通過python應用程序發佈Twitter消息時出錯

Question

#!/usr/bin/python 
# -*- coding: iso-8859-1 -*- 

import Tkinter 
import twitter 

class simpleapp_tk(Tkinter.Tk): 
    def __init__(self,parent): 
     Tkinter.Tk.__init__(self,parent) 
     self.parent = parent 
     self.initialize() 


    def initialize(self): 
     self.grid() 

    api=twitter.api()(username='-----',password='----') 

     self.entryVariable = Tkinter.StringVar() 
     self.entry = Tkinter.Entry(self,textvariable=self.entryVariable) 
     self.entry.grid(column=0,row=0,sticky='EW') 
     self.entry.bind("<Return>", self.OnPressEnter) 
     self.entryVariable.set(u"Enter text here.") 

     button = Tkinter.Button(self,text=u"POST IT !", 
           command=self.OnButtonClick) 
     button.grid(column=1,row=0) 

     self.labelVariable = Tkinter.StringVar() 
     label = Tkinter.Label(self,textvariable=self.labelVariable, 
           anchor="w",fg="white",bg="blue") 
     label.grid(column=0,row=1,columnspan=2,sticky='EW') 
     self.labelVariable.set(u"Hello !") 

     self.grid_columnconfigure(0,weight=1) 
     self.resizable(True,False) 
     self.update() 
     self.geometry(self.geometry())  
     self.entry.focus_set() 
     self.entry.selection_range(0, Tkinter.END) 

    def OnButtonClick(self): 
     self.labelVariable.set(self.entryVariable.get()+" (You clicked the button)") 
     self.entry.focus_set() 
     self.entry.selection_range(0, Tkinter.END) 

    status=api.PostUpdate(self.entry) 

    def OnPressEnter(self,event): 
     self.labelVariable.set(self.entryVariable.get()+" (You pressed ENTER)") 
     self.entry.focus_set() 
     self.entry.selection_range(0, Tkinter.END) 

if __name__ == "__main__": 
    app = simpleapp_tk(None) 
    app.title('POSTit') 
    app.initialize() 
    app.mainloop() 

顯示的錯誤是。 AttributeError：'模塊'對象沒有屬性'api'

Answer 1

錯誤消息很明顯，不是嗎？ twitter模塊沒有名爲「api」的屬性。

快速谷歌向我展示了一些具有「.Api（）」方法（大寫A）的示例。也許這是你的問題。