belongsToMany relationship in Laravel across multiple databases - php

I have model A and model B which lie in two different databases.
Now I have a pivot_table called a_bs in the same database as model A.
I've setup the belongsToMany relatinoship like this in model A
public function bs()
{
return $this->belongsToMany('B', 'a_bs', 'a_id', 'b_id');
}
When I try to access this relationship like so:
$a = A::find($id);
print_r($a->bs->lists('id'));
I get an error that my pivot table doesn't exist in model B's database. Which is obviously correct since the pivot table is in model A's database. How can I let Laravel know that?
Do not suggest to put the pivot table in model B's database

Very simply:
public function bs()
{
$database = $this->getConnection()->getDatabaseName();
return $this->belongsToMany('B', "$database.a_bs", 'a_id', 'b_id');
}
I'm obtaining the database name dynamically because my connection is configured based off an environment variable. Laravel seems to assume the pivot table to exist in the same database as the target relation, so this will force it to look instead to the database corresponding to the model that this method is in, your 'A' realm.
If you're not worried about SQLite databases, i.e. in the scope of a unit-test, that's all you need. But if you are, keep reading.
Firstly, the previous example isn't sufficient on its own. The value of $database would end up being a file-path, so you need to alias it to something that won't break an SQL statement, and make it accessible to the current connection. "ATTACH DATABASE '$database' AS $name" is how you do that:
public function bs()
{
$database = $this->getConnection()->getDatabaseName();
if (is_file($database)) {
$connection = app('B')->getConnection()->getName();
$name = $this->getConnection()->getName();
\Illuminate\Support\Facades\DB::connection($connection)->statement("ATTACH DATABASE '$database' AS $name");
$database = $name;
}
return $this->belongsToMany('B', "$database.a_bs", 'a_id', 'b_id');
}
Warning: Transactions muck this up: If the current connection is using transactions, the ATTACH DATABASE statement will fail. You can use transactions on it after executing that statement though.
Whereas, if the related connection uses transactions, the resulting data will be silently rendered invisible to the current one. This drove me nuts for longer than I'd care to admit, because my queries ran without error, but kept coming up empty. It seems only data truly written to the attached database is actually accessible to the one it's attached to.
So, after being forced to write to your attached database, you may still want your test to clean up after itself. A simple solution there would be to just use $this->artisan('migrate:rollback', ['--database' => $attachedConnectionName]);. But if you have multiple tests that need the same tables, this is not very efficient, as it forces them to have to rebuild them each time.
A better option would be to truncate the tables, but leave their structure in tact:
//Get all tables within the attached database
collect(DB::connection($database)->select("SELECT name FROM sqlite_master WHERE type = 'table'"))->each(function ($table) use ($name) {
//Clear all entries for the table
DB::connection($database)->delete("DELETE FROM '$table->name'");
//Reset any auto-incremented index value
DB::connection($database)->delete("DELETE FROM sqlite_sequence WHERE name = '$table->name'");
});
}
This will wipe all data from that connection, but there's no reason you couldn't apply some kind filter to that however you see fit. Alternatively, you could take advantage of the fact that SQLite DBs are easily-accessible files, and just copy the attached one to a temp file, and use it to overwrite the source after the test is done executing. The result would be functionally identical to a transaction.

You can set the database of the table in the model class:
protected $table = 'A.a_s';
And You have to use singular form when create a pivot table.
/app/model/A.php
class A extends Eloquent {
// Set table name (plural) with database name
protected $table = 'A.a_s';
// Many to many relation
public function b_s() {
return $this->belongsToMany('B');
}
}
/app/model/B.php
class B extends Eloquent {
// Set table name (plural) with database name
protected $table = 'B.b_s';
}
Query
print_r(A::with('b_s')->where('id', 1)->get()->toArray());
MySQL
CREATE TABLE `A`.`a_s` (
`id` INT UNSIGNED NOT NULL AUTO_INCREMENT,
PRIMARY KEY (`id`)
);
CREATE TABLE `B`.`b_s` (
`id` INT UNSIGNED NOT NULL AUTO_INCREMENT,
PRIMARY KEY (`id`)
);
CREATE TABLE `A`.`a_b` (
`a_id` int(10) unsigned NOT NULL,
`b_id` int(10) unsigned NOT NULL,
PRIMARY KEY (`a_id`,`b_id`)
) ENGINE=InnoDB;
INSERT INTO A.a_s VALUES (NULL);
INSERT INTO A.a_s VALUES (NULL);
INSERT INTO B.b_s VALUES (NULL);
INSERT INTO A.a_b VALUES (1,1);
INSERT INTO A.a_b VALUES (1,2);

If the databases/schemas are on the same host server like #NiRR said just do this to override the default schema of the second connection:
return $this->belongsToMany('B', 'real-schema-name.a_bs');
or alternatively
return $this->belongsToMany('A', 'real-schema-name.a_bs');
Depending on which model (A or B) is defined with the connection that isn't using the default schema.
Remember that it's impossible to do a join query that spans across two servers; which server is it going to get executed on? Each is missing some of the needed data needed to preform the request.

