How can I get the standard URL format from a string like an url?-CodePudding

I want to convert a string into the standard URL format in Python.

For example:

example.com --> http://example.com
example.com:80 --> http://example.com
example.com:443 --> https://example.com
example.com:8000 --> http://example.com:8000
examplecom --> invalid url

How can I do for it?

Thank you!

CodePudding user response：

I found the solution:

import validators

valid_links() = list()
for link in links: # links is a list of string to test
    if str(link).endswith(':80'):
        link = 'http://'   link[:-3]
    elif str(link).endswith(':443'):
        link = 'https://'   link[:-4]
    else:
        link = 'http://'   link
    if validators.url(link):
        valid_links.append(link)

CodePudding user response：

Another alternative:

def convert_url(domain):
    valid = False
    has_port = False
    for letter in domain:
        if letter == ".":
            valid = True
        if letter == ":":
            has_port = True

    if valid:
        if has_port:
            domain_name, port = domain.split(":")

            if port == "80":
                return f"http://{domain_name}"
            elif port == "443":
                return f"https://{domain_name}"
            else:
                return f"http://{domain_name}:{port}"
        else:
            return f"http://{domain}"

    else:
        return "invalid url"


print(convert_url("example.com"))
print(convert_url("example.com:80"))
print(convert_url("example.com:443"))
print(convert_url("example.com:8000"))
print(convert_url("examplecom"))