下面的代码成功将图像发送到服务器。
但是我需要的是向数据库添加图像描述和图像日期等参数。
我不知道如何在 HttURLconnection 中添加参数。
String fileName = sourceFileUri;
HttpURLConnection conn = null;
DataOutputStream dos = null;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1 * 1024 * 1024;
File sourceFile = new File(sourceFileUri);
FileInputStream fileInputStream = new FileInputStream(sourceFile_imagepath);
URL url = new URL(upLoadServerUri);
conn = (HttpURLConnection) url.openConnection();
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("uploaded_file", 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);
// create a buffer of maximum size
bytesAvailable = fileInputStream.available();
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("uploadFile", "HTTP Response is : "
+ serverResponseMessage + ": " + serverResponseCode);
if (serverResponseCode == 200) {
runOnUiThread(new Runnable() {
public void run() {
String msg = "File Upload Completed.\n\n See uploaded file here : \n\n"
+ " c:/wamp/www/echo/uploads";
messageText.setText(msg);
Toast.makeText(MainActivity.this,
"File Upload Complete.", Toast.LENGTH_SHORT)
.show();
}
});
}
// close the streams //
fileInputStream.close();
dos.flush();
dos.close();
和 PHP 代码:
<?php
error_reporting(0);
$file_path = "uploads/";
$file_path = $file_path . basename( $_FILES['uploaded_file']['name']);
if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $file_path)) {
echo "success";
} else{
echo "fail";
}
?>
上面的代码可以很好地将代码上传到服务器。
必须将图像描述存储在数据库中。
如何向上述代码添加参数。
以及如何在php服务器端接收它。
任何相关教程或文档。
提前致谢。
最佳答案
检查这个link
Android 代码:
字符串参数=“值”;
dos.writeBytes("Content-Disposition: form-data; name=\"parameter\"" + lineEnd);
//dos.writeBytes("Content-Type: text/plain; charset=UTF-8" + lineEnd);
//dos.writeBytes("Content-Length: " + parameter.length() + lineEnd);
dos.writeBytes(lineEnd);
dos.writeBytes(parameter); // mobile_no is String variable
dos.writeBytes(lineEnd);
PHP 代码:
$parameter =$_POST['parameter'];
关于java - 如何在httpurlconnection android中添加参数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26608433/