Add migration to cleanup duplicate dropdown options
The staging database has duplicate sub-channels (5 "Meta", 2 "Magazine") which causes the last duplicate (with 0 proof types) to overwrite the correct one in the API response. This migration: 1. Identifies duplicate sub-channels and channels 2. Keeps the one with the most children (proof types) 3. Deletes the duplicates 4. Adds unique partial indexes to prevent future duplicates Co-Authored-By: Claude Opus 4.5 <noreply@anthropic.com>
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michael
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"""Cleanup duplicate dropdown options and add unique constraint
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Revision ID: 004_cleanup_duplicates
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Revises: 003_add_meta_proof_types
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Create Date: 2026-01-23
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This migration fixes an issue where duplicate sub-channels (e.g., multiple "Meta" entries)
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were created, causing proof types to not display correctly because the last duplicate
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(with no children) would overwrite the correct one in the API response.
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"""
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from typing import Sequence, Union
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from alembic import op
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import sqlalchemy as sa
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# revision identifiers, used by Alembic.
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revision: str = '004_cleanup_duplicates'
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down_revision: Union[str, None] = '003_add_meta_proof_types'
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branch_labels: Union[str, Sequence[str], None] = None
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depends_on: Union[str, Sequence[str], None] = None
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def upgrade() -> None:
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"""Remove duplicate dropdown options, keeping the ones with children (proof types)."""
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conn = op.get_bind()
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# Step 1: Find and remove duplicate sub-channels (keeping the one with proof types)
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# For each (option_type, parent_id, value) combination, keep the one that has children
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# or the oldest one if none have children
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# First, identify duplicates - sub-channels with same parent and value
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duplicates = conn.execute(
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sa.text("""
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SELECT option_type, parent_id, value, COUNT(*) as cnt
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FROM dropdown_options
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WHERE option_type = 'sub_channel'
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GROUP BY option_type, parent_id, value
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HAVING COUNT(*) > 1
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""")
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).fetchall()
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print(f"[Migration 004] Found {len(duplicates)} duplicate sub-channel groups")
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for dup in duplicates:
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option_type, parent_id, value = dup[0], dup[1], dup[2]
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print(f"[Migration 004] Processing duplicate: {value} (parent_id={parent_id})")
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# Get all entries for this duplicate
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entries = conn.execute(
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sa.text("""
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SELECT d.id,
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(SELECT COUNT(*) FROM dropdown_options WHERE parent_id = d.id) as child_count
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FROM dropdown_options d
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WHERE d.option_type = :option_type
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AND d.parent_id = :parent_id
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AND d.value = :value
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ORDER BY child_count DESC, d.created_at ASC
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"""),
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{"option_type": option_type, "parent_id": parent_id, "value": value}
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).fetchall()
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if len(entries) > 1:
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# Keep the first one (has most children or oldest), delete the rest
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keep_id = entries[0][0]
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print(f"[Migration 004] Keeping {keep_id} with {entries[0][1]} children")
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for entry in entries[1:]:
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delete_id = entry[0]
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print(f"[Migration 004] Deleting duplicate {delete_id} with {entry[1]} children")
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# First delete any orphaned children (shouldn't exist but be safe)
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conn.execute(
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sa.text("DELETE FROM dropdown_options WHERE parent_id = :id"),
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{"id": delete_id}
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)
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# Then delete the duplicate
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conn.execute(
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sa.text("DELETE FROM dropdown_options WHERE id = :id"),
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{"id": delete_id}
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)
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# Step 2: Do the same for channels (just in case)
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channel_duplicates = conn.execute(
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sa.text("""
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SELECT option_type, value, COUNT(*) as cnt
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FROM dropdown_options
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WHERE option_type = 'channel' AND parent_id IS NULL
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GROUP BY option_type, value
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HAVING COUNT(*) > 1
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""")
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).fetchall()
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print(f"[Migration 004] Found {len(channel_duplicates)} duplicate channel groups")
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for dup in channel_duplicates:
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option_type, value = dup[0], dup[1]
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print(f"[Migration 004] Processing duplicate channel: {value}")
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entries = conn.execute(
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sa.text("""
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SELECT d.id,
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(SELECT COUNT(*) FROM dropdown_options WHERE parent_id = d.id) as child_count
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FROM dropdown_options d
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WHERE d.option_type = :option_type
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AND d.parent_id IS NULL
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AND d.value = :value
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ORDER BY child_count DESC, d.created_at ASC
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"""),
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{"option_type": option_type, "value": value}
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).fetchall()
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if len(entries) > 1:
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keep_id = entries[0][0]
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print(f"[Migration 004] Keeping channel {keep_id} with {entries[0][1]} children")
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for entry in entries[1:]:
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delete_id = entry[0]
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print(f"[Migration 004] Deleting duplicate channel {delete_id}")
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# Delete sub-channels and their proof types first
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sub_channels = conn.execute(
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sa.text("SELECT id FROM dropdown_options WHERE parent_id = :id"),
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{"id": delete_id}
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).fetchall()
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for sc in sub_channels:
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conn.execute(
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sa.text("DELETE FROM dropdown_options WHERE parent_id = :id"),
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{"id": sc[0]}
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)
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conn.execute(
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sa.text("DELETE FROM dropdown_options WHERE id = :id"),
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{"id": sc[0]}
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)
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conn.execute(
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sa.text("DELETE FROM dropdown_options WHERE id = :id"),
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{"id": delete_id}
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)
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# Step 3: Add unique constraint to prevent future duplicates
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# For sub-channels: unique on (option_type, parent_id, value)
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# For channels: unique on (option_type, value) where parent_id IS NULL
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# Create a partial unique index for sub-channels and proof types (where parent_id is NOT NULL)
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op.execute("""
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CREATE UNIQUE INDEX IF NOT EXISTS idx_dropdown_options_unique_with_parent
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ON dropdown_options (option_type, parent_id, value)
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WHERE parent_id IS NOT NULL
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""")
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# Create a partial unique index for channels and brand guidelines (where parent_id IS NULL)
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op.execute("""
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CREATE UNIQUE INDEX IF NOT EXISTS idx_dropdown_options_unique_without_parent
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ON dropdown_options (option_type, value)
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WHERE parent_id IS NULL
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""")
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print("[Migration 004] Unique indexes created successfully")
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def downgrade() -> None:
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"""Remove unique constraints (cannot restore deleted duplicates)."""
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op.execute("DROP INDEX IF EXISTS idx_dropdown_options_unique_with_parent")
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op.execute("DROP INDEX IF EXISTS idx_dropdown_options_unique_without_parent")
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