apache/cassandra · error · InvalidRequestException

Source Table '%s.%s' doesn't exist

Error message

Source Table '%s.%s' doesn't exist

What it means

CREATE TABLE LIKE could not find the source table within the (existing) source keyspace. apply() calls keyspaceMetadata.getTableOrViewNullable(sourceTableName) and throws this InvalidRequestException when null. The source must be an actual base table; missing names are rejected before type checks.

Source

Thrown at src/java/org/apache/cassandra/cql3/statements/schema/CopyTableStatement.java:133

    @Override
    public boolean compatibleWith(ClusterMetadata metadata)
    {
        return metadata.directory.commonSerializationVersion.isAtLeast(Version.V5);
    }

    @Override
    public Keyspaces apply(ClusterMetadata metadata)
    {
        Keyspaces schema = metadata.schema.getKeyspaces();
        KeyspaceMetadata sourceKeyspaceMeta = schema.getNullable(sourceKeyspace);

        if (null == sourceKeyspaceMeta)
            throw ire("Source Keyspace '%s' doesn't exist", sourceKeyspace);

        TableMetadata sourceTableMeta = sourceKeyspaceMeta.getTableOrViewNullable(sourceTableName);

        if (null == sourceTableMeta)
            throw ire("Source Table '%s.%s' doesn't exist", sourceKeyspace, sourceTableName);

        if (sourceTableMeta.isIndex())
            throw ire("Cannot use CREATE TABLE LIKE on an index table '%s.%s'.", sourceKeyspace, sourceTableName);

        if (sourceTableMeta.isView())
            throw ire("Cannot use CREATE TABLE LIKE on a materialized view '%s.%s'.", sourceKeyspace, sourceTableName);

        KeyspaceMetadata targetKeyspaceMeta = schema.getNullable(targetKeyspace);
        if (null == targetKeyspaceMeta)
            throw ire("Target Keyspace '%s' doesn't exist", targetKeyspace);

        if (targetKeyspaceMeta.hasTable(targetTableName))
        {
            if (ifNotExists)
                return schema;

            throw new AlreadyExistsException(targetKeyspace, targetTableName);
        }

View on GitHub (pinned to 88fd0f6a0e)

Solutions

1. List tables with `DESCRIBE TABLES` or system_schema.tables to confirm the source name
2. Correct the source table name (quote mixed-case identifiers)
3. Create/recreate the source table before running CREATE TABLE LIKE
4. If the intent was to copy a view/index structure, manually author the CREATE TABLE instead

Example fix

// before
CREATE TABLE ks2.t2 LIKE ks.tabel1; -- typo

// after
CREATE TABLE ks2.t2 LIKE ks.table1;

// Ensure the source table exists
const rows = await session.execute(
  "SELECT table_name FROM system_schema.tables WHERE keyspace_name=? AND table_name=?",
  [srcKs, srcTable]);
if (rows.rows.length === 0) throw new Error(`Source table ${srcKs}.${srcTable} does not exist`);

null

try {
  session.execute(`CREATE TABLE ks2.t2 LIKE ${srcKs}.${srcTable}`);
} catch (e) {
  if (/Source Table .* doesn't exist/.test(e.message)) {
    // fix the name or create the table first
  } else throw e;
}

When it happens

Trigger: `CREATE TABLE t2 LIKE <ks>.<missing>` where the table name is misspelled, the table was dropped, or the target is actually a view/index name in another lookup path; also when quoting/case differs.

Common situations: Typos in table names in migration scripts; schemas drifted between environments; assuming LIKE works against a materialized view or index (separate errors exist for those).

Understand the failure class

Background: 'Could not be found', 'does not exist', 'not found in database': the resource-not-found family when an ID, slug, key, or URI lookup comes back empty — this error's family across 20 libraries.

Related errors

• Table '%s.%s' doesn't exist
• Source Keyspace '%s' doesn't exist
• Cannot use CREATE TABLE LIKE on an index table '%s.%s'.
• Cannot use CREATE TABLE LIKE on a materialized view '%s.%s'.
• Target Keyspace '%s' doesn't exist

AI-assisted analysis of apache/cassandra@88fd0f6a0e (2026-09-10). Data as JSON: /api/errors/82e7c30b133f5ceb. Report an issue: GitHub.