How to replace spaces in URL having path parameter, with a hyphen or an underscore

Question

I am sorry if this Q makes no sense , but is there a way to replace spaces with hyphen in URL's(only) having used path parameter to build it ?

My scenario is as :

I have a view method as below:

from app.service import *

@app.route('/myapp/<search_url>',methods=['GET','POST'])
def search_func(search_url):
    print(search_url) // This prints 'hi-martin king' and i need to pass 'hi-martin king' to below with preserving space
    search_q = search(search_url) 
    return render_template('wordsearch.html',data=search_q['data'])

here search_url I am passing from template

I have the search function which takes search_url as argument(def search(search_url): .....) and does all operations (for ref) which i have imported from service above.

Now when I run , i have the sample URL as,

....myapp/hi-martin%20king

Here I am preserving space to perform query in database (In db it is stored as martin king), but I don't want to show the same in URL instead replace it with a hyphen

I have other way that changing all the values in database (removing spaces but this is tmk not a appropriate solution)

expected o/p:

....myapp/hi-martin-king  (or with an underscore) ....myapp/hi-martin_king

Here how can i preserve spaces to pass it as argument to the function and at the same time I want to replace only in URL ? Is this possible ?

Any help is appreciated ....TIA

about expected output: `myapp/hi-martin-king` and `myapp/hi-martin_king` are different outputs ). could you add input string `search_url` and expected one output? — Danila Ganchar, Jan 23 '20 at 08:08
Yeah they both are different but I put that as to replace space with a hyphen or an underscore — KcH, Jan 23 '20 at 08:47
As you can see you've got 3 the same answers ))) I mean(`replace`, `split` etc). So I think your question isn't valid. Try to rewrite question. By the way try to read about `nginx`, `rewrite rules` etc and how's it works with server side. — Danila Ganchar, Jan 23 '20 at 10:59
You didn't answer Danila's first question, and I think its the most important: *Why do you need to change the code(`%20`) to space(`' '`)*? — Lord Elrond, Jan 28 '20 at 22:54

score 2 · Answer 1 · answered Jan 23 '20 at 10:55

If you want to query your db while preserving spaces you can just url-unescape them for the query using urllib.parse.unquote and preserve them escaped in the url as follows:

import urllib.parse

from app.service import *

@app.route('/myapp/<search_url>',methods=['GET','POST'])
def search_func(search_url):
    unquoted_search = urllib.parse.unquote(search_url)
    search_q = search(unquoted_search) 
    return render_template('wordsearch.html',data=search_q['data'])

score 2 · Answer 2 · answered Jan 28 '20 at 09:07

The string with % in needs unquoting if you want the raw version.

from werkzeug import url_unquote

url_unquote('hi-martin%20king')  # => 'hi-martin king'

Now you have the unquoted string, you can replace spaces with either a hyphen or an underscore.

replace_character = '-'

def fix_string(s):
    return s.replace(' ', replace_character)

fix_string('hi-martin king')  # => 'hi-martin-king'

How to replace spaces in URL having path parameter, with a hyphen or an underscore

expected o/p:

2 Answers2