我在android上传。为此,我正在关注此tutorial。我当前的代码在我的logcat中给了我这个错误
06-23 10:10:22.990: D/dalvikvm(25853): GC_EXTERNAL_ALLOC freed 48K, 50% free 2723K/5379K, external 0K/0K, paused 33ms
06-23 10:10:23.030: E/log_tag(25853): Error in http connection java.net.UnknownHostException: www.example.info
06-23 10:10:23.115: D/CLIPBOARD(25853): Hide Clipboard dialog at Starting input: finished by someone else... !
以下是我的代码的样子
public class UploadFileActivity extends Activity {
/** Called when the activity is first created. */
Button bUpload;
EditText etParam;
InputStream is;
@Override
public void onCreate(Bundle icicle) {
super.onCreate(icicle);
setContentView(R.layout.main);
Bitmap bitmapOrg = BitmapFactory.decodeResource(getResources(), R.drawable.ic_launcher);
ByteArrayOutputStream bao = new ByteArrayOutputStream();
bitmapOrg.compress(Bitmap.CompressFormat.JPEG, 90, bao);
byte [] ba = bao.toByteArray();
String ba1=Base64.encodeToString(ba, 0);
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("image",ba1));
try {
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://www.example.info/androidfileupload/index.php");
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
is = entity.getContent();
} catch(Exception e) {
Log.e("log_tag", "Error in http connection "+e.toString());
}
}
}
以下是我在服务器端使用的内容
<?php
$base=$_REQUEST['image'];
echo $base;
// base64 encoded utf-8 string
$binary=base64_decode($base);
// binary, utf-8 bytes
header('Content-Type: bitmap; charset=utf-8');
// print($binary);
//$theFile = base64_decode($image_data);
$file = fopen('test.jpg', 'wb');
fwrite($file, $binary);
fclose($file);
echo '<img src=test.jpg>';
?>
答案 0 :(得分:2)
UnknownHostException
与您的网址域相关,或者没有互联网/网络连接,
因此,请确保您已在AndroidManifest文件中添加了Internet权限。
<uses-permission android:name="android.permission.INTERNET" />
希望这有助于
答案 1 :(得分:2)
我遵循了同样的教程。只要图像足够小就可以工作。
否则您可以在以下位置获得内存不足问题:
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
回答你的问题 -
您能告诉我如何将图像名称保留为原始图像 名称?我的意思是,我想在此时使用相同的图像名称 上传。
您可以将变量添加到包含原始名称的URL请求中。否则,您将无法将其嵌入到64Byte编码的字符串中。
这样的事情:
"http://www.example.info/androidfileupload/index.php?name=filename"
PHP方面看起来像
$FileName = $_GET['name'];