Sorry, I forgot to include the Notification.user_id => $me: 'conditions' => array( 'Notification.user_id' => $me, 'OR' => array( 'Policy.status' => 1, 'Policy.status IS NULL' ) )
or for MySQL: 'conditions' => array( 'Notification.user_id' => $me, 'IFNULL(Policy.status, 1)' => 1 ) On Feb 16, 8:47 pm, ShadowCross <[email protected]> wrote: > Try: > > 'conditions' => array( > 'OR' => array( > 'Policy.status' => 1, > 'Policy.status IS NULL' > ) > ) > > or for MySQL: > > 'conditions' => array( > 'IFNULL(Policy.status, 1)' => 1 > ) > > On Feb 16, 10:54 am, "Krissy Masters" <[email protected]> > wrote: > > > > > Sorry I did not explain the Notifications better. > > If a Notification is made thru the form (external notification, not related > > to a Policy on the site) then it has no policy_id Simply a record in the > > table with > > id so when the JOINS with POLICY happens all the external are automatically > > forgot about since there are no POLICY records of any kind since its > > external. Nothing to join with. > > > EXTERNAL: > > Id = > 123 > > Policy_id = null > > Name => sdfsf > > ...other files > > > INTERNAL > > Id = > 456 > > Policy_id = 856-954-8 > > ..every other field blank since the data is pulled from the policy_id > > > Thanks, > > > K > > > -----Original Message----- > > From: [email protected] [mailto:[email protected]] On Behalf > > > Of John Andersen > > Sent: Wednesday, February 16, 2011 4:23 AM > > To: CakePHP > > Subject: Re: Joins Conditions Contain Question > > > Would it be possible to assign a status value of "Not applicable" (0) > > to the B notifications, thus you would be able to use a condition like > > Policy.status => array(0,1) AND Notification.user_id => $me > > > or something similar :) > > Enjoy, > > John > > > On 15 Feb., 22:27, "Krissy Masters" <[email protected]> > > wrote: > > > This WILL sound complicated. > > > > Policy hasMany Notification / Notification belongsTo Policy > > > > Now the Notification can be > > > > A. internal (relating to specific content in the site, so the data > > is > > > gathered directly , created by logged in user, what page Policy it > > pertains, > > > other policy info, notification reason..) > > > > B. external (then the user fills out a form) and this is represented > > > by Notification.local table field bool > > > > So now paginate Notifications easy enough until the JOINS condition > > > Notification.user_id => $me, Policy.status => 1. > > > > Since external Notifications have no Policy.status those are not being > > > pulled obviously. This is where the problem is. I have to keep the > > > Policy.status for archive / record keeping legal mumbo jumbo so I simply > > can > > > not delete a Notification, has to be marked read, fixed, outstanding, > > > overdue. > > > > And all Nofifications must be in 1 page (pagination). I tried making an > > > ExternalNotification model / controller but then I run into the issues of > > > paginating 2 separate models as 1. > > > > If there is a way to find by( Notification.user_id => $me, Policy.status > > => > > > 1) or (Notification.user_id => $me, Policy.status => does not exist?) > > > > Some other method suggestion, point me in a general direction. > > > > Thanks guys, > > > > K > > > -- > > Our newest site for the community: CakePHP Video > > Tutorialshttp://tv.cakephp.org > > Check out the new CakePHP Questions sitehttp://ask.cakephp.organdhelp > > others with their CakePHP related questions. > > > To unsubscribe from this group, send email to > > [email protected] For more options, visit this group > > athttp://groups.google.com/group/cake-php -- Our newest site for the community: CakePHP Video Tutorials http://tv.cakephp.org Check out the new CakePHP Questions site http://ask.cakephp.org and help others with their CakePHP related questions. To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/cake-php
