I'm building a laravel application where i have a Reservation model. After the new record is created, the user may want to update it.
I'm looking for a solution to check if the UpdateRequest values are the same of the actual records and if there are updates only update the modified values.
Example, the original record has :
{'id' => 1, 'name' = 'Jhon', surname = 'Doe'}
While the UpdateRequest has
{'id' => 1, 'name' = 'Jhon', surname = 'Kirby'}
So, my questions are :
How do i check which values are different?
How do i update only the
specified values?
I've tried to look at the documentation of laravel but it doesn't seem to refer to this situation, is there any solution or should I just update all the values?
Thank you
Related
I'm a little surprised I haven't found any information on the following question, so please excuse if I've missed it somewhere in the docs. Using SQL Server (2016 locally and Azure) and EFCore Code First we're trying to create a computed table column with a persisted value. Creating the column works fine, but I don't have a clue how to persist the value. Here's what we do:
modelBuilder.Entity<SomeClass>(entity =>
{
entity.Property(p => p.Checksum)
.HasComputedColumnSql("(checksum([FirstColumnName], [SecondColumnName]))");
});
And here is what we'd actually like to get in T-SQL:
CREATE TABLE [dbo].[SomeClass]
(
[FirstColumnName] [NVARCHAR](10)
, [SecondColumnName] [NVARCHAR](10)
, [Checksum] AS (CHECKSUM([FirstColumnName], [SecondColumnName])) PERSISTED
);
Can anyone point me in the right direction?
Thanks in advance, Tobi
UPDATE: Based on a good idea by #jeroen-mostert I also tried to just pass the PERSISTED string as part of the formula:
modelBuilder.Entity<SomeClass>(entity =>
{
entity.Property(p => p.Checksum)
.HasComputedColumnSql("(checksum([FirstColumnName], [SecondColumnName]) PERSISTED)");
});
And also outside of the parentheses:
modelBuilder.Entity<SomeClass>(entity =>
{
entity.Property(p => p.Checksum)
.HasComputedColumnSql("(checksum([FirstColumnName], [SecondColumnName])) PERSISTED");
});
However und somehow surprisingly, the computed column is still generated with Is Persisted = No, so the PERSISTED string simply seems to be ignored.
Starting with EF Core 5, the HasComputedColumnSql method has a new optional parameter bool? stored to specify that the column should be persisted:
modelBuilder.Entity<SomeClass>()
.Property(p => p.Checksum)
.HasComputedColumnSql("checksum([FirstColumnName], [SecondColumnName])", stored: true);
After doing some reading and some tests, I ended up trying the PERSISTED inside the SQL query and it worked.
entity.Property(e => e.Duration_ms)
.HasComputedColumnSql("DATEDIFF(MILLISECOND, 0, duration) PERSISTED");
The generated migration was the following:
migrationBuilder.AddColumn<long>(
name: "duration_ms",
table: "MyTable",
nullable: true,
computedColumnSql: "DATEDIFF(MILLISECOND, 0, duration) PERSISTED");
To check on the database whether it is actually persisted I ran the following:
select is_persisted, name from sys.computed_columns where is_persisted = 1
and the column that I've created is there.
" You may also specify that a computed column be stored (sometimes called persisted), meaning that it is computed on every update of the row, and is stored on disk alongside regular columns:"
modelBuilder.Entity<SomeClass>(entity =>
{
entity.Property(p => p.Checksum)
.HasComputedColumnSql("(checksum([FirstColumnName], [SecondColumnName]), stored: true);
});
This is taken (and slightly modified) from Microsoft Docs.: https://learn.microsoft.com/en-us/ef/core/modeling/generated-properties?tabs=data-annotations#computed-columns
I am following Link to integrate cloudant no sql-db.
There are methods given create DB, Find all, count, search, update. Now I want to update one key value in one of my DB doc file. how Can i achieve that. Document shows like
updateDoc (name, doc)
Arguments:
name - database name
docID - document to update
but when i pass my database name and doc ID its throwing database already created can not create db. But i wanted to updated doc. So can anyone help me out.
Below is one of the doc of may table 'employee_table' for reference -
{
"_id": "0b6459f8d368db408140ddc09bb30d19",
"_rev": "1-6fe6413eef59d0b9c5ab5344dc642bb1",
"Reporting_Manager": "sdasd",
"Designation": "asdasd",
"Access_Level": 2,
"Employee_ID": 123123,
"Employee_Name": "Suhas",
"Project_Name": "asdasd",
"Password": "asda",
"Location": "asdasd",
"Project_Manager": "asdas"
}
So I want to update some values from above doc file of my table 'employee_table'. So what parameters I have to pass to update.
first of all , there is no concept named table in no sql world.
second, to update document first you need to get document based on any input field of document. you can also use Employee_ID or some other document field. then use database.get_query_result
db_name = 'Employee'
database = client.create_database(db_name,throw_on_exists=False)
EmployeeIDValue = '123123'
#here throw_on_exists=False meaning dont throw error if DB already present
def updateDoc (database, EmployeeIDValue):
results = database.get_query_result(
selector= { 'Employee_ID': {'$eq': EmployeeIDValue} }, )
for result in results:
my_doc_id = result["_id"]
my_doc = database[my_doc_id] #===> this give you your document.