This is only possible if the two databases are on the same connection (server).
Its not possible to do this on two different connections (servers) since you'll need all three tables on a single server that will perform the join command.

Explicitly setting or defining the connection for both models works just fine.
protected $connection = 'connection1'; //inside model A
protected $connection = 'connection2'; //inside model B
EDIT
When Laravel is fetching a model from the DB, provided a database/connection has been defined (as a property) within the model, Laravel will use the database name for that connection when constructing the SQL. So when working with multiple connections, it is best to define the connection for every model.

Related

Laravel Virtual Columns fail to save

I added a couple of virtual columns to my database tables using Laravels virtualAs column modifier:
$table->decimal('grand_total')->virtualAs( '(total_value + (total_value*tax_rate))');
Basically it keeps a mysql virtual column that automatically calculates the grand total based on the total and tax rate stored in another column.
However, Laravel does not seem to play nice with virtual columns at all. When saving a record, it attempts to INSERT or UPDATE the virtual column, which is obviously not allowed in mySQL. I could not find a way to configure in the Eloquent model which fields are actually written to the database on an update or insert.
I've tried adding the field to the models $hidden, and $appends but nothing seems to work.
Looking at the Laravel Source code for an insert (https://github.com/laravel/framework/blob/5.6/src/Illuminate/Database/Eloquent/Model.php#L733), it seems to just insert whatever attributes are in $this->attributes. When the record is read from the database the grand_total field is read from the table and set as an attribute and then it is tried to be written again once the record is saved.
Is there any way to get this Laravel to stop trying to save columns that are virtual?
Here's a quick trait I wrote to solve your problem that will filter fields residing in the $virtualFields property before saving. It requires a select (refresh) after the save to get the new value for the virtual field. If you don't need to query this virtual field, I'd highly recommend you look into a mutator instead.
trait HasVirtualFields
{
public function save(array $options = [])
{
if (isset($this->virtualFields)) {
$this->attributes = array_diff_key($this->attributes, array_flip($this->virtualFields));
}
$return = parent::save($options);
$this->refresh(); // Refresh the model for the new virtual column values
return $return;
}
}
class YourModel
{
use HasVirtualFields;
protected $virtualFields = ['grand_total'];
}

