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Create appp.py #9

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21 changes: 21 additions & 0 deletions appp.py
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Original file line number Diff line number Diff line change
@@ -0,0 +1,21 @@
import sqlite3
from flask import Flask, request

app = Flask(__name__)

@app.route('/search')
def search():
# Get the user input from the query parameter
query = request.args.get('query', '')

# Vulnerable: User input is directly concatenated into the SQL query
connection = sqlite3.connect('example.db')
cursor = connection.cursor()
cursor.execute(f"SELECT * FROM users WHERE name = '{query}'") # SQL injection vulnerability

Check failure

Code scanning / CodeQL

SQL query built from user-controlled sources High

This SQL query depends on a
user-provided value
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.

Copilot Autofix

AI about 1 year ago

To fix the SQL injection vulnerability, we should use parameterized queries instead of directly concatenating user input into the SQL query string. Parameterized queries ensure that user input is properly escaped and quoted by the database driver, preventing SQL injection attacks.

In this specific case, we will modify the cursor.execute call to use a parameterized query. This involves replacing the f-string with a query string that contains placeholders (?), and passing the user input as a separate argument to the execute method.

Suggested changeset 1
appp.py

Autofix patch

Autofix patch
Run the following command in your local git repository to apply this patch
cat << 'EOF' | git apply
diff --git a/appp.py b/appp.py
--- a/appp.py
+++ b/appp.py
@@ -13,3 +13,3 @@
     cursor = connection.cursor()
-    cursor.execute(f"SELECT * FROM users WHERE name = '{query}'")  # SQL injection vulnerability
+    cursor.execute("SELECT * FROM users WHERE name = ?", (query,))  # Use parameterized query to prevent SQL injection
     results = cursor.fetchall()
EOF
@@ -13,3 +13,3 @@
cursor = connection.cursor()
cursor.execute(f"SELECT * FROM users WHERE name = '{query}'") # SQL injection vulnerability
cursor.execute("SELECT * FROM users WHERE name = ?", (query,)) # Use parameterized query to prevent SQL injection
results = cursor.fetchall()
Copilot is powered by AI and may make mistakes. Always verify output.
results = cursor.fetchall()
connection.close()

return str(results)

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

Check failure

Code scanning / CodeQL

Flask app is run in debug mode High

A Flask app appears to be run in debug mode. This may allow an attacker to run arbitrary code through the debugger.

Copilot Autofix

AI about 1 year ago

To fix the problem, we need to ensure that the Flask application does not run in debug mode in a production environment. The best way to achieve this is by using an environment variable to control the debug mode. This way, we can easily switch between development and production environments without changing the code.

  1. Import the os module to access environment variables.
  2. Use an environment variable to determine whether to run the app in debug mode.
  3. Update the app.run call to use this environment variable.
Suggested changeset 1
appp.py

Autofix patch

Autofix patch
Run the following command in your local git repository to apply this patch
cat << 'EOF' | git apply
diff --git a/appp.py b/appp.py
--- a/appp.py
+++ b/appp.py
@@ -2,2 +2,3 @@
 from flask import Flask, request
+import os
 
@@ -20,2 +21,3 @@
 if __name__ == '__main__':
-    app.run(debug=True)
+    debug_mode = os.getenv('FLASK_DEBUG', 'False').lower() in ['true', '1', 't']
+    app.run(debug=debug_mode)
EOF
@@ -2,2 +2,3 @@
from flask import Flask, request
import os

@@ -20,2 +21,3 @@
if __name__ == '__main__':
app.run(debug=True)
debug_mode = os.getenv('FLASK_DEBUG', 'False').lower() in ['true', '1', 't']
app.run(debug=debug_mode)
Copilot is powered by AI and may make mistakes. Always verify output.