parser stack overflow

This error occurs when your SQL statement is too deeply nested for SQLite's parser. Common with complex expressions or programmatically generated queries.

The parser stack overflow error means your SQL statement is too complex for SQLite to parse.

Understanding the Error

Error: parser stack overflow

The SQL has too many levels of nesting for SQLite's parser stack.

Common Causes

1. Deeply Nested Expressions

SQL

-- Too many nested parentheses
SELECT ((((((((((((((((((((col))))))))))))))))))));

2. Long OR/AND Chains

SQL

-- Generated query with hundreds of conditions
SELECT * FROM users
WHERE id = 1 OR id = 2 OR id = 3 OR ... -- 500+ OR clauses

3. Nested Subqueries

SQL

SELECT * FROM (
  SELECT * FROM (
    SELECT * FROM (
      SELECT * FROM (
        -- Many more levels
      )
    )
  )
);

4. Complex CASE Expressions

SQL

SELECT
  CASE WHEN a = 1 THEN
    CASE WHEN b = 1 THEN
      CASE WHEN c = 1 THEN
        -- Deeply nested CASE
      END
    END
  END
FROM table1;

How to Fix It

Solution 1: Use IN Instead of OR

SQL

-- Instead of:
SELECT * FROM users WHERE id = 1 OR id = 2 OR id = 3 OR ...

-- Use:
SELECT * FROM users WHERE id IN (1, 2, 3, ...);

Solution 2: Use Temp Table

SQL

-- Create temp table with values
CREATE TEMP TABLE filter_ids (id INTEGER);
INSERT INTO filter_ids VALUES (1), (2), (3), ...;

-- Join instead of long WHERE
SELECT u.* FROM users u
JOIN filter_ids f ON u.id = f.id;

DROP TABLE filter_ids;

Solution 3: Flatten Subqueries

SQL

-- Instead of nested subqueries
SELECT *
FROM a
JOIN b ON a.id = b.a_id
JOIN c ON b.id = c.b_id
JOIN d ON c.id = d.c_id;

Solution 4: Use CTEs

SQL

-- Common Table Expressions are flatter
WITH step1 AS (
  SELECT * FROM a WHERE condition1
),
step2 AS (
  SELECT * FROM step1 JOIN b ON step1.id = b.a_id
),
step3 AS (
  SELECT * FROM step2 JOIN c ON step2.id = c.b_id
)
SELECT * FROM step3;

Solution 5: Split Complex Queries

JAVASCRIPT

// Break into multiple queries
const ids = db.prepare('SELECT id FROM users WHERE active = 1').all();
const filtered = ids.filter(/* complex logic */);
const results = db.prepare(`
  SELECT * FROM users WHERE id IN (${filtered.map(() => '?').join(',')})
`).all(filtered.map(r => r.id));

Solution 6: Simplify Expressions

JAVASCRIPT

// Instead of building complex SQL
// Do logic in application code
const rows = db.prepare('SELECT * FROM data').all();
const results = rows.filter(row => {
  // Complex conditions here
  return someComplexCondition(row);
});

Prevention

JAVASCRIPT

// Limit query complexity when building dynamically
function buildQuery(conditions, maxDepth = 50) {
  if (conditions.length > maxDepth) {
    throw new Error('Too many conditions, use different approach');
  }
  // Build query...
}

Best Practices

Use IN instead of many ORs
Flatten subqueries into JOINs
Use CTEs for readability
Limit dynamic query complexity
Move complex logic to application code