Live Demos & Playground¶
Welcome to the General SQL Parser Live Demos! Test the parsing capabilities of GSP directly in your browser with real-time results and interactive visualizations.
๐ Quick SQL Playground¶
Try parsing any SQL statement instantly. Select your database vendor and see the results in real-time.
๐ฏ Specialized Demos¶
Explore specific aspects of SQL parsing with our specialized interactive demos:
๐ SQL Playground¶
Comprehensive SQL parsing environment - Multi-database support (25+ vendors) - Real-time syntax validation - Advanced parsing options - Export results in multiple formats
๐ณ AST Explorer¶
Visual Abstract Syntax Tree exploration - Interactive tree visualization - Node inspection and details - AST manipulation tools - Code generation from AST
โ๏ธ Database Comparison¶
Cross-database parsing comparison - Parse same SQL across different databases - Highlight vendor-specific differences - Compatibility analysis - Migration assistance tools
๐ผ Use Cases¶
Real-world examples and applications - Code migration scenarios - Query optimization examples - Data lineage analysis - Integration patterns
๐ Popular Examples¶
Try these commonly used SQL patterns:
Complex JOIN Query
SELECT u.username, p.title, c.content
FROM users u
LEFT JOIN posts p ON u.id = p.user_id
LEFT JOIN comments c ON p.id = c.post_id
WHERE u.active = 1
ORDER BY p.created_at DESC;
Window Functions
SELECT employee_id, department_id, salary,
AVG(salary) OVER (PARTITION BY department_id) as dept_avg,
RANK() OVER (ORDER BY salary DESC) as salary_rank
FROM employees;
CTE with Recursion
WITH RECURSIVE employee_hierarchy AS (
SELECT id, name, manager_id, 1 as level
FROM employees WHERE manager_id IS NULL
UNION ALL
SELECT e.id, e.name, e.manager_id, eh.level + 1
FROM employees e
JOIN employee_hierarchy eh ON e.manager_id = eh.id
)
SELECT * FROM employee_hierarchy;
Subquery with Aggregation
SELECT customer_id, customer_name,
(SELECT COUNT(*) FROM orders o
WHERE o.customer_id = c.customer_id) as order_count,
(SELECT MAX(order_total) FROM orders o
WHERE o.customer_id = c.customer_id) as max_order
FROM customers c
WHERE customer_id IN (
SELECT customer_id FROM orders
GROUP BY customer_id
HAVING COUNT(*) > 5
);
๐ง API Integration¶
All demos use the same REST API that you can integrate into your applications:
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Response:
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๐ Getting Started¶
- Try the Quick Playground above with your own SQL
- Explore specialized demos for deep-dive features
- Check our tutorials for integration guidance
- Read the API documentation for programmatic access
๐ Learning Path¶
New to SQL Parsing? Follow this recommended path:
- ๐ Getting Started Tutorial
- ๐งช Try the Quick Playground (above)
- ๐ SQL Playground for comprehensive testing
- ๐ณ AST Explorer for understanding structure
- ๐ Reference Documentation for detailed API