I dont understand how I get a different output for lesson_date field if in the select I use Lessons.lesson_date' or lessons.lesson_date. I thought I am supposed to use Lessons and not lessons and that the for a single name it doesnt really matter.
with Lessons.lesson_date I get on the debug :
'lesson_date' => object(Cake\I18n\FrozenDate) {
'time' => '2015-07-09T00:00:00+00:00',
'timezone' => 'UTC',
'fixedNowTime' => false
},
with lessons.lesson_date I get a better output:
'lessons' => [
'id' => '5399',
'lesson_date' => '2015-07-09'
//this is the code below I am talking about. The Lessons.lesson_date gives a different output than if I change this to lessons.lesson_date
$query3 = $this->Lessons->find()
->contain(['TutoringTypes'])
->select(['lessons.id','Lessons.lesson_date','Lessons.tutoring_type_id',
'TutoringTypes.value'])
->where(['Lessons.lesson_date >' => $a3,'Lessons.lesson_date <' => $a4, .....
That is the expected/correct behavior when following the conventions, the date gets casted according to its type.
The latter output might be better suited for what you are doing with the data, but generally the former is "better" since a date object gives you more freedom of manipulating dates, handling localization, output formatting, etc...
Why?
As to the why, the output is different because the ORM will not cast the value in case of a non-conventional column alias, as it's not present in the type map that holds the information about which column is of which type.
Using lessons.lesson_date will create an alias of
lessons__lesson_date
which is not following the conventions, where as using Lessons.lesson_date will create an alias of
Lessons__lesson_date
which does follow the conventions, and will match the field in the type map, causing the ORM to cast the data.
Changing the behavior
If you need YYYY-MM-DD output, then you could simply output it formatted it in your view like
echo $lesson->lesson_date->i18nFormat('yyyy-MM-dd')
or change the default output format (which is used when the date gets casted to a string)
\Cake\I18n\FrozenDate::setToStringFormat('yyyy-MM-dd');
or maybe even change the type in the type map in order to keep the date string as is
$query = $this->Lessons
->find()
// ...
$types = ['Lessons__lesson_date' => 'string'] + $query->selectTypeMap()->defaults();
$query->selectTypeMap()->defaults($types);
See also
Cookbook > Date & Time > Formatting
Cookbook > Date & Time > Setting the Default Locale and Format String
\Cake\ORM\Query::_addDefaultSelectTypes()
\Cake\Database\Query::selectTypeMap()
\Cake\Database\Query::typeMap()
Related
Even in src/Database/StatementInterface.php the functions fetch and fetchAll have these comment:
$statement = $connection->prepare('SELECT id, title from articles');
$statement->execute();
print_r($statement->fetchAll('assoc')); // will show [0 => ['id' => 1, 'title' => 'a title']]
the functions return ALL fields as string, even the fields are defined as numeric:
array(4) {
["lat"]=>
string(11) "38.49580000"
["lng"]=>
string(11) "-6.95301000"
["taken_at"]=>
string(19) "2021-03-30 16:24:30"
["id"]=>
string(1) "1
The fields lat and lng are defined as DECIMAL(10,8)|(11,8) and id is INTEGER. CakePHP 4.3.1 [and also former versions] return the fields as string which breaks the code.
Am I doing something wrong or is this a bug?
You are using the lowest level of database operations that CakePHP provides, this is almost directly operating with PDO, no casting by CakePHP will apply here, and what exactly you'll receive will entirely depend on the DBMS and the PDO driver that you're using.
Most drivers will return everything as a string, others do cast some values, for example with Postgres you can get integers, while with MySQL you'll get everything as strings.
I am generating a case statement to return either 0 or 1:
$desc_case = $q->newExpr()
->addCase(
[$q->newExpr()->add(["description IS" => NULL])],
[0,1],
["integer","integer"]
);
$q = $q->select(["has_desc" => $desc_case]);
Which results in the following correct SQL:
SELECT [fields removed for clarity], (CASE WHEN (description) IS NULL THEN :c0 ELSE :c1 END) AS `has_desc` FROM skills Skills
I've turned hydration off, and retrieved the result with
->hydrate(false)->toArray();
The result of the CASE statement is returned as a string- either "0" or "1" - which is messing up logic downstream.
