Fix introspection of table options in non-default schemas #6753
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morozov
commented
Feb 1, 2025
| if (str_contains($tableName, '.')) { | ||
| [$schemaName, $tableName] = explode('.', $tableName); | ||
| } else { | ||
| $schemaName = 'dbo'; |
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We still have to default to the dbo schema here because the platform doesn't know of the current schema, which is scoped to a connection.
UPD: technically, we can use CURRENT_SCHEMA(), but I'll leave dbo for now for consistency with the rest of the code.
morozov
commented
Feb 1, 2025
| @@ -227,9 +227,15 @@ protected function _getPortableTableForeignKeysList($tableForeignKeys) | |||
| $name = $tableForeignKey['ForeignKey']; | |||
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| if (! isset($foreignKeys[$name])) { | |||
| $referencedTableName = $tableForeignKey['ReferenceTableName']; 8000 | |||
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| if ($tableForeignKey['ReferenceSchemaName'] !== 'dbo') { | |||
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We still have to default to the dbo schema here because the schema manager is unaware of the current schema in 3.x (see #6677).
greg0ire
approved these changes
Feb 2, 2025
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When introspecting table options, the PostgreSQL and SQL Server schema managers do not take the schema name into account, so all selected options are mistakenly attributed to the table with the same unqualified name in the default schema.
Additionally, the SQL Server platform cannot create tables with comments in the non-default schema.