How do I add a foreign key to an existing SQLite table?

Question

You can't.

Although the SQL-92 syntax to add a foreign key to your table would be as follows:

ALTER TABLE child ADD CONSTRAINT fk_child_parent
                  FOREIGN KEY (parent_id) 
                  REFERENCES parent(id);

SQLite doesn't support the ADD CONSTRAINT variant of the ALTER TABLE command (sqlite.org: SQL Features That SQLite Does Not Implement).

Therefore, the only way to add a foreign key in sqlite 3.6.1 is during CREATE TABLE as follows:

CREATE TABLE child ( 
    id           INTEGER PRIMARY KEY, 
    parent_id    INTEGER, 
    description  TEXT,
    FOREIGN KEY (parent_id) REFERENCES parent(id)
);

Unfortunately you will have to save the existing data to a temporary table, drop the old table, create the new table with the FK constraint, then copy the data back in from the temporary table. (sqlite.org - FAQ: Q11)

You can add the constraint if you alter table and add the column that uses the constraint.

First, create table without the parent_id:

CREATE TABLE child( 
  id INTEGER PRIMARY KEY,  
  description TEXT);

Then, alter table:

ALTER TABLE child ADD COLUMN parent_id INTEGER REFERENCES parent(id);

Please check https://www.sqlite.org/lang_altertable.html#otheralter

The only schema altering commands directly supported by SQLite are the "rename table" and "add column" commands shown above. However, applications can make other arbitrary changes to the format of a table using a simple sequence of operations. The steps to make arbitrary changes to the schema design of some table X are as follows:

1. If foreign key constraints are enabled, disable them using PRAGMA foreign_keys=OFF.
2. Start a transaction.
3. Remember the format of all indexes and triggers associated with table X. This information will be needed in step 8 below. One way to do this is to run a query like the following: SELECT type, sql FROM sqlite_master WHERE tbl_name='X'.
4. Use CREATE TABLE to construct a new table "new_X" that is in the desired revised format of table X. Make sure that the name "new_X" does not collide with any existing table name, of course.
5. Transfer content from X into new_X using a statement like: INSERT INTO new_X SELECT ... FROM X.
6. Drop the old table X: DROP TABLE X.
7. Change the name of new_X to X using: ALTER TABLE new_X RENAME TO X.
8. Use CREATE INDEX and CREATE TRIGGER to reconstruct indexes and triggers associated with table X. Perhaps use the old format of the triggers and indexes saved from step 3 above as a guide, making changes as appropriate for the alteration.
9. If any views refer to table X in a way that is affected by the schema change, then drop those views using DROP VIEW and recreate them with whatever changes are necessary to accommodate the schema change using CREATE VIEW.
10. If foreign key constraints were originally enabled then run PRAGMA foreign_key_check to verify that the schema change did not break any foreign key constraints.
11. Commit the transaction started in step 2.
12. If foreign keys constraints were originally enabled, reenable them now.

The procedure above is completely general and will work even if the schema change causes the information stored in the table to change. So the full procedure above is appropriate for dropping a column, changing the order of columns, adding or removing a UNIQUE constraint or PRIMARY KEY, adding CHECK or FOREIGN KEY or NOT NULL constraints, or changing the datatype for a column, for example.

Yes, you can, without adding a new column. You have to be careful to do it correctly in order to avoid corrupting the database, so you should completely back up your database before trying this.

for your specific example:

CREATE TABLE child(
  id INTEGER PRIMARY KEY,
  parent_id INTEGER,
  description TEXT
);

--- create the table we want to reference
create table parent(id integer not null primary key);

--- now we add the foreign key
pragma writable_schema=1;
update SQLITE_MASTER set sql = replace(sql, 'description TEXT)',
    'description TEXT, foreign key (parent_id) references parent(id))'
) where name = 'child' and type = 'table';

--- test the foreign key
pragma foreign_keys=on;
insert into parent values(1);
insert into child values(1, 1, 'hi'); --- works
insert into child values(2, 2, 'bye'); --- fails, foreign key violation

or more generally:

pragma writable_schema=1;

// replace the entire table's SQL definition, where new_sql_definition contains the foreign key clause you want to add
UPDATE SQLITE_MASTER SET SQL = new_sql_definition where name = 'child' and type = 'table';

// alternatively, you might find it easier to use replace, if you can match the exact end of the sql definition
// for example, if the last column was my_last_column integer not null:
UPDATE SQLITE_MASTER SET SQL = replace(sql, 'my_last_column integer not null', 'my_last_column integer not null, foreign key (col1, col2) references other_table(col1, col2)') where name = 'child' and type = 'table';

pragma writable_schema=0;

Either way, you'll probably want to first see what the SQL definition is before you make any changes:

select sql from SQLITE_MASTER where name = 'child' and type = 'table';

If you use the replace() approach, you may find it helpful, before executing, to first test your replace() command by running:

select replace(sql, ...) from SQLITE_MASTER where name = 'child' and type = 'table';

As @Daniel Vassallo has said, you can't do it. The code that you have to use is something like this:

Given the table:

CREATE TABLE child( 
id INTEGER PRIMARY KEY, 
parent_id INTEGER, 
description TEXT);

I assume that you want to add the following Foreignk Key:

FOREIGN KEY (parent_id) REFERENCES parent(id);

So I would create a temporary table based on that table, then I would create a new table as the first one but with the Foreign Key and finally I would add the data of the temporary table to it:

CREATE TEMPORARY TABLE temp AS
SELECT 
    id,
    parent_id,
    description
FROM child;

DROP TABLE child;

CREATE TABLE child (
    id INTEGER PRIMARY KEY, 
    parent_id INTEGER, 
    description TEXT,
    FOREIGN KEY(parent_id) REFERENCES parent(id));

INSERT INTO child
 (  id,
    parent_id,
    description)
SELECT
    id,
    parent_id,
    description
FROM temp;