SQL to ActiveRecord criteria - php

I am trying to reproduce the following SQL in an ActiveRecord Criteria:
SELECT COALESCE(price, standardprice) AS price
FROM table1
LEFT JOIN table2
ON (pkTable1= fkTable1)
WHERE pkTable1= 1
So far I have the following:
$price = Table1::model()->find(array(
"select" => "COALESCE(price, standardprice) AS price",
'with' => array(
'table2' => array(
'joinType' => 'LEFT JOIN',
'on' => 'pkTable1= fkTable1',
)
),
'condition' => 'pkTable1=:item_id',
'params' => array(':item_id' => 1)
));
But this results into the following error: 'Active record "Table1" is trying to select an invalid column "COALESCE(price". Note, the column must exist in the table or be an expression with alias.
The column should exist though, here are the 2 table structures:
Table1
pkTable1 int(11) - Primary key
standardprice decimal(11,2)
name varchar(255) //not important here
category varchar(255) //not important here
Table2
pkTable2 int(11) - Primary key //not important here
fkType int(11) - Foreign key //not important here
fkTable1 int(11) - Foreign key, linking to Table1
price decimal(11,2)
What exactly am I doing wrong?

You will need to use a CDbExpression for the COALESCE() expression:
$price=Table1::model()->find(array(
'select'=>array(
new CDbExpression('COALESCE(price, standardprice) AS price'),
),
'with' => array(
'table2' => array(
'joinType'=>'LEFT JOIN',
'on'=>'pkTable1=fkTable1',
),
),
'condition'=>'pkTable1=:item_id',
'params'=>array(':item_id'=>1)
));
I further believe if table2 has been linked in the relations() method in your Table1 model, the following line should be sufficient:
'with'=>array('table2'),

I've managed to solve the issue as following: Wrap the COALESCE expression in an array, and change the alias to an existing columnname in the table.
$price = Table1::model()->find(array(
"select" => array("COALESCE(price, standardprice) AS standardprice"),
'with' => array(
'table2' => array(
'joinType' => 'LEFT JOIN',
'on' => 'pkTable1= fkTable1',
)
),
'condition' => 'pkTable1=:item_id',
'params' => array(':item_id' => 1)
));
Thank you to Willemn Renzema for helping me with the array part. I'm still not entirely sure why the alias needs to be an existing column name (in this case the error was that price doesn't exist in Table1).

Related

Get all rows where the field is not NULL in CakePHP 3

How can I get all the instances of a Table where the fields are not NULL ?
Here is the configuration:
I have a Table 1 where the instances have a relationship "hasmany" with a Table 2. I want to get all the instances of Table 1 linked with a Table 2 instance not NULL.
The CakePHP doc helped me finding the exists() and isNotNull() conditions but I did not achieve.
Here is how I imagined:
$Table1 = TableRegistry::get('Table1')->find('all')->contain([
'Table2' => [
'sort' => ['Table2.created' => 'desc']
]
])->where([
'Table1.id' => $id,
'Table2 IS NOT NULL'
]);
$this->set(compact('Table1'));
But it obviously does not work.
edit : I expect to get all the line of the Table1 which contain existing Not NULL Table2 line(s) linked. The problem is in the 'where' array with the 'Table2 IS NOT NULL', it does not work.
And without this line 'Table2 IS NOT NULL', I get all the Table1 line which contain a Table2 line or not (because some line of Table1 are not linked at all and I don't want to get these lines).
Assuming the tables follow convention and use "id" as the primary key, I suggest the easiest fix would be testing that field for NOT NULL.
I.e., replace this:
'Table2 IS NOT NULL'
with this:
'Table2.id IS NOT NULL'
or:
'Table2.id !=' => null
or:
'Table2.id >' => 0
I've successfuly get the Table1 lines with its existing Table2 line(s) associated.
query = TableRegistry::get('Table1')->find();
$query->select(['Table1.id', 'count' => $query->func()->count('Table2.id')])->matching('Table2')->group(['Table1.id'])->having(['count
>' => 0]);
$table1Ids = [];
foreach ($query as $z)
{
$table1Ids[] = $z->id;
}
$table1= TableRegistry::get('Table1')->find('all')->contain([
'Table2' => [
'sort' => ['Table2.created' => 'desc']
]
])->where([
'id IN' => $table1Ids,
]);

