user validation for opening urls

[Ashley]

HI All,

I' mlearning python and writing few programs to automate my testing. I am trying to do url validation based on what user has entered.If user inputs the version as 21.1.0 then I should be able to search for 21.1.10 in the base_url and if match found then only open the link. I'm struggling to find the match with the version user enters.
If I enter 1.1.0 it scans the website for 1.1.0 and if there's 21.1.0, it finds the match and prints vresion found. I want if it matches only with 21.1.0 then only open the link or if user enters 2.2.0 and there's 2.2.0 then I should find the match for 2.2.0 and not for 12.2.0.
Please note this is not the correct base_url.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

import urllib2 import re base_url = "http://artifactory.int.abc.com:9090/artifactory/simple/test-local/com/abc/test/test.far/" user_input = raw_input("Enter the test version you want to upgrade to: ")   #connect to a URL website = urllib2.urlopen(base_url)   #read html code html = website.read() print(html)   test2 = re.search(user_input, html) if user_input in html:     print("version found") else:     print("version not found")

[metulburr]

What character is before the version sequence in the url? Can you give a similar url where these versions are in the url?

[noisefloor]

Hi,

I' m learning python

Based on your code, <you are using Python 2 - which is in less than 6 month EOL. Today, there is no reason when learning Python to use 2.7. Use a supported 3.x version instead.

Although it does not solve your original problem, it makes your code future proof :-)

On your problem: using re.match instead of re.search should do what you want (Explanation).
By the way: what does the page holding the versions look like? Probably you could do the whole sting by simple string comparison. Your current code may next to no use of the power of regular expressions.

Regards, noisefloor

[Ashley]

I'm getting the following output. As a user if I enter the user_input as 1.1.0, it should go find the match for "1.1.0" and not "11.1.0"
Basically I want web crawling without installing any additional modules.

Output:
1.1.0
1.1.2
1.2.1
1.2.2
1.3.1
11.1.0
12.1.2

[perfringo]

As metulburr already asked: what characters are before version? Can you give sample url? Are they always with same structure?

[Ashley]

There's no charachter. These vresions number are folders. I want to do a user validation. When user enters, 1.1.0 and there's no folder 1.1.0 then it should throw error for usert to notify to enter correct version.
Example would be:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

import urllib2 import re base_url = "https://www.python.org/doc/versions/" user_input = raw_input("Enter the test version you want to Download: ")    #connect to a URL website = urllib2.urlopen(base_url)    #read html code html = website.read() print(html) x= user_input.strip() test2 = re.match(x, html) if user_input in html:     print("version found") else:     print("version not found")

[metulburr]

If your matching the exact input then you dont need regex. Parse down the url to the version and compare to the input given.

1 2 3 4

>>> '1.1.0' in '21.1.0' True >>> '1.1.0' == '21.1.0' False