Question

我有以下代码：

def saveFile(self, master = None):
    f = asksaveasfile(mode='w',defaultextension='.csv')
    if f is None: # asksaveasfile returns `None` if dialog closed with "cancel".
        return
    f.close()
    cords2save = globalCords # coordinates from another csv file
    csvOpen = open(f, 'w')
    w = csv.writer(fp)
    w.writerow(cords2save)

当我跑步时，我得到：

Exception in Tkinter callback
Traceback (most recent call last):
File "C:\Users\Jem\AppData\Local\Programs\Python\Python35-     32\lib\tkinter\__init__.py", line 1550, in __call__
return self.func(*args)
File "C:\Users\Jem\Documents\bbk\1FINisd\LineSimplification.py", line 143, in saveFile
csvOpen = open(f, 'w')
TypeError: invalid file: <_io.TextIOWrapper     name='C:/Users/Jem/Documents/bbk/1FINisd/ard.csv' mode='w' encoding='cp1252'>

我真的被卡住，因为其他线程解决方案不起作用 - 我做错了什么？

由于

Answer 1

f = asksaveasfile(mode='w',defaultextension='.csv')

asksaveasfile返回一个文件对象。然后尝试在该文件对象上调用open。但open并不期望文件，它需要文件名。

请尝试使用asksaveasfilename。

def saveFile(self, master = None):
    filename = asksaveasfilename(mode='w',defaultextension='.csv')
    if not filename:
        return
    with open(filename, 'w') as file:
        w = csv.writer(file)
        w.writerow(globalCords)

或者，继续使用asksaveasfile，但不要关闭该文件，也不要尝试打开新文件。

def saveFile(self, master = None):
    f = asksaveasfile(mode='w',defaultextension='.csv')
    if not f:
        return
    w = csv.writer(f)
    w.writerow(globalCords)
    f.close()

Python中的CSV写入错误

1 个答案: