Re: Escaping the semicolon?

[Sergio Correia]

On Dec 4, 2007 10:40 AM, Nick [EMAIL PROTECTED] wrote:


 Is this expected behavior?

  s = '123;abc'
  s.replace(';', '\;')
 '123\\;abc'


Everything is Ok. It's still a single backslash. Try:

 print  s.replace(';', '\;')

Or
 x = s.replace(';', '\;')
 print x

Best,
Sergio
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Re: Escaping the semicolon?

[Bruno Desthuilliers]

Nick a écrit :
 Hi all,
 
 Is this expected behavior?
 
 s = '123;abc'
 s.replace(';', '\;')
 '123\\;abc'

  print s.replace(';', '\;')
123\;abc

 I just wanted a single backslash.

You got it - even if it's not obvious !-)

 I can see why this probably happens
 but i wondered if it is definitely intentional.

  s2 = '123\;abc'
  s2
'123\\;abc'
  print s2
123\;abc
  list(s2)
['1', '2', '3', '\\', ';', 'a', 'b', 'c']

As you can see, '\\' is counted as a single character !-)
Since the backslash is the escape character, you need to escape it to 
have a litteral backslash:

  s3 = '\'
   File stdin, line 1
 s3 = '\'
^
SyntaxError: EOL while scanning single-quoted string
 
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Re: Escaping the semicolon?

[Diez B. Roggisch]

Nick wrote:

 Hi all,
 
 Is this expected behavior?
 
 s = '123;abc'
 s.replace(';', '\;')
 '123\\;abc'
 
 I just wanted a single backslash. I can see why this probably happens
 but i wondered if it is definitely intentional.

There is only a single backslash. But the interactive prompt will use the
repr()-function to print out returned values. Which will for strings print
their escaped syntax.

Try the above with a 

print s.replace(...)

and you will see your desired outcome.

diez
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Re: Escaping the semicolon?

[Tim Chase]

 Is this expected behavior?
 
 s = '123;abc'
 s.replace(';', '\;')
 '123\\;abc'

You're asking the interpreter to print a representation of your 
string, so it does so.  Representations wrap the results in 
quotes and escape characters within that need escaping.

  s.replace(';', '\;')
'123\\;abc'
  print repr(s.replace(';', '\;'))
'123\\;abc'
  print s.replace(';', '\;')
123\;abc

Additionally, it's best-practice to use raw strings or literal 
backslashes, making your replacement either

   r'\;'

or

   '\\;'

because depending on the character following the back-slash, it 
may be translated as a character you don't intend it to be.

-tkc


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Re: Escaping the semicolon?

[Mel]

Nick wrote:
 Is this expected behavior?
 
 s = '123;abc'
 s.replace(';', '\;')
 '123\\;abc'
 
 I just wanted a single backslash. I can see why this probably happens
 but i wondered if it is definitely intentional.

What you're seeing on the screen is a literalization of the string 
value for the sake of the display.  Consider

Python 2.5.1 (r251:54863, May  2 2007, 16:56:35)
[GCC 4.1.2 (Ubuntu 4.1.2-0ubuntu4)] on linux2
Type help, copyright, credits or license for more information.
  s = '123;abc'
  b = s.replace(';', '\;')
  b
'123\\;abc'
  len(b)
8


The length suggests that there's only one backslash in the string.


Mel.

On the other hand

  repr(b)
'123;abc'

Isn't what I expected.  No, wait, it is.  It's the value of repr(b) 
repred by the Python display logic.

MPW
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Re: Escaping the semicolon?

[Nick]

Thanks guys, you answered that interactive prompt question really
clearly however, whats going on here. This works now -

 working_string = '123;abc'
 search_string = ';'
 print working_string.replace(search_string, '\\' + search_string)
123\;abc

But this doesn't -

---
import sys
import string

input = string.join(sys.argv[1:], '')
escape_char_list = [ '(', '[]', ';', '^', '\\', '/', '|', '*', '$',
'', '[', ']', ')', '?' ]

for ch in escape_char_list:
input = input.replace(ch, '\\' + ch)

print input
---

Try 123 *?/ abc d;o /$' as the argument... and you get -

123 \*\?\/ abc d\\;o \/\$

Still two back slashes, did i miss something very obvious? Sorry,
sometimes these things are exasperating and just need more eyes or a
head screwed on (if thats the case!).

Nick
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Re: Escaping the semicolon?

[Jerry Hill]

On Dec 4, 2007 11:33 AM, Nick [EMAIL PROTECTED] wrote:
 Try 123 *?/ abc d;o /$' as the argument... and you get -

 123 \*\?\/ abc d\\;o \/\$

That's because of the order you're doing the replacement.  Put a print
statement inside your for loop and you'll see something like this:

input starts as 123 *?/ abc d;o /$'
Then when you replace ';' with '\;' you get input = 123 *?/ abc d\;o /$'
Then the next replacement replaces '\' with '\\' so you get input =
123 *?/ abc d\\;o /$'

If you move '\\' to the front of your list of replacement characters,
things will probably work as you expect.

-- 
Jerry
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Re: Escaping the semicolon?

[Nick]

 If you move '\\' to the front of your list of replacement characters,
 things will probably work as you expect.

 --
 Jerry

I knew it would be something like that! Thanks for your help.
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