), comment, author, null, null);
}
}
HTTPWorker只是一个构造HTTP方法的异步任务。
...
/* Add arguments */
arguments.add(new BasicNameValuePair("idrestaurant", params[0]));
arguments.add(new BasicNameValuePair("comment", params[1]));
arguments.add(new BasicNameValuePair("author", params[2]));
if (params.length > 3) {
arguments.add(new BasicNameValuePair("image", params[3]));
arguments.add(new BasicNameValuePair("bigimage", params[4]));
}
...
然后我把它发送到这样的服务器。
/**
* Executes a httppost to a server instance with the given POST arguments
* and returns a String response from the server.
*/
private String httppost(String url, ArrayList<NameValuePair> args) {
/* Create the channel for communicaton */
InputStream is = null;
/* Send request to server */
try {
/* Create the POST */
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);
/* Add the login information "POST" variables in the php */
httppost.setEntity(new UrlEncodedFormEntity(args));
/* Execute the http POST and get the response */
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
is = entity.getContent();
} catch (Exception e) {
Log.e(TAG, "Error in http connection " + e.toString());
return null;
}
/* Read response from server */
try {
/* Read the response stream */
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
/* Copy the response to StringBuilder */
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
/* Return the response as string */
return sb.toString();
} catch (Exception e) {
Log.e(TAG, "Error converting result " + e.toString());
return null;
}
}
答案 1 :(得分：1)
非常感谢你。实际上，问题来自服务器。 base64被发送到一个php脚本，该脚本将字符串存储到一个字段...这是一个longtext类型（mysql）。事情是破坏了数据，我把它改成blob并且工作正常。典型的不是吗？
非常感谢你的帮助。
我的第一个主题和一个伟大的社区在这里很棒。
稍后见。