Accessing query parameters in Flask

Introduction

Query parameters let you send extra information in a web address. Accessing them helps your app respond differently based on what the user asks.

When you want to filter a list of items on a webpage, like showing only red shirts.

When you want to remember user choices, like sorting order or page number.

When you want to pass small bits of data without changing the page content.

When building search features that take input from the URL.

When you want to share links that include specific settings or filters.

Syntax

Flask

from flask import request

value = request.args.get('parameter_name')

request.args holds all query parameters as a dictionary-like object.

Use .get() to safely get a parameter value or None if missing.

Examples

Gets the value of the 'color' query parameter.

Flask

color = request.args.get('color')

Gets 'page' parameter or defaults to '1' if not provided.

Flask

page = request.args.get('page', '1')

Converts all query parameters into a regular Python dictionary.

Flask

all_params = request.args.to_dict()

Sample Program

This Flask app has a route '/search' that reads query parameters 'q' and 'page'. It shows them in the response. If 'page' is missing, it uses '1' as default.

Flask

from flask import Flask, request

app = Flask(__name__)

@app.route('/search')
def search():
    query = request.args.get('q', '')
    page = request.args.get('page', '1')
    return f"Search query: {query}, Page: {page}"

if __name__ == '__main__':
    app.run(debug=True)

Important Notes

Query parameters are always strings. Convert them if you need numbers.

If a parameter appears multiple times, request.args.get() returns the first one.

Use request.args.getlist('param') to get all values if repeated.

Summary

Query parameters add extra info to URLs for flexible web responses.

Use request.args.get() to read them safely in Flask.

Always handle missing parameters by providing defaults or checks.

Practice

1. In Flask, which method is used to safely access a query parameter named user from the URL?

easy

A. request.form['user']

B. request.args.get('user')

C. request.query['user']

D. request.params.get('user')

Solution

1. Step 1: Understand Flask request object

   Flask's request object holds incoming request data, including query parameters in args.

2. Step 2: Use the correct method to access query parameters

   request.args.get('user') safely retrieves the value of the user parameter or returns None if missing.

3. Final Answer:

   request.args.get('user') -> Option B

4. Quick Check:

   Use args.get() for query parameters [OK]

Hint: Use request.args.get('param') to read query parameters [OK]

Common Mistakes:

• Using request.form for query parameters
• Trying to access request.query which doesn't exist
• Using request.params instead of request.args

2. Which of the following is the correct syntax to get a query parameter page with a default value of 1 in Flask?

easy

A. request.args.get('page', default=1)

B. request.args.get('page', 1)

C. request.args['page'] or 1

D. request.args.get('page', fallback=1)

Solution

1. Step 1: Check the get() method parameters

   The get() method accepts a default keyword argument to provide a fallback value if the parameter is missing.

2. Step 2: Identify correct syntax for default

   Using request.args.get('page', default=1) correctly sets 1 as default. Passing 1 as second positional argument is invalid.

3. Final Answer:

   request.args.get('page', default=1) -> Option A

4. Quick Check:

   Use default= for fallback in get() [OK]

Hint: Use default= in get() to set fallback value [OK]

Common Mistakes:

• Passing default value as second positional argument
• Using fallback= which is not valid in Flask
• Accessing with [] which raises error if missing

3. What will be the output of this Flask route when accessed with URL /search?term=flask?

from flask import Flask, request
app = Flask(__name__)

@app.route('/search')
def search():
    term = request.args.get('term', 'none')
    return f"Search term: {term}"

medium

A. Error: term parameter missing

B. Search term: none

C. Search term:

D. Search term: flask

Solution

1. Step 1: Understand the URL query parameter

   The URL contains ?term=flask, so the query parameter term has value 'flask'.

2. Step 2: Check how term is retrieved in code

   request.args.get('term', 'none') returns 'flask' because the parameter exists.

3. Final Answer:

   Search term: flask -> Option D

4. Quick Check:

   Parameter present returns its value, not default [OK]

Hint: If parameter exists, get() returns its value, else default [OK]

Common Mistakes:

• Assuming default is returned even if parameter exists
• Confusing query parameters with form data
• Expecting empty string instead of actual value

4. Identify the error in this Flask code snippet that tries to get a query parameter id:

from flask import Flask, request
app = Flask(__name__)

@app.route('/item')
def item():
    id = request.args['id']
    return f"Item ID: {id}"

medium

A. Using request.args['id'] raises KeyError if 'id' missing

B. Missing import of request module

C. Route decorator syntax is incorrect

D. Using f-string inside return is invalid

Solution

1. Step 1: Analyze how query parameter is accessed

   Using request.args['id'] accesses the parameter directly and raises KeyError if missing.

2. Step 2: Identify safer alternative

   Using request.args.get('id') avoids error by returning None if missing.

3. Final Answer:

   Using request.args['id'] raises KeyError if 'id' missing -> Option A

4. Quick Check:

   Direct dict access can cause KeyError [OK]

Hint: Use get() to avoid KeyError when parameter might be missing [OK]

Common Mistakes:

• Assuming request.args['id'] never fails
• Ignoring possibility of missing parameters
• Confusing import errors with runtime errors

5. You want to create a Flask route /filter that reads two optional query parameters: category (default 'all') and limit (default 10, integer). Which code correctly handles both parameters safely?

hard

A. category = request.args.get('category') or 'all'
limit = int(request.args.get('limit', 10))

B. category = request.args['category'] or 'all'
limit = int(request.args['limit'] or 10)

C. category = request.args.get('category', default='all')
limit = int(request.args.get('limit', default=10))

D. category = request.args.get('category', 'all')
limit = int(request.args.get('limit', '10'))

Solution

1. Step 1: Handle category parameter with default

   Using request.args.get('category', default='all') safely returns 'all' if missing.

2. Step 2: Handle limit parameter with default and convert to int

   request.args.get('limit', default=10) returns string or int 10; wrapping with int() converts string to integer safely.

3. Step 3: Check other options for errors

   category = request.args['category'] or 'all' limit = int(request.args['limit'] or 10) risks KeyError; category = request.args.get('category') or 'all' limit = int(request.args.get('limit', 10)) uses positional default incorrectly; category = request.args.get('category', 'all') limit = int(request.args.get('limit', '10')) converts default string '10' which works but less clear than int 10.

4. Final Answer:

   category = request.args.get('category', default='all')\nlimit = int(request.args.get('limit', default=10)) -> Option C

5. Quick Check:

   Use get() with default and convert types explicitly [OK]

Hint: Use get() with default and convert types explicitly [OK]

Common Mistakes:

• Using [] which raises error if missing
• Passing default as positional argument incorrectly
• Not converting string query param to int