Oozie工作流程错误E0501:无法执行授权操作

时间:2020-10-22 11:39:11

标签: hadoop authorization workflow oozie impersonation

我正在将 oozie 4.3.0 hadoop 2.7.3 一起使用。

当我要创建和运行工作流程时,出现以下错误: 错误:E0501:E0501:无法执行授权操作,用户:st_jsgane.gane不能模拟st_jsgane.gane

这是我的hadoop的core-site.xml

def EntryRow(left_wl, counter,dx):
        counter = counter
        dx = dx
        
        entry_box_1 = tk.Entry(left_wl)#,width=15,height = 10) # bg="lightblue")
        entry_box_2 = tk.Entry(left_wl,)# width=18,height = 10) # bg="lightblue")
        entry_box_3 = tk.Entry(left_wl,)# width=20,height = 10) # bg="lightblue")
        entry_box_4 = tk.Entry(left_wl,)# width=18,height = 10) # bg="lightblue")
        entry_box_5 = tk.Entry(left_wl,)# width=18,height = 10) # bg="lightblue")
        entry_box_6 = tk.Entry(left_wl,)# width=18,height = 10) #bg="lightblue")
        entry_box_7 = tk.Entry(left_wl, )#width=18,height = 10) #bg="lightblue")
        
        entry_box_1.place(x = dx, y = 130)# width 15, height = 10) 
        entry_box_2.place(x = dx, y = 160)
        entry_box_3.place(x = dx, y = 220)
        entry_box_4.place(x = dx, y = 250)
        entry_box_5.place(x = dx, y = 310)
        entry_box_6.place(x = dx, y = 340)
        entry_box_7.place(x = dx, y = 370 )

def submit():
    W_no2 = W_no.get()
    
    
    tk.Label(left_wl, text = 'A',bd = 2, width = 30,bg = 'white',font = ('Arial',10,'bold'),relief = "solid").place(x = 10, y = 100)
    tk.Label(left_wl, text = 'B',bd = 2, width = 30,bg = 'white',font = ('Arial',10,'bold'),relief = "solid").place(x = 10, y = 130)    
    tk.Label(left_wl, text = 'C',bd = 2, width = 30,bg = 'white',font = ('Arial',10,'bold'),relief = "solid").place(x = 10, y = 160)    
    tk.Label(left_wl, text = 'D',bd = 2, width = 30,bg = 'white',font = ('Arial',10,'bold'),relief = "solid").place(x = 10, y = 190)    
    tk.Label(left_wl, text = 'E',bd = 2, width = 28,bg = 'white',font = ('Arial',10,'bold'),relief = "solid").place(x = 18, y = 220)    
    tk.Label(left_wl, text = 'F',bd = 2, width = 28,bg = 'white',font = ('Arial',10,'bold'),relief = "solid").place(x = 18, y = 250)    
    tk.Label(left_wl, text = 'G',bd = 2, width = 30,bg = 'sky blue',font = ('Arial',10,'bold'),relief = "solid").place(x = 10, y = 280)    
    tk.Label(left_wl, text = 'H',bd = 2, width = 30,bg = 'white',font = ('Arial',10,'bold'),relief = "solid").place(x = 10, y = 310)    
    tk.Label(left_wl, text = 'I',bd = 2, width = 30,bg = 'white',font = ('Arial',10,'bold'),relief = "solid").place(x = 10, y = 340)    
    tk.Label(left_wl, text = 'J',bd = 2, width = 30,bg = 'white',font = ('Arial',10,'bold'),relief = "solid").place(x = 10, y = 370)    
    
    
    dx = 300 
    
    for i in range(W_no2):
        
        tk.Label(left_wl, text = i+1,bd = 2, width = 15,bg = 'white',font = ('Arial',10,'bold'),relief = "sunken").place(x = dx, y = 100)
        
        dx = dx + 150
    
    global columns
    
    counter = 0
    columns = []  
    
    dx = 300
    while W_no2 > counter:
        columns.append(EntryRow(left_wl, counter,dx))
        print(columns)
        counter += 1
        dx += 150
    

W_Button = tk.Button ( left_wl, text = 'Sumbit', bg ='skyblue',font = ('Arial',10,'bold'), width = 15, command = submit)
W_Button.place( x = 300, y= 50)

data_dict = {}
data_list = []

def extract_data(columns):
    for row in columns:
        # pick the format you like more
        data_list.append((row.entry_box_1.get(), row.entry_box_2.get(), row.entry_box_3.get(),row.entry_box_4.get(),row.entry_box_5.get(),row.entry_box_6.get(),row.entry_box_7.get())) # row.entry_box_0.get(), ,row.entry_box_8.get(),row.entry_box_9.get()
        data_dict[row.entry_box_1.get()] = (row.entry_box_2.get(), row.entry_box_3.get(),row.entry_box_4.get(),row.entry_box_5.get(),row.entry_box_6.get(),row.entry_box_7.get()) # row.entry_box_1.get(),,row.entry_box_8.get(),row.entry_box_9.get()
        print(data_list)
        print(data_dict)



btn = Button(left_wl, text="Finished", command=lambda: extract_data(columns))
btn.place(x = 15, y = 400)


0 个答案:

没有答案