文件上传数据

Question

我有一个excel文件与各种各样的行和列。我上传这个文件并将数据 保存在数据库中。但我在解决这个问题时遇到问题并需要帮助： -

1）上传具有相同名称但具有不同内容的文件将上一个文件数据。

所以说，我上传abc.xls有3行它uplods 3行，但如果我上传另一个文件与10行具有相同的名称abc.xls它显示以前上传的文件的结果，并只显示3结果。

我试过了独特的文件，但后来我没有这样的文件或目录。

# Models.py 
    class ExcelFile(models.Model): 

     excel_file = models.FileField(upload_to='documents/') 
     class Meta: 
      verbose_name = 'ExcelFile' 
      verbose_name_plural = 'ExcelFiles' 


     def __unicode__(self): 
      return self.excel_file.name 

在管理员中，我看到文件已经是独一无二的，比如abc_1.xls，abc_2.xls等等。 这里是我的代码

#views.py 
    if request.method == "POST": 
      form_data = ImportExcelForm(request.POST, request.FILES) 
      if form_data.is_valid(): 
       cd = form_data.cleaned_data 
       file_obj, created = ExcelFile.objects.get_or_create(excel_file=cd['excel_file']) 
       try: 
        data_list = excel_parser(cd['excel_file'].name.replace(" ", "_")) 
       except: 
        data_list = excel_parser(get_correct_filename(cd['excel_file'])) 
########### and so on ####################### 

def get_correct_filename(filename): 
    replacements = {"(": "", ")": ""," ":"_"} 
    new_file = "".join(replacements.get(c, c) for c in filename.name) 
    return new_file 


def excel_parser(filename): 
    """ 
    Excel file will first come here. It will be read sheetwise. 
    This functions will return a data list. 
    """ 

    file_path = settings.MEDIA_ROOT + 'documents/' + filename 
    #Here it is reading abc.xls only as the filename is abc.xls 
    book = open_workbook(file_path) 
    data_list = [] 
    sheet_list = [] 

    total_sheets = book.nsheets 
    for sheet in range(total_sheets): 
     sheet_counter = book.sheet_by_index(sheet) 
     data_list = extract_data(book,sheet) 
     sheet_list.append(data_list) 
    return sheet_list 

Answer 1

你必须创建目录“上传/标识”到媒体目录。你可以手动或自动完成：

How to check if a directory exists and create it if necessary?