Sometimes, while web development, we can come across a task in which we may require to perform a split of query parameters from URLs which is done by ‘?’ character. This has application over web development as well as other domains which involve URLs. Lets discuss certain ways in which this task can be performed.
Method #1 : Using split()
This is one of the way in which we can solve this problem. We split by ‘?’ and return the first part of split for result.
Python3
# Python3 code to demonstrate working of # Split URL from Query Parameters# Using split()# initializing stringtest_str = 'www.geeksforgeeks.org?is = best'# printing original stringprint("The original string is : " + str(test_str))# Split URL from Query Parameters# Using split()res = test_str.split('?')[0]# printing result print("The base URL is : " + res) |
The original string is : www.geeksforgeeks.org?is=best The base URL is : www.geeksforgeeks.org
Time Complexity: O(n) -> (split function)
Auxiliary Space: O(n)
Method #2 : Using rfind()
This is another way in which we need to perform this task. In this, we find the first occurrence of ‘?’ from right and slice the string.
Python3
# Python3 code to demonstrate working of # Split URL from Query Parameters# Using rfind()# initializing stringtest_str = 'www.geeksforgeeks.org?is = best'# printing original stringprint("The original string is : " + str(test_str))# Split URL from Query Parameters# Using rfind()res = test_str[:test_str.rfind('?')]# printing result print("The base URL is : " + res) |
The original string is : www.geeksforgeeks.org?is=best The base URL is : www.geeksforgeeks.org
Time Complexity: O(n)
Auxiliary Space : O(n)
Method #3 : Using index().Finding index of ‘?’ and then used string slicing
Python3
# Python3 code to demonstrate working of# Split URL from Query Parameters# Using index()# initializing stringtest_str = 'www.geeksforgeeks.org?is = best'# printing original stringprint("The original string is : " + str(test_str))# Split URL from Query Parameters# Using index()res = test_str[0:test_str.index('?')]# printing resultprint("The base URL is : " + res) |
The original string is : www.geeksforgeeks.org?is = best The base URL is : www.geeksforgeeks.org
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #4 : Using operator.getitem(),index() methods
Approach
- Found index of ? using index() method
- Used operator.getitem(),slice() to extract the sliced string from start(0) to the index of ? and assigned to res variable
- Displayed the res variable
Python3
# Python3 code to demonstrate working of# Split URL from Query Parameters# Using index()# initializing stringtest_str = 'www.geeksforgeeks.org?is = best'# printing original stringprint("The original string is : " + str(test_str))# Split URL from Query Parameters# Using index()import operatorres = operator.getitem(test_str,slice(0, test_str.index('?')))# printing resultprint("The base URL is : " + res) |
The original string is : www.geeksforgeeks.org?is = best The base URL is : www.geeksforgeeks.org
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #5 : Using urlparse function:
1.Import the urlparse function from the urllib.parse module.
2.Define the input URL string as test_str.
3.Use the urlparse function to parse the test_str URL into a ParseResult object.
4.Use the _replace method to create a new ParseResult object with the query parameter set to None.
5.Use the geturl method to generate a new URL string from the modified ParseResult object.
6.Print the new URL string to the console.
Python3
# Importing the urlparse function from the urllib.parse modulefrom urllib.parse import urlparse# Defining the input URL string# Using the urlparse function to parse the input URL into its component partsparsed_url = urlparse(test_str)# printing original stringprint("The original string is : " + str(test_str))# Using the _replace method to create a new parsed URL object with the query parameter set to None# This effectively removes the query parameter from the URLnew_parsed_url = parsed_url._replace(query=None)# Using the geturl method to generate a new URL string from the modified parsed URL objectnew_url_str = new_parsed_url.geturl()# Printing the new URL stringprint(new_url_str) |
The original string is : http://www.geeksforgeeks.org?is=best http://www.geeksforgeeks.org
Time complexity:
Parsing the URL using the urlparse function has a time complexity of O(n), where n is the length of the input string.
Using the _replace method has a time complexity of O(1), as it simply creates a new ParseResult object with a modified query parameter.
Using the geturl method has a time complexity of O(n), where n is the length of the output URL string.
Overall, the time complexity of this code is O(n), where n is the length of the input and output strings.
Auxiliary Space:
The space complexity of this code is O(n), where n is the length of the input and output strings.
This is because the urlparse function creates a new ParseResult object that stores the various components of the URL (such as the scheme, netloc, path, query, and fragment).
The _replace method creates a new ParseResult object with a modified query parameter, and the geturl method generates a new URL string from this object.
Thus, the amount of space required by this code is proportional to the length of the input and output strings.
Method #6: Using re.split()
- Import the re module.
- Define a regular expression pattern to match the query parameters section of the URL (the part after the ? character).
- Use the re.split() function to split the URL using the regular expression pattern.
- The first element of the resulting list will be the base URL.
Python3
# Python3 code to demonstrate working of # Split URL from Query Parameters# Using re.split()# import re moduleimport re# initializing stringtest_str = 'www.geeksforgeeks.org?is = best'# printing original stringprint("The original string is : " + str(test_str))# Split URL from Query Parameters# Using re.split()pattern = r'\?' # regular expression pattern to match the query parameters sectionres = re.split(pattern, test_str)[0]# printing result print("The base URL is : " + res) |
The original string is : www.geeksforgeeks.org?is = best The base URL is : www.geeksforgeeks.org
Time complexity: The time complexity of this method is O(n), where n is the length of the input string.
Auxiliary space: The space complexity of this method is O(n), where n is the length of the input string.