Yii active record relation limit to one record

I am using PHP Yii framework's Active Records to model a relation between two tables. The join involves a column and a literal, and could match 2+ rows but must be limited to only ever return 1 row.
I'm using Yii version 1.1.13, and MySQL 5.1.something.
My problem isn't the SQL, but how to configure the Yii model classes to work in all cases. I can get the classes to work sometimes (simple eager loading) but not always (never for lazy loading).
First I will describe the database. Then the goal. Then I will include examples of code I've tried and why it failed.
Sorry for the length, this is complex and examples are necessary.
The database:
TABLE sites
columns:
id INT
name VARCHAR
type VARCHAR
rows:
id name type
-- ------- -----
1 Site A foo
2 Site B bar
3 Site C bar
TABLE field_options
columns:
id INT
field VARCHAR
option_value VARCHAR
option_label VARCHAR
rows:
id field option_value option_label
-- ----------- ------------- -------------
1 sites.type foo Foo Style Site
2 sites.type bar Bar-Like Site
3 sites.type bar Bar Site
So sites has an informal a reference to field_options where:
field_options.field = 'sites.type' and
field_options.option_value = sites.type
The goal:
The goal is for sites to look up the relevant field_options.option_label to go with its type value. If there happens to be more than one matching row, pick only one (any one, doesn't matter which).
Using SQL this is easy, I can do it 2 ways:
I can join using a subquery:
SELECT
sites.id,
f1.option_label AS type_label
FROM sites
LEFT JOIN field_options AS f1 ON f1.id = (
SELECT id FROM field_options
WHERE
field_options.field = 'sites.type'
AND field_options.option_value = sites.type
LIMIT 1
)
Or I can use a subquery as a column reference in the select clause:
SELECT
sites.id,
(
SELECT id FROM field_options
WHERE
field_options.field = 'sites.type'
AND field_options.option_value = sites.type
LIMIT 1
) AS type_label
FROM sites
Either way works great. So how do I model this in Yii??
What I've tried so far:
1. Use "on" array key in relation
I can get a simple eager lookup to work with this code:
class Sites extends CActiveRecord
{
...
public function relations()
{
return array(
'type_option' => array(
self::BELONGS_TO,
'FieldOptions', // that's the class for field_options
'', // no normal foreign key
'on' => "type_option.id = (SELECT id FROM field_options WHERE field = 'sites.type' AND option_value = t.type LIMIT 1)",
),
);
}
}
This works when I load a set of Sites objects and force it to eager load type_label, e.g. Sites::model()->with('type_label')->findByPk(1).
It does not work if type_label is lazy-loaded.
$site = Sites::model()->findByPk(1);
$label = $site->type_option->option_label; // ERROR: column t.type doesn't exist
2. Force eager loading always
Building on #1 above, I tried forcing Yii to always to eager loading, never lazy loading:
class Sites extends CActiveRecord
{
public function relations()
{
....
}
public function defaultScope()
{
return array(
'with' => array( 'type_option' ),
);
}
}
Now everything always works when I load Sites, but it's no good because there are other models (not pictured here) that have relations that point to Sites, and those result in errors:
$site = Sites::model()->findByPk(1);
$label = $site->type_option->option_label; // works now
$other = OtherModel::model()->with('site_relation')->findByPk(1); // ERROR: column t.type doesn't exist, because 't' refers to OtherModel now
3. Make the reference to the base table somehow relative
If there was a way that I could refer to the base table, other than "t", that was guaranteed to point to the correct alias, that would work, e.g.
'on' => "type_option.id = (SELECT id FROM field_options WHERE field = 'sites.type' AND option_value = %%BASE_TABLE%%.type LIMIT 1)",
where %%BASE_TABLE%% always refers to the correct alias for table sites. But I know of no such token.
4. Add a true virtual database column
This way would be the best, if I could convince Yii that the table has an extra column, which should be loaded just like every other column, except the SQL is a subquery -- that would be awesome. But again, I don't see any way to mess with the column list, it's all done automatically.
So, after all that... does anyone have any ideas?
EDIT Mar 21/15: I just spent a long time investigating the possibility of subclassing parts of Yii to get the job done. No luck.
I tried creating a new type of relation based on BELONGS_TO (class CBelongsToRelation), to see if I could somehow add in context sensitivity so it could react differently depending on whether it was being lazy-loaded or not. But Yii isn't built that way. There is no place where I can hook in code during query buiding from inside a relation object. And there is also no way I can tell even what the base class is, relation objects have no link back to the parent model.
All of the code that assembles these queries for active records and their relations is locked up in a separate set of classes (CActiveFinder, CJoinQuery, etc.) that cannot be extended or replaced without replacing the entire AR system pretty much. So that's out.
I then tried to see if I can create "fake" database column entries that would actually be a subquery. Answer: no. I figured out how I could add additional columns to Yii's automatically generated schema data. But,
a) there's no way to define a column in such a way that it can be a derived value, Yii assumes it's a column name in way too many places for that; and
b) there also doesn't appear to be any way to avoid having it try to insert/update to those columns on save.
So it really is looking like Yii (1.x) just does not have any way to make this happen.
Limited solution provided by #eggyal in comments: #eggyal has a suggestion that will meet my needs. He suggests creating a MySQL view table to add extra columns for each label, using a subquery to look up the value. To allow editing, the view would have to be tied to a separate Yii class, so the downside is everywhere in my code I need to be aware of whether I'm loading a record for reading only (must use the view's class) or read/write (must use the base table's class, does not have the extra columns). That said, it is a workable solution for my particular case, maybe even the only solution -- although not an answer to this question as written, so I'm not going to put it in as an answer.
OK, after a lot of attempts, I have found a solution. Thanks to #eggyal for making me think about database views.
As a quick recap, my goal was:
link one Yii model (CActiveRecord) to another using a relation()
the table join is complex and could match more than one row
the relation must never join more than one row (i.e. LIMIT 1)
I got it to work by:
creating a view from the field_options base table, using SQL GROUP BY to eliminate duplicate rows
creating a separate Yii model (CActiveRecord class) for the view
using the new model/view for the relation(), not the original table
Even then there were some wrinkles (maybe a Yii bug?) I had to work around.
Here are all the details:
The SQL view:
CREATE VIEW field_options_distinct AS
SELECT
field,
option_value,
option_label
FROM
field_options
GROUP BY
field,
option_value
;
This view contains only the columns I care about, and only ever one row per field/option_value pair.
The Yii model class:
class FieldOptionsDistinct extends CActiveRecord
{
public function tableName()
{
return 'field_options_distinct'; // the view
}
/*
I found I needed the following to override Yii's default table data.
The view doesn't have a primary key, and that confused Yii's AR finding system
and resulted in a PHP "invalid foreach()" error.
So the code below works around it by diving into the Yii table metadata object
and manually setting the primary key column list.
*/
private $bMetaDataSet = FALSE;
public function getMetaData()
{
$oMetaData = parent::getMetaData();
if (!$this->bMetaDataSet) {
$oMetaData->tableSchema->primaryKey = array( 'field', 'option_value' );
$this->bMetaDataSet = TRUE;
}
return $oMetaData;
}
}
The Yii relation():
class Sites extends CActiveRecord
{
// ...
public function relations()
{
return (
'type_option' => array(
self::BELONGS_TO,
'FieldOptionsDistinct',
array(
'type' => 'option_value',
),
'on' => "type_option.field = 'sites.type'",
),
);
}
}
And all that does the trick. Easy, right?!?

