python - replace values in url

Question

I have a URL like the following -

url = "http://servername:8080/api/environment/component.get?query=(name LIKE '%componentname%')"

I want to replace the following values with argument parameters I have as strings -

servername
environment
component
componentname

I want it then to be passed into requests -

r = requests.get(url, auth=('domain\user', 'password')

The problem is I can't figure out how to replace the values. I can't do %s, %d because it's possible that the value in %componentname% could start with an s or d which would break the encode.

Use string.format with { } to escape. You can even use {0} {1} etc to avoid accidental escaping — Tim
– Tim, Commented Aug 17, 2015 at 18:01
I actually did <servername> and then did url.replace('<servername>', 'server') seems inefficient though to have to have 4 separate url.replace... — whoisearth
– whoisearth, Commented Aug 17, 2015 at 18:05

primero · Accepted Answer · 2015-08-17 18:32:10Z

2

From what I understand the following solution should work for you, and it's more straight forward(It was also sugested in the first comment by user Tim Castelijns):

url = "http://{0}:8080/api/{1}/{2}.get?query=(name LIKE '%{3}%')".format(servername, environment, component, componentname)

If you want more readability you could even do it like this:

url = "http://{servername}:8080/api/{environment}/{component}.get?query=(name LIKE '%{componentname}%')".format(servername, environment, component, componentname)

answered Aug 17, 2015 at 18:32

primero

5911 gold badge6 silver badges17 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Mikkel Over a year ago

The second option does not work for me (using python 3.9), actually what works is

"http://{servername}:8080/api/{environment}/{component}.get?query=(name LIKE '%{componentname}%')".format(servername=servername, environment=environment, component=component, componentname=componentname)

Meet Mehta · Accepted Answer · 2015-08-17 18:06:58Z

0

You can use python's inbuilt string concatenation.

url = "http://servername:8080/api/environment/component.get?query=(name LIKE '"+ componentname +"')"

It will append number or string as it is and you don't need to take care of %s or %d.

answered Aug 17, 2015 at 18:06

Meet Mehta

4,9193 gold badges22 silver badges27 bronze badges

3 Comments

whoisearth Over a year ago

I tried this with all the replacements and it didn't like it because of the internal '' and the outside ""

Meet Mehta Over a year ago

Can you pls clarify. You can always escape internal quotes.

Tim Over a year ago

@whoisearth just use ' and " then. They mix well in python. Example 'this is "a string" that is valid'

Collectives™ on Stack Overflow

python - replace values in url

2 Answers 2

1 Comment

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related