0
我在獲取文件路徑時遇到了一些問題,以便可以從指定的(文本)文件打開並執行我的數據。下面是迄今爲止我所編寫的代碼:在Python中獲取文件路徑2.7
def pickfile():
options={}
options['defaultextension'] = '.txt'
options['filetypes'] = [('all files','.*'), ('text files', '.*txt')]
options['initialfile'] = 'sample.txt'
options['initialdir'] = 'C:\Users\profcs\Desktop'
filename=open(tkFileDialog.askopenfilename(**options))
if filename:
print(filename)
return
with open(filename, 'rb') as f:
reader = csv.reader(f)
try:
for row in reader:
print row
except csv.Error as e:
sys.exit('file %s, line %d: %s' % (filename, reader.line_num,e))
but1 = Button(widget1, text='Pick Your File', command=pickfile)
but1.pack(side=BOTTOM, padx=10, pady=1, anchor=SE)
but1.config(relief=RAISED, bd=2)
當我顯示一個文件名,我現在得到這種形式的路徑:
================ RESTART: C:\Users\profcs\Desktop\BD TEST.py ================
<open file u'C:/Users/profcs/Desktop/sample.txt', mode 'r' at 0x01EFF128>
如何過濾這條道路,只有獲得'C:/Users/profcs/Desktop/sample.txt'
這樣我可以打開我的文件?
在此先感謝。