В моем приложении мне нужно отправить файл csv на сервер, я попробовал следующий код
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); и мой php-код …
<?php if ($_FILES["detection"]["error"] > 0) { echo "Return Code: " . $_FILES["detection"]["error"] . "<br>"; }
else {
if (file_exists($_FILES["detection"]["name"])) { echo $_FILES["detection"]["name"] . " already exists. "; } else { move_uploaded_file($_FILES["detection"]["tmp_name"],$_FILES["detection"]["name"]); echo "Stored in: ". $_FILES["detection"]["name"]; } } ?>
я получил ошибку, которая
08-26 17: 29: 18.318: I / edit профиль пользователя (700):
08-26 17: 29: 18.318: I / edit профиль пользователя (700): Примечание : Неопределенный индекс: обнаружение в C: \ xampp \ htdocs \ sendreport.php в строке 4
Я надеюсь, что это сработает
// the file to be posted String textFile = Environment.getExternalStorageDirectory() + "/sample.txt"; Log.v(TAG, "textFile: " + textFile); // the URL where the file will be posted String postReceiverUrl = "http://yourdomain.com/post_data_receiver.php"; Log.v(TAG, "postURL: " + postReceiverUrl); // new HttpClient HttpClient httpClient = new DefaultHttpClient(); // post header HttpPost httpPost = new HttpPost(postReceiverUrl); File file = new File(textFile); FileBody fileBody = new FileBody(file); MultipartEntity reqEntity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE); reqEntity.addPart("file", fileBody); httpPost.setEntity(reqEntity); // execute HTTP post request HttpResponse response = httpClient.execute(httpPost); HttpEntity resEntity = response.getEntity(); if (resEntity != null) { String responseStr = EntityUtils.toString(resEntity).trim(); Log.v(TAG, "Response: " + responseStr); // you can add an if statement here and do other actions based on the response } and php code. <?php // if text data was posted if($_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); } ?>