CakePHP join with comma separated ids

Here I want to join two table with comma separated ids
For example my data is like:
[Restaurant] => Array
(
[RST_ID] => 171
[RST_NAME] => oneone
[RST_IMAGE] =>
[RST_CAT_ID] => 2,4,6
[RST_CT_ID] => 27
[RST_IS_TOP] => 3
[RST_QR_CODE] =>
[RST_CREATED_DATE] => 1394536725
[RST_MODIFIED_DATE] => 1394536725
[RST_STATUS] => 1
)
[Category] => Array
(
[CAT_ID] => 2
[CAT_NAME] => Vegetarian
[CAT_CREATED_DATE] => 1375175962
[CAT_MODIFIED_DATE] => 1375175962
[CAT_STATUS] => 1
)
My Model Code:
var $belongsTo = array(
'Category' => array(
'className' => 'Category',
'foreignKey' => 'RST_CAT_ID',
'conditions' => array('Category.CAT_ID IN ( Restaurant.RST_CAT_ID)')
)
);
Real Query:
SELECT
`Restaurant`.`RST_ID`, `Restaurant`.`RST_NAME`, `Restaurant`.`RST_IMAGE`,
`Restaurant`.`RST_CAT_ID`, `Restaurant`.`RST_CT_ID`, `Restaurant`.`RST_IS_TOP`,
`Restaurant`.`RST_QR_CODE`, `Restaurant`.`RST_CREATED_DATE`,
`Restaurant`.`RST_MODIFIED_DATE`, `Restaurant`.`RST_STATUS`,
`Category`.`CAT_ID`, `Category`.`CAT_NAME`, `Category`.`CAT_CREATED_DATE`,
`Category`.`CAT_MODIFIED_DATE`, `Category`.`CAT_STATUS`, `City`.`CT_ID`,
`City`.`CT_NAME`, `City`.`CT_CREATED_DATE`, `City`.`CT_MODIFIED_DATE`,
`City`.`CT_STATUS`
FROM `dailybit_dailybites`.`restaurant` AS `Restaurant`
LEFT JOIN `dailybit_dailybites`.`category` AS `Category`
ON (`Restaurant`.`RST_CAT_ID` = `Category`.`CAT_ID`
AND `Category`.`CAT_ID` IN ( `Restaurant`.`RST_CAT_ID`))
LEFT JOIN `dailybit_dailybites`.`city` AS `City`
ON (`Restaurant`.`RST_CT_ID` = `City`.`CT_ID`)
WHERE 1 = 1
So what’s the solution here?
It's giving me just one category data that for first id only.
First have a look at this question: MySQL search in comma list
As you can see the belongsTo query is just generating a join on the single id, CakePHP by default doesn't respect this special case. You will have to alter your query and pass all the ids manually, but your DB design is bad and it doesn't follow the CakePHP conventions at all.
How do you prevent duplicates (which would waste space)
How do you remove a given value (Requires custom function, leading to possibility of errors?
How do you respond to performance issues as the size of my tables increase?
Instead of changing the query you should change this awkward DB design. You want to use HABTM here and a join table: Restaurant hasAndBelongsToMany Categoryy.
restaurants <-> restaurants_categories <-> categories
If you insist on using this bad DB design you'll have to use bindModel() and set the conditions manually:
'conditions' => array('FIND_IN_SET (Category.CAT_ID, ' . $listOfIds. ')')
I haven't tested this, try it yourself, see FIND_IN_SET() vs IN()
You'll have to have another method that gets you all the ids you want here before. Like I said, this is ineffectice and bad design.
You have to set your foreign Key false and find_in_set condition
var $belongsTo = array(
'Category' => array(
'className' => 'Category',
'foreignKey' => false,
'conditions' => array('FIND_IN_SET(Category.CAT_ID,Restaurant.RST_CAT_ID)')
)
);
// you can pass an array at the place of 'Restaurant.RST_CAT_ID'

