我想創建某種prio生成器。這保持非常簡單。我有 8 個不同的問題，并根據答案輸出優先級。到目前為止，這也很有效。但是，我的“其他”條件似乎不起作用的錯誤。示例：如果選擇了問題 1 和問題 8，則 Else 條件應輸出“不可能”。不幸的是，這不起作用。我已經定義了可能性，并且超出定義規則的所有內容都應該符合 else 條件。這是代碼：
from tkinter import *
ws = Tk()
ws.title('Prio Guide')
ws.geometry('600x680')
ws.configure(bg="white")
text = StringVar()
label = Label( ws,
text='empty',
textvariable=text,
bg="green")
prio1 = StringVar()
prio1_text = Label( ws,text='empty',textvariable=prio1, bg="white")
prio1.set("Prio 1 - Description")
prio2 = StringVar()
prio2_text = Label( ws, text='empty', textvariable=prio2, bg="white")
prio2.set("Prio 2- Description")
prio3 = StringVar()
prio3_text = Label( ws, text='empty', textvariable=prio3, bg="white")
prio3.set("Prio 3- Description")
prio4 = StringVar()
prio4_text = Label( ws, text='empty', textvariable=prio4, bg="white")
prio4.set("Prio 4- Description")
#define the Prio
def prio():
if cb.get() == 1 or cb1.get() == 1 or cb2.get() == 1 or cb.get() & cb1.get() == 1 or cb2.get() & cb1.get() == 1 or cb.get() & cb1.get() & cb2.get() == 1:
text.set("Prio 1")
elif cb3.get() == 1 or cb4.get() == 1 or cb3.get() & cb4.get() == 1:
text.set("Prio 2")
elif cb5.get() == 1 or cb5.get() & cb6.get() == 1:
text.set("Prio 3")
elif cb7.get() == 1 or cb6.get() & cb7.get() == 1:
text.set("Prio 4")
else:
text.set("not possible")
#create IntVar fot the value of checkboxes
cb = IntVar()
cb1 = IntVar()
cb2 = IntVar()
cb3 = IntVar()
cb4 = IntVar()
cb5 = IntVar()
cb6 = IntVar()
cb7 = IntVar()
#create checkbutton
Checkbutton(ws,
bg="white",
text="q1",
variable=cb,
onvalue=1,
offvalue=0,
command=prio,).pack(anchor=W,padx=(100, 10))
Checkbutton(ws,
bg="white",
text="q2",
variable=cb1,
onvalue=1,
offvalue=0,
command=prio).pack(anchor=W,padx=(100, 10))
Checkbutton(ws,
bg="white",
text="q3",
variable=cb2,
onvalue=1,
offvalue=0,
command=prio).pack(anchor=W,padx=(100, 10))
Checkbutton(ws,
bg="white",
text="q4",
variable=cb3,
onvalue=1,
offvalue=0,
command=prio).pack(anchor=W,padx=(100, 10))
Checkbutton(ws,
bg="white",
text="q5",
variable=cb4,
onvalue=1,
offvalue=0,
command=prio).pack(anchor=W,padx=(100, 10))
Checkbutton(ws,
bg="white",
text="q6",
variable=cb5,
onvalue=1,
offvalue=0,
command=prio).pack(anchor=W,padx=(100, 10))
Checkbutton(ws,
bg="white",
text="q7",
variable=cb6,
onvalue=1,
offvalue=0,
command=prio).pack(anchor=W,padx=(100, 10))
Checkbutton(ws,
bg="white",
text="q8",
variable=cb7,
onvalue=1,
offvalue=0,
command=prio).pack(anchor=W,padx=(100, 10))
label.pack(anchor=S)
prio1_text.pack(anchor=S,padx=(100, 10))
prio2_text.pack(anchor=S,padx=(100, 10))
prio3_text.pack(anchor=S,padx=(100, 10))
prio4_text.pack(anchor=S,padx=(100, 10))
ws.mainloop()
謝謝你的幫助！
uj5u.com熱心網友回復：
您的“或”有問題，只要您的 or 條件之一等于 true，它就不會回傳您的 else 條件。
我想提出一個棘手的解決方案，即使用模式來獲得優先級。
from tkinter import *