$model->save() adding 'where id = 1' condition

I'm trying to update a model, I load the model, take all the data from the POST and then save it, easy... But my record was never updating so went to the log and discovered that the update query is adding a weird condition. FYI, MD_ID is my primary key.
So, I load the model, the next line is the SQL produced by Yii:
$model = Ositems::model()->findByPk($id);
SELECT * FROM "MTODETALLADO_INV" "t" WHERE "t"."MD_ID"=249217
If echo the json_encode of the loaded model I get that dictionary in my browser:
echo json_encode($model->getAttributes());
{""MD_BODEGA":"01","MD_PRODUCTO":"0031253","MD_CANTIDAD":"1","MD_PRECIOTOTAL":"1466",,"MD_PORCENTAJEDESCUENTO":"0","MD_IDCABECERA":"97403","MD_ID":"249217","MD_OBSERVACION":null}
At this point everything looks right, now I take the values from post:
$model->attributes = $_POST;
And here if echo the values of the model I get the new values right, now here is the problem: I save the model and this is the SQL Yii runs (I replaced the :yp_ values to make it more readable)
$model->save();
UPDATE "MTODETALLADO_INV" SET
MD_BODEGA"='01'
MD_PRODUCTO"='0020514
MD_CANTIDAD"='10'
MD_PORCENTAJEDESCUENTO"='0
MD_IDCABECERA"=97403
MD_ID"=249218
MD_PRECIOTOTAL"='36210'
MD_OBSERVACION"=''
WHERE "MTODETALLADO_INV"."MD_ID"=1
And there is the problem! WHERE "MTODETALLADO_INV"."MD_ID"=1, Why would it make it 1 if all this time my model id has been 249218 ?
A few considerations:
My model only takes some columns that I need from the actual table, Yii sets the other columns as null and I omitted them in the previous code.
The table is in a foreign db, I use have a custom ActiveRecord which manages the CDbConnection to a database according to the user. (It's a webservice app)
I followed what the function save() did and could finally find the problem was when it tried to get the primary key. I had this method in my model:
public function primaryKey()
{
return array('MS_ID');
}
}
But it had to be:
public function primaryKey()
{
return 'MS_ID';
}
}
Somehow that was causing the problem.

How to create an object of "unknown" class?

