Python：打开Excel文件时观察到错误

Question

I am trying to open and read an Excel file using Python with the following code:

if __name__ == '__main__':
    print len(sys.argv)
    if len(sys.argv) > 1:
        print len(sys.argv)
        fp = sys.argv[1]  
        fn = sys.argv[2]
    else:
        fn = raw_input('Please enter the file path of your excel file: ')
        print 'ok'
    fw = fn[0:12] + 'Translated' + fn[15:]
    print 'ok1'

    fr = open('%s\%s' % (fp, fn), 'r')
    fw = open('%s\%s' % (fp, fw), 'wb')

Unfortunately, I get the following error:

Traceback (most recent call last):
  File "C:\EclipseWorkspaces\PY2.7\PXBCM1.5\src\Translator.py", line 43, in <module>
    fr = open('%s\%s'% (fp,fn),'r')
NameError: name 'fp' is not defined

I do not understand what's wrong with my code.

Answer 1

As the error message suggests, the name fp is not defined. Looking at your code, I assume that your code branched in the else condition and thus fp was literally not defined.

If we look at an extract of your code:

if len(sys.argv) > 1:
    print len(sys.argv)
    fp = sys.argv[1]  # <-- fp is defined: GOOD
    fn = sys.argv[2]
else:
    fn = raw_input('Please enter the file path of your excel file: ')
    print 'ok'
    # <-- fp was not defined: BAD

So just make sure that by the time you call the open function, both fp and fn were defined, with statements like fp = … and fn = … .

---

Apart from that, your code presents one major flaw: you use the condition if len(sys.argv) > 1 but it just tells you that sys.argv is at least of length 2. Thus, it allows you to access safely sys.argv[0] (the 1st element) and sys.argv[1] (the 2nd element) but you cannot assume that sys.argv[2] exists.