I've traced the execution code as best I can, and it looks like CakePHP is using the type names passed to correctly bind the values, but nowhere does the type make its way into the TypeMap used for mapping output.
An easy workaround is to adjust the values after the fact (which I'm doing), but I'd like to make this work as expected on principle... :)
The type information passed to addCase() is only ment to be used for input casting, ie the values passed will be bound as the given types. The return values, ie the values being selected via the compiled CASE statement, are in no way being affected.
If you want to affect the type used for casting selected values of columns that do not exist in the schema (note that changing the schema will also affect other parts of the ORM and the query builder), then you have to change the type map accordingly, for example:
$query
->getSelectTypeMap()
->addDefaults([
'has_desc' => 'integer'
]);
See also
Cakephp-3.x: How to change the data type of a selected alias?
In CakePHP 3.x I can do this:
$Substances = TableRegistry::get('Substances');
$query = $Substances->find()->where($where_conditions)->select(['id']);
debug($query);
This will show me the Query object.
If I want to get the SQL string I can use debug($query->sql());. This will give me the SQL string with placeholders for any parameters, e.g.
SELECT ... WHERE ... in (:c0,:c1,:c2))
When using debug($query) I can see the values for :c0, :c1, etc:
'params' => [
':c0' => [
'value' => (int) 47,
'type' => 'smallinteger',
'placeholder' => 'c0'
],
':c1' => [
'value' => (int) 93,
'type' => 'smallinteger',
'placeholder' => 'c1'
],
':c2' => [
'value' => (int) 845,
'type' => 'smallinteger',
'placeholder' => 'c2'
],
':c3' => [
'value' => (int) 354,
'type' => 'smallinteger',
'placeholder' => 'c3'
]
]
However, I cannot access them outside the debug statement. For example $query->params() or $query['params'] doesn't give me the parameters - it will error. I want to be able to pass this array to a custom function, so how can I access it?
It's strange because I can use debug($query->sql()) to get the SQL string as above, and params is just another thing in that object, but doesn't seem to be accessible.
I've read How to get params from query object in CakePHP 3 but think that's a different question as it was to do with not seeing the values in the debug statement due to the default depth that debug would provide.
The reason I want to do this is because I want to be able to do a CREATE TABLE AS query that will write the values of the SELECT statement into a new table (Important: see this link for an example of how that works in vanilla MySQL). I can't figure out how to do that with the ORM in Cake, so was planning on writing a custom function. But I need to be able to access both the SQL as well as the parameters bound so that the query can be executed in my own function.
If you know of a solution where I can use the ORM to do the CREATE TABLE AS query, I'm still interested to know about this. However I would like to know if params are accessible outside debug() as well.
Premise: I did not actually understand why you need the params
anyway. The information you need is stored by the query ValueBinder object
so you could simply do
$params = $query->getValueBinder()->bindings();
debug($params);
but for some reason this will get you an empty array. My guess is that the query need some kind of initialization first.
in fact if you run
debug($query);
$params = $query->getValueBinder()->bindings();
debug($params);
you'll see your params. I think someone more expert than me will come and give a full explanation
edit: I noticed that debugging $query calls $query->sql() which in turns calls conection->compileQuery();
so you can do
$query->sql(); // you need this to compile the query
// and generate the bindings
$params = $query->getValueBinder()->bindings();
CakePHP does not provide specific methods for creating such CREATE TABLE AS statements, so you'll have to build that on your own.
Compiling a query as the one shown in your question is simple enough using the query objects sql() method, and as arilia already mentioned, you'll be able to access the parameters bound to that query after is was compiled.
Having the compiled SQL and the associated value binder, you can combine this with a custom raw query to build your CREATE TABLE AS statement. All you need to do is prepare a new statement with the compiled SQL, and attach the value binder via its own attachTo() method.