I have a MySQL database and a table tobjects where each record has its id, parameter, value (something like XML) and one can say that this parameter column determines the "type" of an object.
The objects are used in some other tables, depending on their types, so each of them should be handled in specific way.
Because "handling" is somewhat common (I use the same function) I created a TObject class (not abstract but could be) from which I inherit other classes; this inheritance method is very useful and that's the very reason I use object oriented programming. For example TObject has retrieve() method that gets from db all the necessary data, not those in tobjects table but others too, which are type dependent, so I override it in some classes.
The problem I encountered is that when I create an object I do not know what class should it be. Of course, I can SELECT Parameter FROM tobjects WHERE id=$id, and then (with switch) create object of the proper class, and use its retrieve() method (each class retrieves different data, only those from tobjects are common) to get data from the db, that causes me to run query two times and some part of work outside the class, which works, but is not gentle.
The best solution would be if I can create a TObject and then, upon retrieving, change the class of the object to the one I need and it would be TObject's descendant, but I'm almost sure it's not possible.
Is my solution, that I run the first query just to select one field from tobjects only to determine object's class right? Or is there a trick to change object's class in runtime?
If understand what you are doing correctly, here is the way I would approach this:
Passing PDO::FETCH_CLASS | PDO::FETCH_CLASSTYPE to the first argument of PDOStatement::fetch() will return an object of class PDOStatement::fetchColumn(0) - in other words, it determines the class name to instantiate from the value of the first column of the result set.
To leverage this, you would JOIN tobjects ON targetTable.objectType = tobjects.id and select tobjects.Parameter as the first column in the result set. If the Parameter column already holds a 1:1 mapping of database object types to class names, this is all you need to do, however I'm not sure whether this is the case, and it probably shouldn't be, because it makes it more difficult to substitute another class at a later date.
To overcome this limitation, I suggest you create a temporary table when you first connect the database, which maps Parameter values to class names, which you can JOIN onto the query to obtain the target class name.
So the flow would go something like this:
// Set up the connection
$db = new PDO('mysql:yourDSNhere');
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
// Create a temp table to store the mapping
$db->query("
CREATE TEMPORARY TABLE `objectMappings` (
`Parameter` INT NOT NULL PRIMARY KEY,
`ClassName` VARCHAR(255)
) ENGINE=MEMORY
");
// A mapping of Parameter IDs to class names
$classMap = array(
1 => 'Class1',
2 => 'Class2',
3 => 'Class3',
// ...
);
// Build a query string and insert
$rows = array();
foreach ($classMap as $paramId => $className) {
// this data is hard-coded so it shouldn't need further sanitization
$rows[] = "($paramId, '$className')";
}
$db->query("
INSERT INTO `objectMappings`
(`Parameter`, `ClassName`)
VALUES
".implode(',
', $rows)."
");
// ...
// When you want to retrieve some data
$result = $db->query("
SELECT m.ClassName, t.*
FROM targetTable t
JOIN tobjects o ON t.objectType = o.id
JOIN objectMappings m ON o.Parameter = m.Parameter
WHERE t.someCol = 'some value'
");
while ($obj = $result->fetch(PDO::FETCH_CLASS | PDO::FETCH_CLASSTYPE)) {
// $obj now has the correct type, do stuff with it here
}

Monitor usage of a certain database field

I have a nasty problem. I want to get rid of a certain database field, but I'm not sure in which bits of code it's called. Is there a way to find out where this field is used/called from (except for text searching the code; this is fairly useless seeing as how the field is named 'email')?
Cheers
I would first text search the files for the table name, then only search the tables that contain the table name for the field name.
I wrote a program to do this for my own purposes. It builds an in-memory listing of tables and fields and relates the tables to the fields. Then it loops through tables, searching for the code files that contain the table names, and then searches those files for the fields in the tables found. I'd recommend a similar methodology in your case.
setting mysql to log all queries for some time might help. the queries will give you the tip where to look
brute force - set up a test instance - remove the column - and excercise your test suite.
create a before insert trigger on that table that monitors the insertion on that column.
at the same time create another table called monitor with only one column email
make that table insert the value of NEW.email field into monitor.email as well as in real table.
so you can run your application and check for the existence of any non-null value in monitor table
You should do this in PHP i would expect
For example:
<?php
class Query
{
var $command;
var $resource;
function __construct($sql_command = '')
{
$this->command = $sql_command;
}
public function setResource($resource)
{
$this->resource = $resource;
}
}
//then you would have some kind of database class, but here we would modify the query method.
class Database
{
function query(Query $query)
{
$resource = mysql_query($query->command);
$query->setResource($resource);
//Then you can send the class to the monitor
QueryMonitor::Monitor($query);
}
}
abstract class QueryMonitor
{
public static Monitor(Query $query)
{
//here you use $query->resource to do monitoring of queryies
//You can also parse the query and gather what query type it was:-
//Select or Delete, you can also mark what tables were in the Query
//Even meta data so
$total_found = mysql_num_rows($query->resource);
$field_table = mysql_field_table ($query->resource);
//Just an example..
}
}
?>
Obviously it would be more advanced than that but you can set up a system to monitor every query and every queries meta data in a log file or w.e

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