Question

我有一个包含各种列和行的excel文件。我上传此文件并保存数据在数据库中。但我遇到了问题，需要帮助解决这个问题： -

1）上传具有相同名称但内容不同的文件上传以前的文件数据。

所以，假设我上传abc.xls有3行，它上行3行，但如果我上传另一个文件，其中10行具有相同的名称abc.xls，它会显示以前上传的文件的结果，只显示3个结果。< / p>

我尝试过像我这样的文件或目录。但是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，我看到文件已经是唯一的，如abc_1.xls，abc_2.xls等。这是我的代码

   #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

您必须在媒体目录中创建目录'uploads / logos'。你可以手动或自动完成：

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

文件上传数据

1 个答案: