为什么我得到“模块”对象在 python 3 中不可调用？

Question

一、所有相关代码

主程序

import string
import app
group1=[ "spc", "bspc",",","."]#letters, space, backspace(spans mult layers)
# add in letters one at a time
for s in string.ascii_lowercase:
    group1.append(s)
group2=[0,1,2,3,4,5,6,7,8,9, "tab ","ent","lAR" ,"rAR" , "uAR", "dAR"]
group3= []
for s in string.punctuation:
    group3.append(s)#punc(spans mult layers)
group4=["copy","cut","paste","save","print","cmdW","quit","alf","sWDW"] #kb shortcut
masterGroup=[group1,group2,group3,group4]
myApp =app({"testFKey":[3,2,2]})

应用程序.py

 import tkinter as tk
import static_keys
import dynamic_keys
import key_labels
class app(tk.Frame):

    def __init__(inputDict,self, master=None,):
        tk.Frame.__init__(self, master)
        self.grid(sticky=tk.N+tk.S+tk.E+tk.W)
        self.createWidgets(self, inputDict)
    def createWidgets(self,inDict):
        top=self.winfo_toplevel()
        top.rowconfigure(0, weight=1)
        top.columnconfigure(0, weight=1)
        self.rowconfigure(0, weight=1)
        self.columnconfigure(0, weight=1)
        tempDict = {}
        for k,v in inDict.items():
                if 1<=v[0]<=3:
                    tempDict[k] = static_keys(*v[1:])
                elif v[0] ==4:
                    tempDict[k] = dynamic_keys(k,*v[1:])
                elif  v[0]==5:
                    tempDict[k] = key_labels(*v[1:])
        for o in tempDict:
            tempDict[o].grid()
        return tempDict

static_keys.py

 import tkinter
class static_keys(tkinter.Label):
    """class for all keys that just are initiated then do nothing
    there are 3 options
    1= modifier (shift etc)
    2 = layer
    3 = fkey, eject/esc"""
    def __init__(t,selector,r,c,parent,self ):
        if selector == 1:
            tkinter.Label.__init__(master=parent, row=r, column=c, text= t, bg ='#676731')
        if selector == 2:
            tkinter.Label.__init__(master=parent, row=r, column=c, text= t, bg ='#1A6837')
        if selector == 3:
            tkinter.Label.__init__(master=parent, row=r, column=c, text= t, bg ='#6B6966')

现在对问题进行描述。当我在 python3 中运行 main.py 时，出现错误

File "Desktop/kblMaker/main.py", line 13, in <module>
myApp =app({"testFKey":[3,2,2]})
TypeError: 'module' object is not callable

原文由 fozbstuios 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

您有一个名为 app 的模块，其中包含一个名为 app 的类。如果您只是在 main.py 中执行 import app 则 app 将引用该模块，而 app.app 将引用该类。这里有几个选项：

• 保留您的导入语句，并在 main.py 中使用 myApp = app.app({"testFKey":[3,2,2]})
• 将 import app 替换为 from app import app ，现在 app 将引用该类并且 myApp = app({"testFKey":[3,2,2]}) work

原文由 Andrew Clark 发布，翻译遵循 CC BY-SA 3.0 许可协议

Answer 2

在 main.py 将第二行更改为：

 from app import app

问题是您有 app 模块和 app 类。但是您正在导入模块，而不是其中的类：

 myApp = app({"testFKey": [3, 2, 2]})

（您也可以将上面的内行“ app ”替换为“ app.app ”）

原文由 Tadeck 发布，翻译遵循 CC BY-SA 3.0 许可协议