'''Now you can do update'''
my_doc['Employee_Name'] = 'XYZ'
my_doc.save() ====> this command updates current document
Employees table has a field named current_address_id. I'm adding a new address to Addresses like:
$updatedEntity = $this->patchEntity($employee, [
//some other fields
'user' => $userData,
'employees_phones' => $phonesData,
'employees_addresses' => $addressesData,
], [
'associated' => ['Users', 'EmployeesPhones', 'EmployeesAddresses']
]);
$this->save($updatedEntity);
I'm inserting the new address successfully but now I need to update Employees.current_address_id with the new address ID. How can I do this?
Table::save() returns the entity with updated ids. So store the return value in a variable and use appropriate property of the entity.
Try this one. It works for me.
$this->ModelName->save($updatedEntity);
$lastInsertedId = $updatedEntity->id;
The save method will update the entity with the last insert id as long as the entity id is not already set.
In CakePHP I have two models: Clients & Tickets. A client can have many tickets and a ticket can only have 1 client.
When adding a new ticket I want to automatically create a new client by only entering a name. So the form would be:
Name = "Name client" >> Name should be save in Client table en new client_ID in Ticket table
Info = Ticket information >> Save in Ticket table.
"Save"
I'm not sure how this works. I have associations in the model and tried to saveAll but there is no data stored in the Client table. And how do I get the ID in the Ticket table?
Hope someone can point me in the right direction. I've searched for other answers but cant seem to find a solution. Is saveAll the right way to do this?
You should set a php condition before insert values into the table, the sql request 4 exemple return the numbers of repetition of the same ticket_id so if the returned value is greater than 1 the request can't execute else the request insert the values
The function to count number of rows : mysql_num_rows('request');
You can use the callback methods.
If you have defined your relation in the Ticket model with belongsTo Client, it will be accessible from that model.
public $belongsTo = array(
'Client' => array(
'className' => 'Client',
'foreignKey' => 'client_id',
),
);
So, if you want to make a new client before saving the ticket, you can do:
public function beforeSave ($options = array()) {
$this->data['Client']['name'] = $someVar;
$this->Client->save($this->data);
$this->data[$this->alias]['client_id'] = $this->Client->getInsertID();
return true;
}
I have a Post model which hasMany PostField
every post can have several fields stored in the post_fields table..
post_fields has this structure: (id, post_id, name, value)
posts table has some common fields for all posts, but any additional fields should be stored in post_fields table..
I created a search form that is used to filter the posts
when specifying filters for the fields in the posts table, it works fine..
but I want to make the filter to work even on the other fields found in post_fields ..
I can retrieve the posts first then filter them manually, but i want something more efficient !
EXAMPLE: let's suppose that posts are describing some products..
post (id, title, created, price)
post_fields (id, post_id, name, value)
in this case, all posts have title, created and price..
but if a post (id=3) wants to have a weight field, we should do that by creating a record in post_fields, the record should be :
{ id: .. , post_id: 3, name: weight, value: .. }
it's easy now to filter posts according to price (e.g. price between min & max)..
but, what if i want to filter posts according to weight ??
e.g. i want all posts that have weight greater than 10 !!
I would like to achieve this preferably in one query, using joins maybe or subqueries ..
I don't know how to do that in cakePHP, so if any one has an idea, plz HELP !!
even if someone just has an idea but doesn't have details, that could help ...
thanx in advance !
There is no way to search against the children of a hasMany relationship. You will need to run your query against the PostFields model. ie: $this->PostField->find('all', array('conditions'=>array('PostField.name' => 'weight', 'PostField.value' > 10)));
If you want to do a query against both the PostField and Post models at the same time (ie: price < $1.00 and weight > 10, you will need to do a custom query, as CakePHP has no built-in solution for doing so TMK. Should look something like this:
$query = "SELECT ... FROM posts as Post, post_fields as PostField WHERE PostField.name = 'weight' AND PostField.value > 10 AND POST.price < 1.0 AND PostField.post_id = Post.id;"
$posts = $this->Post->query($query);
EDIT:
I would do this. You're not going to get away with doing a single call, but this is still a clean solution.
$postIds = null;
if(/*we need to run query against PostFields*/) {
$conditions = array(
'OR' => array(
array(
'AND' => array(
'PostField.name' => 'weight',
'PostField.value' > 10
)
),
array(
'AND' => array(
'PostField.name' => 'height',
'PostField.value' < 10
)
)
)
);
$fields = array('PostField.id', 'PostField.post_id');
$postIds = $this->Post->PostField->find('list', array('conditions'=>$conditions, 'fields'=>$fields));
}
$conditions = array('Post.price' < 1.0);
if($postIds) {
$conditions['Post.id'] = $postIds;
}
$posts = $this->Post->find('all', array('conditions'=>$conditions));
You should look into using the Containable behavior for your models. This way, you can filter the returned columns as you like. (I think this is the type of filtering you want to do)