-2
因此,在生成以下錯誤時運行代碼。 它表示NameError未被用戶代碼處理。如何在Python中解決未處理的NameError問題
錯誤:
Traceback (most recent call last):
File "D:\3rd sem\Object Oriented Programming\Lab\VS\PythonApplication1\PythonApplication1\PythonApplication1.py", line 50, in <module>
main()
File "D:\3rd sem\Object Oriented Programming\Lab\VS\PythonApplication1\PythonApplication1\PythonApplication1.py", line 10, in main
grade=str(input("Enter the grade: "))
File "<string>", line 1, in <module>
NameError: name 'a' is not defined
的代碼是這樣的:
classnum=int(input("Enter the num of classes: "))
def main():
totalcredit=0
totalgpa=0
for i in range(1,classnum+1):
print "class", i
credit=int(input("Enter the credit: "))
grade=str(input("Enter the grade: "))
totalgpa+=coursePoints(credit,grade)
totalcredit+=credit
totalcourse=classnum
semestergpa=totalgpa/totalcredit
print("Semester summary")
print("courses taken: ", classnum)
print("credits taken: ", totalcredit)
print("GPA points: ", totalgpa)
print("Semester GPA: ", semestergpa)
def coursePoints(Credit,Grade):
if Grade == 'A+' or Grade == 'a+':
return 4*Credit
elif Grade == 'A' or Grade == 'a':
return 4*Credit
elif Grade == 'A-' or Grade == 'a-':
return 3.67*Credit
elif Grade == 'B+' or Grade == 'b+':
return 3.33*Credit
elif Grade == 'B' or Grade =='b':
return 3*Credit
elif Grade == 'B-' or Grade == 'b-':
return 2.67*Credit
elif Grade == 'C+' or Grade == 'c+':
return 2.33*Credit
elif Grade == 'C' or Grade == 'c':
return 2*Credit
elif Grade == 'C-' or Grade == 'c-':
return 1.67*Credit
elif Grade =='D+' or Grade == 'd+':
return 1.33*Credit
elif Grade == 'D' or Grade == 'd':
return 1*Credit
elif Grade == 'D-' or Grade == 'd-':
return 0.33*Credit
else:
return 0
main()
能
的解決方案,任何人的幫助。
在此先感謝。
@霸wonSunuwar堆棧溢出是不是真的設計的背部和反覆的討論。它是爲小的具體問題和答案而設立的。如果您有新問題,請創建一個新帖子。然而,這就是說,我建議你在提出一個新問題之前,谷歌「在賦值之前引用的python變量」,因爲這是一個常見的錯誤。 – CoryKramer
謝謝大家的建議 –