這是我的代碼,我在這里接受 Excel 檔案并讀取它。然后我讀取了行值的 Excel 檔案并將它們插入到資料庫中,然后插入到相應的列中。最后,我將檔案放在我在計算機上確定的檔案夾中。但是我在這里遇到了問題,當我想在api中多次下載同名檔案時,它表明我只是第一次將它上傳到檔案夾,但它會將值列印到資料庫中。我無法對此做出任何邏輯,你認為我能在這里做什么?
@Override
public ResponseEntity<? extends Response> acceptExcelFileAndSenderName(MultipartFile filePath, String senderName) throws IOException {
String fileName = filePath.getOriginalFilename();
if (fileName.substring(fileName.length() - 5).equals(".xlsx")) {
try (InputStream excelFile = filePath.getInputStream()) {
String phoneNumber = "";
String textMessage = "";
Workbook workbook = new XSSFWorkbook(excelFile);
Sheet datatypeSheet = workbook.getSheetAt(0);
Iterator<Row> iterator = datatypeSheet.iterator();
while (iterator.hasNext()) {
Row currentRow = iterator.next();
Iterator<Cell> cellIterator = currentRow.iterator();
while (cellIterator.hasNext()) {
Cell currentCell = cellIterator.next();
if (currentCell.getCellType() == CellType.NUMERIC) {
phoneNumber = NumberToTextConverter.toText(currentCell.getNumericCellValue());
} else if (currentCell.getCellType() == CellType.STRING) {
textMessage = String.valueOf(currentCell.getStringCellValue());
}
}
FileDetail fileDetail = new FileDetail();
fileDetail.setPhoneNumber(phoneNumber);
fileDetail.setTextMessage(textMessage);
fileDetail.setSender(senderName);
this.fileDetailRepository.save(fileDetail);
}
excelFile.close();
String destination = "C:\\Users\\anar.memmedov\\Desktop\\app\\" filePath.getOriginalFilename();
File file1 = new File(destination);
filePath.transferTo(file1);
return new ResponseEntity<>(new SuccessResponse(MessageCase.FILE_SUCCESSFULLY_WRITTEN_TO_DATABASE.getMessage(), 200), HttpStatus.OK);
} catch (IOException e) {
e.printStackTrace();
}
}
return new ResponseEntity<>(new ErrorResponse(MessageCase.FAILED_HAPPEND_WHEN_FILE_WRITTEN_TO_DATABASE.getMessage(), 400), HttpStatus.BAD_REQUEST);
}
uj5u.com熱心網友回復:
嘗試為每個檔案創建一個唯一的檔案名,例如:
String destination = "C:\\Users\\anar.memmedov\\Desktop\\app\\" System.nanoTime() "_" filePath.getOriginalFilename();
uj5u.com熱心網友回復:
我改變了一些東西并作業,它看起來像這樣
DateTimeFormatter dateTimeFormatter = DateTimeFormatter.ofPattern("yyyy-MM-dd-HH-mm-ss");
String destination = "C:\\Users\\anar.memmedov\\Desktop\\app\\" "_" LocalDateTime.now().format(dateTimeFormatter) "_" filePath.getOriginalFilename();
轉載請註明出處,本文鏈接:https://www.uj5u.com/ruanti/325084.html
下一篇:如何在Swagger中解決這個問題https://stackoverflow.com/questions/16015548/how-to-send-multipart-form-data-reques
