-3

私のアプリケーションでは、csvファイルをサーバーに送信する必要があります。次のコードを試しました

            HttpPost httppost = new HttpPost(url);

        InputStreamEntity reqEntity = new InputStreamEntity(
                new FileInputStream(file), -1);
        reqEntity.setContentType("binary/octet-stream");
        reqEntity.setChunked(true); // Send in multiple parts if needed
        httppost.setEntity(reqEntity);
        HttpResponse response = httpclient.execute(httppost);

私のphpコードは..

<?php


if ($_FILES["detection"]["error"] > 0)
{
echo "Return Code: " . $_FILES["detection"]["error"] . "<br>";
 }

そうしないと {

 if (file_exists($_FILES["detection"]["name"]))
  {
   echo $_FILES["detection"]["name"] . " already exists. ";
  }
 else
   {
  move_uploaded_file($_FILES["detection"]["tmp_name"],$_FILES["detection"]["name"]);
  echo "Stored in: ". $_FILES["detection"]["name"];
   }
 }

?>

エラーが発生しました

08-26 17:29:18.318: I/edit user profile(700):
08-26 17:29:18.318: I/edit user profile(700):注意: 未定義のインデックス: C:\xampp\htdocs\での検出sendreport.phpの4行目

4

1 に答える 1

4

うまくいくことを願っています

 // the file to be posted
String textFile = Environment.getExternalStorageDirectory() + "/sample.txt";
 Log.v(TAG, "textFile: " + textFile);

 // the URL where the file will be posted
 String postReceiverUrl = "http://yourdomain.com/post_data_receiver.php";
Log.v(TAG, "postURL: " + postReceiverUrl);

 // new HttpClient
 HttpClient httpClient = new DefaultHttpClient();

// post header
HttpPost httpPost = new HttpPost(postReceiverUrl);

File file = new File(textFile);
FileBody fileBody = new FileBody(file);

MultipartEntity reqEntity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
reqEntity.addPart("file", fileBody);
httpPost.setEntity(reqEntity);

// execute HTTP post request
HttpResponse response = httpClient.execute(httpPost);
HttpEntity resEntity = response.getEntity();

if (resEntity != null) {

String responseStr = EntityUtils.toString(resEntity).trim();
Log.v(TAG, "Response: " +  responseStr);

// you can add an if statement here and do other actions based on the response
}

and php code.
<?php
// if text data was posted
if($_POST){
print_r($_POST);
}

 // if a file was posted
 else if($_FILES){
 $file = $_FILES['file'];
 $fileContents = file_get_contents($file["tmp_name"]);
 print_r($fileContents);
 }
 ?>
于 2013-08-27T07:57:53.510 に答える