CakePHP: adding DISTINCT to find causes an associated to get omitted

We build a fairly complex set of $conditions, then we search using Paginator:
$this->Paginator->settings = array( 'conditions'=>$conditions,
'joins'=>$joins,
'limit'=>25,
//'fields' => array('DISTINCT (User.id)')
);
$resume_display_array = $this->Paginator->paginate('User');
You'll see the 'fields' is commented out. When we uncomment that field, it causes associated model Resume to get dropped from the results array (without it, each result returned includes a [Resume] key.
Why is this happening?
Here's the SQL dump WITH the DISTINCT option in use:
SELECT DISTINCT (`User`.`id`), `User`.`id`
FROM `sw`.`users` AS `User`
LEFT JOIN `sw`.`taggings_users` AS `TaggingUser`
ON (`User`.`id`= `TaggingUser`.`user_id`)
LEFT JOIN `sw`.`taggings` AS `Tagging`
ON (`TaggingUser`.`tagging_id`= `Tagging`.`id`)
LEFT JOIN `sw`.`resumes` AS `Resume`
ON (`Resume`.`user_id` = `User`.`id`)
WHERE `User`.`first_name` LIKE '%jordan%' AND NOT (`Tagging`.`tag_name` = ('done'))
LIMIT 25
Here's the SQL dump WITHOUT the DISTINCT option:
SELECT `User`.`id`, `User`.`username`, `User`.`password`, `User`.`role`, `User`.`created`, `User`.`modified`, `User`.`email`, `User`.`wordpress_user_id`, `User`.`first_name`, `User`.`last_name`, `User`.`group_id`, `Resume`.`id`, `Resume`.`user_id`, `Resume`.`wordpress_resume_id`, `Resume`.`wordpress_user_id`, `Resume`.`has_file`, `Resume`.`is_stamped`, `Resume`.`is_active`, `Resume`.`file_extension`, `Resume`.`created`, `Resume`.`modified`, `Resume`.`is_deleted`
FROM `sw`.`users` AS `User`
LEFT JOIN `sw`.`taggings_users` AS `TaggingUser`
ON (`User`.`id`= `TaggingUser`.`user_id`)
LEFT JOIN `sw`.`taggings` AS `Tagging`
ON (`TaggingUser`.`tagging_id`= `Tagging`.`id`)
LEFT JOIN `sw`.`resumes` AS `Resume`
ON (`Resume`.`user_id` = `User`.`id`)
WHERE `User`.`first_name` LIKE '%jordan%' AND NOT (`Tagging`.`tag_name` = ('done'))
LIMIT 25
Update
Here is a var_dump of $joins:
array(
(int) 0 => array(
'table' => 'taggings_users',
'type' => 'LEFT',
'alias' => 'TaggingUser',
'conditions' => array(
(int) 0 => 'User.id= TaggingUser.user_id'
)
),
(int) 1 => array(
'table' => 'taggings',
'type' => 'LEFT',
'alias' => 'Tagging',
'conditions' => array(
(int) 0 => 'TaggingUser.tagging_id= Tagging.id'
)
)
)
If you specify the fields option, you need to specify all fields that you want to retrieve.
By setting it ONLY to User.id, you're telling it that that is the ONLY field you want returned.
You could also try something like below as a catchall (not sure if that works, but if not, you get the idea - you're limiting your own results):
'fields' => array('*', 'DISTINCT (User.id)')

Cakephp 2.0: complex find query

I'm trying to pull some info out of two tables linked by the hasMany and belongsTo associations.
requisitions hasMany locations and locations belongsTo requisitions
TABLE `requisitions` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`user_id` int(11) NOT NULL,
`fecha_generacion` date NOT NULL,
`solicitado_a` varchar(60) NOT NULL,
`proyecto` varchar(150) NOT NULL,
`obra_no` varchar(11) NOT NULL,
`observaciones` text NOT NULL,
PRIMARY KEY (`id`)
)
and
TABLE `locations` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`requisition_id` int(11) NOT NULL,
`fecha` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`name` enum('pendiente','tecnico','existencia','cotizando','generar_o','archivada')
NOT NULL DEFAULT 'pendiente',
`image_path` varchar(150) NOT NULL DEFAULT 'estado0.png',
`note` text NOT NULL,
PRIMARY KEY (`id`)
)
Requisition goes from one Location to another and I need to keep track of its current Location looking by a given Location as 'pendiente','tecnico'...
So I need to generate a list with the last Location for each Requisition and then filter that list by the Location.name
I believe the only way to do this is with a query around another query, so I'm trying to understand cakephp syntax with more simple queries first.
I was trying to search for the last 'pendiente' Location with the next code from my RequisitionsController.
$lastPendiente = $this->Requisition->Location->find('all', array(
'conditions' => array('Location.name' => 'pendiente'),
'fields' => array('MAX(Location.id) AS olderLocation', 'Location.requisition_id'),
'group' => 'Requisition.id',
));
I have the query
SELECT MAX(`Location`.`id`) AS olderLocation, `Location`.`requisition_id` FROM `petrofil_demo`.`locations` AS `Location` LEFT JOIN `petrofil_demo`.`requisitions` AS `Requisition` ON (`Location`.`requisition_id` = `Requisition`.`id`) WHERE `Location`.`name` = 'pendiente' GROUP BY `Requisition`.`id`
output...
array(
(int) 0 => array(
(int) 0 => array(
'olderLocation' => '22'
),
'Location' => array(
'requisition_id' => '29'
)
),
(int) 1 => array(
(int) 0 => array(
'olderLocation' => '5'
),
'Location' => array(
'requisition_id' => '30'
)
),
(int) 2 => array(
(int) 0 => array(
'olderLocation' => '13'
),
'Location' => array(
'requisition_id' => '31'
)
)
)
...which is great because those are exactly the last requisitions with a 'pendiente' location but here comes the second query or the condition where I'm clueless. I need to be sure my requisition last state was 'pendiente' and not another possible locations. For example my requisition_id =>30 last location is really 'tecnico' so I need to find a way to exclude it from showing on my results.
You could quit the condition from the query, put the 'name' column in the 'fields' and add an order by sentence:
$lastPendiente = $this->Requisition->Location->find('all', array(
'fields' => array('MAX(Location.id) AS olderLocation', 'Location.requisition_id', 'name'),
'group' => 'Requisition.id',
'order' => 'fecha DESC'
));
This way you only get the last status from the Location table. Then, you could iterate the results, filtering by the 'name' column and deleting those whit a 'name' diferent from 'pendiente':
foreach($lastPendiente as $k => $location){
if($location['Location']['name'] != 'pendiente'){
unset($lastPendiente[$k]);
}

