alessandro-nori commented on code in PR #1629:
URL: https://github.com/apache/iceberg-python/pull/1629#discussion_r1948580037
##########
pyiceberg/catalog/sql.py:
##########
@@ -610,15 +610,26 @@ def list_namespaces(self, namespace: Union[str,
Identifier] = ()) -> List[Identi
table_stmt =
select(IcebergTables.table_namespace).where(IcebergTables.catalog_name ==
self.name)
namespace_stmt =
select(IcebergNamespaceProperties.namespace).where(IcebergNamespaceProperties.catalog_name
== self.name)
if namespace:
- namespace_str = Catalog.namespace_to_string(namespace,
NoSuchNamespaceError)
- table_stmt =
table_stmt.where(IcebergTables.table_namespace.like(namespace_str))
- namespace_stmt =
namespace_stmt.where(IcebergNamespaceProperties.namespace.like(namespace_str))
+ namespace_like = Catalog.namespace_to_string(namespace,
NoSuchNamespaceError) + "%"
Review Comment:
empirically, I found out that not adding `%` only returns the namespace
itself.
I believe the like() function doesn't make any assumptions about where the
fuzzy match character should appear in the string you provide, so it's up to
the user to include the % where needed.
I found this
[example](https://docs.sqlalchemy.org/en/20/core/sqlelement.html#sqlalchemy.sql.expression.ColumnOperators.like)
in the doc.
Alternatively we could use
[startswith](https://docs.sqlalchemy.org/en/20/core/sqlelement.html#sqlalchemy.sql.expression.ColumnOperators.startswith)
which does the concatenation of the string with `%` for us.
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