One thing you might also have to do, is to define custom aliases in your select(), as otherwise you'd end up with columns selected (and created) in the form of Substances_id.
$Substances = TableRegistry::get('Substances');
$selectQuery = $Substances
->find()
->where($where_conditions)
->select(['id' => 'id']); // < create aliases as required
// compile the ORM query, this will populate the value binder
$selectSql = $selectQuery->sql();
// combine table creation SQL and compiled ORM query in a single statement
$createStatement = $Substances
->getConnection()
->prepare('CREATE TABLE dynamic_table AS ' . $selectSql);
// attach the ORM querys value binder, binding all its values to the given statement
$selectQuery->getValueBinder()->attachTo($createStatement);
$success = $createStatement->execute();
This should create SQL similar to:
CREATE TABLE dynamic_table AS
SELECT
id AS id
FROM
substances Substances
WHERE
...
See also
Cookbook > Database Access & ORM > Database Basics > Interacting with Statements
API > \Cake\ORM\Association::attachTo()
I'm trying to get a query working using a case statement, and can't figure out how to get the case to return a column value instead of a constant. I have the query working perfectly, except that the column names I'm providing for the results are being quoted or otherwise mishandled by Cake or maybe PDO somewhere down in a layer that I can't dig my way through. I got as far down as bindValue, but none of the documentation I encountered along the way tells me how to do this.
I have found this example comment:
$statement->bindValue(1, 'a title');
$statement->bindValue(2, 5, PDO::INT);
$statement->bindValue('active', true, 'boolean');
$statement->bindValue(5, new \DateTime(), 'date');
but in all these cases, the value provided is a constant. I need to pass in a string that is the name of the column that I want returned.
I tried both 'string' (resulted in quoted column name) and 'integer' (resulted in 0). I tried PDO::FETCH_COLUMN (seemed highly unlikely, but looked like the next best bet from http://php.net/manual/en/pdo.constants.php, and easy to try it...). I tried 'literal', inspired by the way you can put literal strings into expressions (resulted in Error: unknown type "literal"). That error message led me to src/Database/Type.php, but nothing in there helped me either.
So, I'm pretty much stumped. Here's a simple version of the code I have (leaving out a couple of conditions and unrelated columns):
$query = $this->Games->find();
$team_id = $query->newExpr()->addCase(
[$query->newExpr()->eq('Games.status', 'home_default')],
['home_team_id', 'away_team_id'],
['string', 'string']
);
$defaulting = $query
->select([
'id' => $team_id,
'count' => 'COUNT(Games.id)',
])
->where([
'Games.status IN' => ['home_default', 'away_default'],
])
->group('id')
->toArray();
This generates this SQL:
SELECT
(CASE WHEN Games.status = 'home_default'
THEN 'home_team_id' ELSE 'away_team_id' END) AS `id`,
COUNT(Games.id) AS `count`
FROM games Games
WHERE Games.status in ('home_default','away_default')
GROUP BY id
Note that THEN 'home_team_id' ELSE 'away_team_id' END should be simply THEN home_team_id ELSE away_team_id END. This will then allow me to read the list of ids of teams that have defaulted games along with the number of games they defaulted.
By default the values passed to the second argument of QueryExpression::addCase() are being treated as to be converted to literal values, not as identifiers. If you need the latter, then you should use an expression, an IdentifierExpression.
use Cake\Database\Expression\IdentifierExpression;
// ...
$team_id = $query->newExpr()->addCase(
[
$query->newExpr()->eq('Games.status', 'home_default')
],
[
new IdentifierExpression('Games.home_team_id'),
new IdentifierExpression('Games.away_team_id')
]
);
Also ditch the third argument in this case, you don't want the values to be string literals (for expressions the types would be ignored anyways).
