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Issue with constrained() method used after foreignIdFor(), instead of table name when $table parameter is not passed uses column name #25830

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Issue with constrained() method used after foreignIdFor(), instead of table name when $table parameter is not passed uses column name

Issue with constrained() method used after foreignIdFor(), instead of table name when $table parameter is not passed uses column name #25830

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Job Run time
MariaDB 10
1m 3s
SQL Server 2017
1m 14s
SQL Server 2019
1m 9s
MySQL 5.7
1m 22s
MySQL 8
1m 36s
PostgreSQL 10
1m 10s
SQLite
17s
PostgreSQL 14
1m 14s
9m 5s