Getting the column 'country_id' in 'where' clause is ambiguous - error in search query in cakephp 1.3

I am working with CakePHP 1.3 version for search functionality using Search Plugin.
I have three models:
Demo,
Country
State
Demo has two foreign keys, country_id and state_id. State has the foreign key country_id.
What I am doing is, I have search form which have country & state drop down which fetch all data from countries & states table. When i search any of country from dropdown & submit the form it will show me below error. If i search using only state dropdown i get the correct result.
When I execute the search query, I get the error
'Column 'country_id' in where clause is ambiguous'
My query is:
SELECT `Demo`.`id`, `Demo`.`demo2`, `Demo`.`desc`, `Demo`.`subject`, `Demo`.`gender`, `Demo`.`country_id`, `Demo`.`state_id`, `Demo`.`image_url`, `Country`.`id`, `Country`.`name`, `State`.`id`, `State`.`country_id`, `State`.`description` FROM `demos` AS `Demo` LEFT JOIN `countries` AS `Country` ON (`Demo`.`country_id` = `Country`.`id`) LEFT JOIN `states` AS `State` ON (`Demo`.`state_id` = `State`.`id`) WHERE `country_id` = 2
Model relationships in Demo table:
var $belongsTo = array(
'Country' => array(
'className' => 'Country',
'foreignKey' => 'country_id',
'conditions' => '',
'fields' => '',
'order' => ''
),
'State' => array(
'className' => 'State',
'foreignKey' => 'state_id',
'conditions' => '',
'fields' => '',
'order' => ''
),
);
The controller query to fetch all Country in dropdown is:
$country=$this->Country->find('list'); //just display the list of country in dropdown
The query search the data from all fields except Country (country_id), because it will not know which country_id it is looking for from table Demo or table State. I need the country_id from the demo table to get the correct result.
As I understand you want to make a find over Demo for a specific country_id.
Well you should define which "country_id" you're using because more than one of those tables
has such a column.
Just use Demo.country_id in the conditions array:
array('conditions' => array('Demo.country_id' => 2));
And you should see some SQL generated by Cake like this:
SELECT `Demo`.`id`, `Demo`.`demo2`, `Demo`.`desc`, `Demo`.`subject`, `Demo`.`gender`, `Demo`.`country_id`, `Demo`.`state_id`, `Demo`.`image_url`, `Country`.`id`, `Country`.`name`, `State`.`id`, `State`.`country_id`, `State`.`description` FROM `demos` AS `Demo` LEFT JOIN `countries` AS `Country` ON (`Demo`.`country_id` = `Country`.`id`) LEFT JOIN `states` AS `State` ON (`Demo`.`state_id` = `State`.`id`) WHERE `Demo`.`country_id` = 2
Try this:
SELECT
Demo.id,
Demo.demo2,
Demo.desc,
Demo.subject,
Demo.gender,
Demo.country_id,
Demo.state_id,
Demo.image_url,
Country.id,
Country.name,
State.id,
State.country_id,
State.description
FROM demos AS Demo
LEFT JOIN countries AS Country ON (Demo.country_id = Country.id)
LEFT JOIN states AS State ON (Demo.state_id = State.id) WHERE Demo.country_id = 2

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