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Implement the information schema with_hierarchy column
In PostgreSQL, this is included in the SELECT privilege, so show YES or NO depending on whether SELECT is granted.
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@ -3684,7 +3684,14 @@ ORDER BY c.ordinal_position;
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<row>
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<entry><literal>with_hierarchy</literal></entry>
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<entry><type>yes_or_no</type></entry>
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<entry>Applies to a feature not available in <productname>PostgreSQL</></entry>
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<entry>
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In the SQL standard, <literal>WITH HIERARCHY OPTION</literal>
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is a separate (sub-)privilege allowing certain operations on
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table inheritance hierarchies. In PostgreSQL, this is included
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in the <literal>SELECT</literal> privilege, so this column
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shows <literal>YES</literal> if the privilege
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is <literal>SELECT</literal>, else <literal>NO</literal>.
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</entry>
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</row>
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</tbody>
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</tgroup>
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@ -5364,7 +5371,14 @@ ORDER BY c.ordinal_position;
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<row>
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<entry><literal>with_hierarchy</literal></entry>
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<entry><type>yes_or_no</type></entry>
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<entry>Applies to a feature not available in <productname>PostgreSQL</></entry>
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<entry>
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In the SQL standard, <literal>WITH HIERARCHY OPTION</literal>
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is a separate (sub-)privilege allowing certain operations on
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table inheritance hierarchies. In PostgreSQL, this is included
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in the <literal>SELECT</literal> privilege, so this column
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shows <literal>YES</literal> if the privilege
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is <literal>SELECT</literal>, else <literal>NO</literal>.
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</entry>
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</row>
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</tbody>
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</tgroup>
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@ -1785,7 +1785,7 @@ CREATE VIEW table_privileges AS
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pg_has_role(grantee.oid, c.relowner, 'USAGE')
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OR c.grantable
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THEN 'YES' ELSE 'NO' END AS yes_or_no) AS is_grantable,
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CAST('NO' AS yes_or_no) AS with_hierarchy
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CAST(CASE WHEN c.prtype = 'SELECT' THEN 'YES' ELSE 'NO' END AS yes_or_no) AS with_hierarchy
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FROM (
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SELECT oid, relname, relnamespace, relkind, relowner, (aclexplode(relacl)).* FROM pg_class
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