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當我在Python中運行下面的程序時，該函數接受變量，但完全跳過其餘部分並重新顯示程序的主菜單，而不執行任何操作。此外，即使選擇了第一個或第二個選項（不需要第三個變量），它也會跳過限定的「if」語句並請求所有變量。順便說一句，它不應該是一個縮進錯誤，我只是縮進顯示它是在stackoverflow中的代碼。正在發生什麼事？ Python在函數中跳過行
編輯：NEVERMIND。我得到它的工作。函數括號中的變量都必須相同。 DUH！ 嫌額頭
option = 1
while option !=0:
print "\n\n\n************MENU************"
print "1. Counting"
print "2. Fibbonacci Sequence"
print "0. GET ME OUTTA HERE!"
print "*" * 28
option = input("Please make a selection: ") #counting submenu
if option == 1:
print "\n\n*******Counting Submenu*******"
print "1. Count up by one"
print "2. Count down by one"
print "3. Count up by different number"
print "4. Count down by different number"
print "*" * 28
countingSubmenu = input("Please make a selection: ")
x=0
y=0
z=0
q=0
def counting (x, y, z, countingSubmenu, q):
x = input("Please choose your starting number: ")
y = input("Please choose your ending number: ")
if countingSubmenu == 1:
for q in range (x, y+1, 1):
print q
elif countingSubmenu == 2:
for q in range (x, y, -1):
print q
elif countingSubmenu == 3:
z = input("Please choose an increment: ")
for q in range (x, y+1, z):
print q
else:
z = input("Please choose an increment: ")
for q in range (x, y, -z):
print q
return x, y, z, q
if countingSubmenu == 1:
counting(countingSubmenu, x, y, z, q)
if countingSubmenu == 2:
counting(countingSubmenu, x, y, z, q)
if countingSubmenu == 3:
counting(countingSubmenu, x, y, z, q)
if countingSubmenu == 4:
counting(countingSubmenu, x, y, z, q)
「NEVERMIND。我得到它的工作。函數括號中的變量都必須是相同的。」這是一種不好的更新。請刪除它。請發佈一個**答案**與其他人需要的信息來弄清楚發生了什麼。回答你的問題沒關係。道歉不好。最好刪除它，而不是提供這樣一個令人困惑的更新。 –