I have been trying to figure out how to do this and it seems that its not something that many people are trying to do in cakephp or I am just completely misunderstanding the documentation.. I am using the following query below to get a field value so I get a value where the "findbycreated" is 0... this part works fine
$unregisteredemail = $this->Testmodel->findBycreated('0');
$emailaddress = $unregisteredemail['Testmodel']['emailaddress'] ;
$emailpassword = $unregisteredemail['Testmodel']['password'] ;
But now, after I do some things with this data that I retrieved, I want to mark a field, in the same row, in the same model / table as a value of '1' to indicate that an action has taken place (email address has been successfully created, for example)... I just can't figure out how to do this in cakephp despite my efforts of going through the documentation and searching, this should be rather simple, I am tempted, at this point, to just use a regular mysql query as its a simple query.. basically the query is (please excuse my syntax as I haven't used direct mysql queries in a while) "update (database / table) set 'created'='1' where 'emailaddress'=$emailaddress"
Or I could use the row ID, if needed, as cakephp seems to prefer this, but still can't get how to do this.. this is my attempt below that is not working:
// update database to show that email address has been created
$this->Testmodel->read('emailaddress', $this->Testmodel->findBycreated('0'))
$this->Testmodel->id = 1;
$this->Testmodel->set(array(
'created' => '1'
));
$this->Testmodel->save();
There are, as you can see from the previous answers, several ways to achieve the same end. I'd just like to explain a little about why your way didn't work.
In the model, CakePHP has abstracted the database row(s) into an array according its implementation of ORM . This provides us with a handy way of manipulating the data and chucking it around the MVC architecture.
When you say:
$this->Testmodel->set(array(
'created' => '1'
));
You are dealing directly with the model, but the data is actually stored, as an array, in a class variable called $data. To access and manipulate this data, you should instead say:
$this->data['Testmodel']['created'] => '1';
The reason for specifying the model name as the first index is that where associated tables have been retrieved, these can be accessed in the same way, so you might have , for instance:
Array([Testmodel] => Array ([id] => 1,
[created] => [1],
...
)
[Relatedmodel] => Array ([id] => 1,
[data] => asd,
...
)
)
...and so on. Very handy.
Now, when you use $this->MyModelName->save() with no parameters, it uses $this->data by default and uses the part of the array of data appropriate to the model you are calling the save method on. You can also pass an array of data, formatted in the same way if, for some reason, you don't (or can't) use $this->data.
Your use of the method read() is incorrect. The first parameter should be null, a string or an array of strings (representing fieldname(s)). The second parameter should be the id of the record you wish to read. Instead, for param 2, you are passing the result of a find, which will be an array. The result, which you are not capturing, will be empty.
I would write your code like:
$this->data = $this->Testmodel->read('emailaddress',1);
$this->data['Testmodel']['created'] = 1;
$this->Testmodel->save();
or more succinctly:
$this->Testmodel->id = 1;
$this->Testmodel->saveField('created', 1);
In this situation I would let Cake deal with the id's and just focus on changing the row data and resaving it to the database
$row = $this->Model->findBycreated(0);
$row['Model']['emailaddress'] = 1;
$this->Model->save($row);
This way, you don't have to worry about the id's, as the id will be in your dataset anyway, so just change what you want and then tell Cake to save it.
Ninja edit, Be sure that you are returning a full row with an id from your findBycreated() method.
There're many ways to do your work.I suggest you to read the cookbook about saving data in cakephp.And besides david's solution another simple way would be
$this->Testmodel->id = 1;
$this->Testmodel->saveField('created' =>'1');
Ok, I think I finally found the solution, I was able to get this to work:
$this->Test->updateAll(
array(
'Test.field' => 'Test.field+100'
),
array(
'Test.id' => 1
)
);
I think you have to use updateAll as anything else will just create a new row.. basically CakePHP, for whatever reason, neglected to include a function for updating just one field so you have to put it into an array with the updateAll to make it work...
the +100 is where the updated info goes, so in this case "100" would be what the field is updated to.
In cakephp 3.x the syntax seems to be different. This is what worked for me in 3.x:
$this->Tests->updateAll(
[
'Tests.field = Tests.field+100'
],
[
'Tests.id' => 1
]
];
The difference is that the entire expression needs to be in the value of the first array.