#5973: When in debugging mode, announce if a model is being made automatically
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Reporter: Aijo | Type: Enhancement
Status: new | Priority: Low
Milestone: 1.2.x.x | Component: Error
Handling
Version: 1.2 Final | Severity: Minor
Keywords: find all left join id primary key | Php_version: n/a
Cake_version: |
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If debug is set to 2 or above ''(via configure::write('debug', 2); )'' and
a model is being made automatically due to a model file not being present
even though a belongsTo relationship exists, then in the debug output
announce that 'Some models are being made automatically'.[[BR]][[BR]]
This way if someone wishes automatic models to be used then it only adds
one line of clutter to the debug. But if a model is being made
automatically because of a naming convention oversight or something
similar then they can be alerted to the fact that
a model is being made automatically.[[BR]][[BR]]
After checking my code several times and it proving to be perfect, I was
still receiving an error with the SQL query, it was looking for model.id
instead of the define primaryKey in my model, I had no idea cakephp could
make models automatically in the case of one missing. It turned out that I
hadn't named my file with the appropriate naming conventions (it was named
clientlocations.php insetad of client_locations.php). This nature of error
wasn't covered by the debugger and I feel it would help new bakers to
diagnose problems with Model::find();[[BR]][[BR]]
Thanks for your time.
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Ticket URL: <https://trac.cakephp.org/ticket/5973>
CakePHP : The Rapid Development Framework for PHP <https://trac.cakephp.org/>
Cake is a rapid development framework for PHP which uses commonly known design
patterns like ActiveRecord, Association Data Mapping, Front Controller and MVC.
Our primary goal is to provide a structured framework that enables PHP users at
all levels to rapidly develop robust web applications, without any loss to
flexibility.
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