ws = Tk()
ws.title('Prio Guide')
ws.geometry('600x680')
ws.configure(bg="white")
text = StringVar()
label = Label(ws,
text='empty',
textvariable=text,
bg="green")
prio1 = StringVar()
prio1_text = Label(ws, text='empty', textvariable=prio1, bg="white")
prio1.set("Prio 1 - Description")
prio2 = StringVar()
prio2_text = Label(ws, text='empty', textvariable=prio2, bg="white")
prio2.set("Prio 2- Description")
prio3 = StringVar()
prio3_text = Label(ws, text='empty', textvariable=prio3, bg="white")
prio3.set("Prio 3- Description")
prio4 = StringVar()
prio4_text = Label(ws, text='empty', textvariable=prio4, bg="white")
prio4.set("Prio 4- Description")
# define the Prio
def getCbValue():
return "".join(str(x) for x in [cb.get(), cb1.get(), cb2.get(), cb3.get(), cb4.get(), cb5.get(), cb6.get(), cb7.get()])
def getPrio(cbValue):
prioConf = {'Prio 1': ['10000000', '01000000', '00100000', '11000000', '01100000'],
'Prio 2': ['00010000', '00001000', '00011000'],
'Prio 3': ['00000100', '00000110'],
'Prio 4': ['00000001', '00000011']}
key_list = list(prioConf.keys())
value_list = list(prioConf.values())
for i, d in enumerate(value_list):
if cbValue in d:
return key_list[i]
def prio():
prio = getPrio(getCbValue())
if prio:
text.set(prio)
else:
text.set("not possible")
# create IntVar fot the value of checkboxes
cb = IntVar()
cb1 = IntVar()
cb2 = IntVar()
cb3 = IntVar()
cb4 = IntVar()
cb5 = IntVar()
cb6 = IntVar()
cb7 = IntVar()
# create checkbutton
Checkbutton(ws,
bg="white",
text="q1",
variable=cb,
onvalue=1,
offvalue=0,
command=prio,).pack(anchor=W, padx=(100, 10))
Checkbutton(ws,
bg="white",
text="q2",
variable=cb1,
onvalue=1,
offvalue=0,
command=prio).pack(anchor=W, padx=(100, 10))
Checkbutton(ws,
bg="white",
text="q3",
variable=cb2,
onvalue=1,
offvalue=0,
command=prio).pack(anchor=W, padx=(100, 10))
Checkbutton(ws,
bg="white",
text="q4",
variable=cb3,
onvalue=1,
offvalue=0,
command=prio).pack(anchor=W, padx=(100, 10))
Checkbutton(ws,
bg="white",
text="q5",
variable=cb4,
onvalue=1,
offvalue=0,
command=prio).pack(anchor=W, padx=(100, 10))
Checkbutton(ws,
bg="white",
text="q6",
variable=cb5,
onvalue=1,
offvalue=0,
command=prio).pack(anchor=W, padx=(100, 10))
Checkbutton(ws,
bg="white",
text="q7",
variable=cb6,
onvalue=1,
offvalue=0,
command=prio).pack(anchor=W, padx=(100, 10))
Checkbutton(ws,
bg="white",
text="q8",
variable=cb7,
onvalue=1,
offvalue=0,
command=prio).pack(anchor=W, padx=(100, 10))
label.pack(anchor=S)
prio1_text.pack(anchor=S, padx=(100, 10))
prio2_text.pack(anchor=S, padx=(100, 10))
prio3_text.pack(anchor=S, padx=(100, 10))
prio4_text.pack(anchor=S, padx=(100, 10))
ws.mainloop()
您可以檢查有 2 個新方法“getCbValue”和“getPrio”。
我還重構了您的 if-else，因此每當模式匹配時，它將回傳優先級，否則將回傳錯誤訊息“不可能”。
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