 – for example, that you have written policies and procedures that describe positive and preventive behaviour management, including behaviour support plans, and that identify the interventions that can and can’t be used.
Allegations of abuse – You must have effective processes for dealing with allegations of abuse and with any incidents that raise safety concerns. This will include, for example, processes for reporting cases of harm to the police or another agency.
If the application form is filled out successfully, then an approvals assessor will arrange a site visit to the community venue. If there are any requirements that are unsatisfied, the assessor will provide written advice. Once the written advice is complied with, they may inform the assessor. The assessor will then grant approval depending on how well the advice was complied with.
<filename>apache/htdocs/slims7_cendana/admin/modules/membership/member_custom_fields.inc.php
<?php
/**
 * Membership data custom fields
 *
 * Copyright (C) 2010  <NAME> (<EMAIL>)
 *
 * This program is free software; you can redistribute it and/or modify
 * it under the terms of the GNU General Public License as published by
 * the Free Software Foundation; either version 3 of the License, or
 * (at your option) any later version.
 *
 * This program is distributed in the hope that it will be useful,
 * but WITHOUT ANY WARRANTY; without even the implied warranty of
 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 * GNU General Public License for more details.
 *
 * You should have received a copy of the GNU General Public License
 * along with this program; if not, write to the Free Software
 * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA  02110-1301  USA
 *
 */

// be sure that this file not accessed directly
if (INDEX_AUTH!= 1) {
    die("can not access this file directly");
}

/**
 * Here you can add custom field to SLiMS Membership database
 * The field you define here must also exists in 'biblio' table in database
 * field 'type' can be on of these: 'text', 'longtext', 'numeric', 'dropdown', 'checklist', 'date' or 'choice'
 *
 */

/*
$member_custom_fields = array(
	'customfield1' => array(
		'dbfield' => 'customfield1', // name of field in 'biblio' table in database, make sure you already define it!
		'label' => __('Custom Field 1'), // label of field in form
		'type' => 'text', // type of field
		'default' => '', // default value of field
		'max' => '50', // maximum character to enter in 'text' field type
		'data' => false, // an array of data for 'dropdown', 'checklist' or 'choice'
		'indexed' => true, // NOT APPLICABLE YET, FOR FUTURE RELEASE USE
		'width' => 50), // width of field in form for 'text' field type, maximum is 100

	'customfield2' => array(
		'dbfield' => 'customfield2',
		'label' => __('Custom Field 2'),
		'type' => 'dropdown',
		'default' => 'value2',
		'data' => array(
			array('value1', 'Value 1'),
			array('value2', 'Value 2')
			),
		'indexed' => true),

	'customfield3' => array(
		'dbfield' => 'customfield3',
		'label' => __('Custom Field 3'),
		'type' => 'checklist',
		'default' => array('value2', 'value3'),
		'data' => array(
			array('value1', 'Value 1'),
			array('value2', 'Value 2'),
			array('value3', 'Value 3')
			),
		'indexed' => true),

	'customfield4' => array(
		'dbfield' => 'customfield4',
		'label' => __('Custom Field 4'),
		'type' => 'choice',
		'default' => 'value2',
		'data' => array(
			array('value1', 'Value 1'),
			array('value2', 'Value 2')
			),
		'indexed' => true),

	'customfield5' => array(
		'dbfield' => 'customfield5',
		'label' => __('Custom Field 5'),
		'default' => date('Y-m-d'),
		'type' => 'date')
);
*/
Description: Mataro is the Spanish name for Mourvedre, a grape grown extensively in the Rhone region of France as well as north of Barcelona in Spain. It came to California with the earliest of settlers since it could be dry farmed. This wine has unique aromas, taste and mouth feel with a long after-taste. Compliments a wide range of light foods up through a pork loin. 85% Mourvedre and 15% Barbera. A People Friendly Wine! Food Pairing Suggestion: Try a glass of Mataro with peaches and brie on crackers - you'll be coming back for seconds (a favorite at our Passport weekends).
Q: SQL query to pull data from three different tables I have to create a SQL query to list all the Nurses in the ‘Sparrow’ Wing ordered by last name, first name.
However, I need to pull Nurse_name and Nurse_surname from the Nurse table, which is linked to another table called Sister by the foreign key Sister_ID, this table is then linked to another table called Wing which has the foreign key Sister_ID.
The nurse is managed by the sister and the sister manages the wing.
Can anyone help me with an SQL query for this? As it is, I am only able to get the data from nurse and sister tables.

A: Since you seem to be aware that you should use the inner join to connect tables (but apparently not that the connection needs to be via the related columns) you should apply that knowledge to connect all the tables you need to answer the query.
If you start at the end result and work your way backwards you first chose the columns you need:
select Nurse.Nurse_name, Nurse.Nurse_surname

and then as they belong to the Nurse table you use that as source in the from clause
from Nurse

to get the Wing you need to connect the Sister table, but to connect that and Nurse you first need the SisterNurse table joined on the shared attribute
join SisterNurse on Nurse.Nurse_ID = SisterNurse.Nurse_ID

now you can join Sister on the attribute shared with SisterNurse
join Sister on Sister.Sister_ID = SisterNurse.Sister_Id

and finally you can join Wing
join Wing on Wing.sister_ID = Sister.Sister_ID

limit the Wings to the one names 'Sparrow'
where Wing.Wing_Name = 'Sparrow'

and order the data
order by Nurse.Nurse_surname, Nurse.Nurse_name

Put it all together and you get:
select Nurse.Nurse_name, Nurse.Nurse_surname
from Nurse
join SisterNurse on Nurse.Nurse_ID = SisterNurse.Nurse_ID
join Sister on Sister.Sister_ID = SisterNurse.Sister_Id
join Wing on Wing.sister_ID = Sister.Sister_ID
where Wing.Wing_Name = 'Sparrow'
order by Nurse.Nurse_surname, Nurse.Nurse_name


A: You don't give much information about the schema involved but maybe this will help:
select
  n.Nurse_name
, n.Nurse_surname
, w.Wing_name
, managing_nurse.Nurse_name
, managing_nurse.surname
from
  Nurse n
  join Sister s on n.Sister_ID=n.Sister_ID
  join Wing w on s.Sister_ID=w.Sister_ID
  join Nurse managing_nurse on w.Nurse_Manager_ID=managing_nurse.Sister_ID


A: A simple way is to use 'union':
select * from (
select nurse_name, nurse_surname, 'nurse' as nurse_type from nurse_table
   union 
select nurse_name, nurse_surname,'sister' as nurse_type from syster join nurse_table on nurse_table.syster_id = syster.syster_id
   union 
select nurse_name, nurse_surname, 'wing' as nurse_type from wing join syster on wing.syster_id = syster.syster_id join nurse_table on nurse_table.syster_id =  syster.syster_id )
order by nurse_surname, nurse_name
