How to replace whitespaces with underscore?

I want to replace whitespace with underscore in a string to create nice URLs. So that for example:

"This should be connected" 

Should become

"This_should_be_connected" 

I am using Python with Django. Can this be solved using regular expressions?

14 Answers

You don't need regular expressions. Python has a built-in string method that does what you need:

mystring.replace(" ", "_") 

Replacing spaces is fine, but I might suggest going a little further to handle other URL-hostile characters like question marks, apostrophes, exclamation points, etc.

Also note that the general consensus among SEO experts is that dashes are preferred to underscores in URLs.

import re def urlify(s): # Remove all non-word characters (everything except numbers and letters) s = re.sub(r"[^\w\s]", '', s) # Replace all runs of whitespace with a single dash s = re.sub(r"\s+", '-', s) return s # Prints: I-cant-get-no-satisfaction" print(urlify("I can't get no satisfaction!")) 

This takes into account blank characters other than space and I think it's faster than using re module:

url = "_".join( title.split() ) 

Django has a 'slugify' function which does this, as well as other URL-friendly optimisations. It's hidden away in the defaultfilters module.

>>> from django.template.defaultfilters import slugify >>> slugify("This should be connected") this-should-be-connected 

This isn't exactly the output you asked for, but IMO it's better for use in URLs.

Using the re module:

import re re.sub('\s+', '_', "This should be connected") # This_should_be_connected re.sub('\s+', '_', 'And so\tshould this') # And_so_should_this 

Unless you have multiple spaces or other whitespace possibilities as above, you may just wish to use string.replace as others have suggested.

use string's replace method:

"this should be connected".replace(" ", "_")

"this_should_be_disconnected".replace("_", " ")

Python has a built in method on strings called replace which is used as so:

string.replace(old, new) 

So you would use:

string.replace(" ", "_") 

I had this problem a while ago and I wrote code to replace characters in a string. I have to start remembering to check the python documentation because they've got built in functions for everything.

Surprisingly this library not mentioned yet

python package named python-slugify, which does a pretty good job of slugifying:

pip install python-slugify 

Works like this:

from slugify import slugify txt = "This is a test ---" r = slugify(txt) self.assertEquals(r, "this-is-a-test") txt = "This -- is a ## test ---" r = slugify(txt) self.assertEquals(r, "this-is-a-test") txt = 'C\'est déjà l\'été.' r = slugify(txt) self.assertEquals(r, "cest-deja-lete") txt = 'Nín hǎo. Wǒ shì zhōng guó rén' r = slugify(txt) self.assertEquals(r, "nin-hao-wo-shi-zhong-guo-ren") txt = 'Компьютер' r = slugify(txt) self.assertEquals(r, "kompiuter") txt = 'jaja---lol-méméméoo--a' r = slugify(txt) self.assertEquals(r, "jaja-lol-mememeoo-a") 

I'm using the following piece of code for my friendly urls:

from unicodedata import normalize from re import sub def slugify(title): name = normalize('NFKD', title).encode('ascii', 'ignore').replace(' ', '-').lower() #remove `other` characters name = sub('[^a-zA-Z0-9_-]', '', name) #nomalize dashes name = sub('-+', '-', name) return name 

It works fine with unicode characters as well.

You can try this instead:

mystring.replace(r' ','-') 

mystring.replace (" ", "_") 

if you assign this value to any variable, it will work

s = mystring.replace (" ", "_") 

by default mystring wont have this

OP is using python, but in javascript (something to be careful of since the syntaxes are similar.

// only replaces the first instance of ' ' with '_' "one two three".replace(' ', '_'); => "one_two three" // replaces all instances of ' ' with '_' "one two three".replace(/\s/g, '_'); => "one_two_three" 

x = re.sub("\s", "_", txt) 

perl -e 'map { $on=$_; s/ /_/; rename($on, $_) or warn $!; } <*>;' 

Match et replace space > underscore of all files in current directory