How do i access the '\n' (newline) character in text widget? because I've a problem doing my project.
I would like to ignore the character but .........
if (A2int[plain]=='\n') do the trick. Any ideas?
KeyError: '\n'
katak_guy 0 Newbie Poster
Ene Uran 638 Posting Virtuoso
Honestly, few of us are mind-readers, you need to give us more information and code.
katak_guy 0 Newbie Poster
bah, im sorry about the insufficient information to my question.
I am doing a program to encrypt a text from plaintext to ciphertext and the other way around, using Caesar Cipher idea. The only difference is this program has two keys. One to encrypt characters in odd position and the second key is to encrypt the characters in even position.
The key is in binary form due to the requirement of my system implementing binary input. It converts binary to decimal then uses the decimal value to add with the ascii value for the result = ciphertext.
When i was using text widget somehow '\n' exist and i've only mapped characters from A-Z and space. So I thought of doing it by mapping the '\n' character with a value.
Is there any better solution to my codings? Any improvement to be done and suggestions?
I've only started learning python 4 days ago. Sorry for the newb post =( .The code is posted as below.
from Tkinter import *
class App(object):
def __init__(self, master):
frame = Frame(master, bg="black", borderwidth=30)
frame.pack(expand=NO)
self.enc=Button(frame, width=12, font = "Verdana 14 bold",
text="Encrypt", command=self.caesar_encrypt)
self.dec=Button(frame, width=12, font = "Verdana 14 bold",
text="Decrypt", command=self.caesar_decrypt)
self.exit=Button(frame, width=12, font = "Verdana 14 bold",
text="Clear All", command=self.clear_all)
Label(frame, font = "Verdana 14 bold", fg='cyan', bg="black", text="Key").grid(row=0, column=1)
Label(frame, font = "Verdana 14 bold", fg='cyan', bg="black", text="Plaintext").grid(row=2, column=1)
Label(frame, font = "Verdana 14 bold", fg='cyan', bg="black", text="Ciphertext").grid(row=4, column=1)
self.entry1=Entry(frame, font = "Verdana 14", width=25, bg = 'orange')
self.entry2=Entry(frame, font = "Verdana 14", width=25, bg = 'orange')
self.text3=Text(frame, font = "Verdana 14", width=50, height=10, bg = 'orange')
self.text4=Text(frame, font = "Verdana 14", width=50, height=10, bg = 'orange')
self.entry5=Entry(frame, font = "Verdana 14", width=25, bg = 'orange')
self.entry6=Entry(frame, font = "Verdana 14", width=25, bg = 'orange')
self.enc.grid(row=2, column=0)
self.dec.grid(row=3, column=0)
self.exit.grid(row=4, column=0)
self.entry1.grid(row=1, column=1, sticky=W)
self.entry2.grid(row=1, column=1, sticky=E)
self.text3.grid(row=3, column=1)
self.text4.grid(row=5, column=1)
self.entry5.grid(row=6, column=1, sticky=W)
self.entry6.grid(row=6, column=1, sticky=E)
def clear_all(self):
self.entry1.delete(0,END)
self.entry2.delete(0,END)
self.text3.delete(1.0,END)
self.text4.delete(1.0,END)
self.entry5.delete(0,END)
self.entry6.delete(0,END)
def binary2dec(self):
s = list(self.entry1.get().strip())
s.reverse()
t = list(self.entry2.get().strip())
t.reverse()
result1 = 0
result2 = 0
for i, n in enumerate(s):
result1 += int(s[i]) * pow(2, i)
self.entry5.delete(0, END)
self.entry5.insert(0, result1)
for i, n in enumerate(t):
result2 += int(t[i]) * pow(2, i)
self.entry6.delete(0, END)
self.entry6.insert(0, result2)
return result1, result2
def caesar_encrypt(self):
A2int = {'A':0, 'B':1, 'C':2, 'D':3, 'E':4, 'F':5, 'G':6,'H':7,'I':8, 'J':9, 'K':10,
'L':11, 'M':12, 'N':13, 'O':14, 'P':15, 'Q':16, 'R':17, 'S':18, 'T':19, 'U':20,
'V':21, 'W':22, 'X':23, 'Y':24, 'Z':25, ' ':26, '\n':999}
int2A = {0:'A', 1:'B', 2:'C', 3:'D', 4:'E', 5:'F', 6:'G', 7:'H', 8:'I', 9:'J', 10:'K',
11:'L', 12:'M', 13:'N', 14:'O', 15:'P', 16:'Q', 17:'R', 18:'S', 19:'T', 20:'U',
21:'V', 22:'W', 23:'X', 24:'Y', 25:'Z', 26:' ', 999:'\n'}
plain = self.text3.get(1.0,END).upper()
key_odd, key_even = self.binary2dec()
ciphertext = ''
for i, n in enumerate(plain):
a='';
a=A2int[plain[i]]
if (a==999): continue
#ODD KEY
if (i%2==1):
n = A2int[plain[i]] + key_odd
if (n == 26): n = n
if (n > 26): n = n % 27
ciphertext += int2A[n]
#EVEN KEY
if (i%2==0):
n = A2int[plain[i]] + key_even
if (n == 26): n = n
if (n > 26): n = n % 27
ciphertext += int2A[n]
self.text4.delete(1.0, END)
self.text4.insert(1.0, ciphertext)
def caesar_decrypt(self):
A2int = {'A':0, 'B':1, 'C':2, 'D':3, 'E':4, 'F':5, 'G':6,'H':7,'I':8, 'J':9, 'K':10,
'L':11, 'M':12, 'N':13, 'O':14, 'P':15, 'Q':16, 'R':17, 'S':18, 'T':19, 'U':20,
'V':21, 'W':22, 'X':23, 'Y':24, 'Z':25, ' ':26, '\n':999}
int2A = {0:'A', 1:'B', 2:'C', 3:'D', 4:'E', 5:'F', 6:'G', 7:'H', 8:'I', 9:'J', 10:'K',
11:'L', 12:'M', 13:'N', 14:'O', 15:'P', 16:'Q', 17:'R', 18:'S', 19:'T', 20:'U',
21:'V', 22:'W', 23:'X', 24:'Y', 25:'Z', 26:' ', 999:'\n'}
cipher = self.text4.get(1.0,END).upper()
key_odd, key_even = self.binary2dec()
plaintext = ''
for i, n in enumerate(cipher):
a='';
a=A2int[cipher[i]]
if (a==999): continue
#ODD KEY
if (i%2==1):
n = A2int[cipher[i]] - key_odd
while (n<0): n = n + 27
plaintext += int2A[n]
#EVEN KEY
if (i%2==0):
n = A2int[cipher[i]] - key_even
while (n<0): n = n + 27
plaintext += int2A[n]
self.text3.delete(1.0, END)
self.text3.insert(1.0, plaintext)
root = Tk()
root.title("Caesar Cipher")
root.maxsize(834,695)
app = App(root)
root.mainloop() katharnakh 7 Posting Whiz in Training
How do i access the '\n' (newline) character in text widget? because I've a problem doing my project.
I would like to ignore the character but .........if (A2int[plain]=='\n') do the trick. Any ideas?
KeyError: '\n'
hey, if you are asking how to get index of '\n' character or to find whether string is having that character, then
from Tkinter import *
root = Tk()
def func():
if '\n' in text.get(1.0, END):
print 'at: ', text.get(1.0, END).index('\n')
text = Text(root, height=10, width=50)
text.grid(row=0, column=0)
b = Button(root, text='click', command=func)
b.grid(row=1)
root.mainloop() Be a part of the DaniWeb community
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