I'm new with postgres database codeigniter
for return last id in mysql
im using this in my model
public function daftar($data){
$this->db->insert('akun', $data);
return $this->db->insert_id();
}
but I'm confuse how to return las id ($this->db->insert_id) in postgres?
From the CodeIgniter documentation:
If using the PDO driver with PostgreSQL, or using the Interbase driver, this function requires a $name parameter, which specifies the appropriate sequence to check for the insert id.
In your case you need return $this->db->insert_id('akun_id_akun_seq'); if "akun_id_akun_seq" is the name of the respective sequence.
If your INSERT is something like this:
INSERT INTO public."MyTable"
(
"SomethingIdFk",
"Date"
)
VALUES
(
8,
CURRENT_TIMESTAMP
);
And MyTable has a serial like MyTableId as primary key, then in your model you can do this:
$id= $this->db->insert_id('public."MyTable_MyTableId_seq"');
to get the last insert id.
That works for me.
More info you can find in this post.
Related
I am using UUID's in PostgreSQL as Primary Key for my tables. I am trying to insert the UUID using PDO and prepared statement. However, I am not able to bind the value using bindValue. These are the steps I am trying to follow:
$sql = "INSERT INTO customers.customers(customer_id, first_name, last_name, email_address)";
$sql .= " VALUES(":customer_id", ":first_name", ":last_name", ":email_address")";
$this->stmt = prepare($sql);
$this->stmt = bindValue(":customer_id", $customer_uuid); //*** WHAT PDO value to use Here??????
$this->stmt = bindValue(":first_name", $first_name, PDO::PARAM_STR);
$this->stmt = bindValue(":last_name", $last_name, PDO::PARAM_STR);
$this->stmt = bindValue(":email_address", "$email_address, PDO::PARAM_STR);
$this->stmt->execute();
As can be seen here, each "bindValue" statement required to have a PDO parameter type value. I have not been able to find what value can be used for a UUID type column. I tried using the PDO::PARAM_STR, but this creates a Data Type error when inserting the UUID int he column as the column type for customer_id is UUID.
Any suggestions from the community here?
I tried converting the UUID to string and then back to UUID, but it did not work. From here, I don't know what else I can try.
I am using DatabaseClient for building a custom Repository. After I insert or update an Item I need that Row data to return the saved/updated Item. I just can´t wrap my head around why .all(), .first(), .one() are not returning the Result Map, although I can see that the data is inserted/updated in the database. They just signal onComplete. But .rowsUpdated() returns 1 row updated.
I observed this behaviour with H2 and MS SQL Server.
I´m new to R2dbc. What am I missing? Any ideas?
#Transactional
public Mono<Item> insertItem(Item entity){
return dbClient
.sql("insert into items (creationdate, name, price, traceid, referenceid) VALUES (:creationDate, :name, :price, :traceId, :referenceId)")
.bind("creationDate", entity.getCreationDate())
.bind("name", entity.getName())
.bind("price", entity.getPrice())
.bind("traceId", entity.getTraceId())
.bind("referenceId", entity.getReferenceId())
.fetch()
.first() //.all() //.one()
.map(Item::new)
.doOnNext(item -> LOGGER.info(String.format("Item: %s", item)));
}
The table looks like this:
CREATE TABLE [dbo].[items](
[creationdate] [bigint] NOT NULL,
[name] [nvarchar](32) NOT NULL,
[price] [int] NOT NULL,
[traceid] [nvarchar](64) NOT NULL,
[referenceid] [int] NOT NULL,
PRIMARY KEY (name, referenceid)
)
Thanks!
This is the behavior of an insert/update statement in database, it does not return the inserted/updated rows.
It returns the number of inserted/updated rows.
It may also return some generated values by the database (such as auto increment, generated uuid...), by adding the following line:
.filter(statement -> statement.returnGeneratedValues())
where you may specify specific generated columns in parameter. However this has limitations depending on the database (for example MySql can only return the last generated ID of an auto increment column even if you insert multiple rows).
If you want to get the inserted/updated values from database, you need to do a select.
I created a "favorite" functionality, which is similar to the common "Like" functionality in many websites.
There are 3 tables:
"User" with primary key UUID
"Photo" with pk UUID
"Favorite" with pk user.UUID and post.UUID
The corresponding SQL is:
CREATE TABLE public."user" (
id uuid DEFAULT public.gen_random_uuid() NOT NULL
);
CREATE TABLE public."photo" (
id uuid DEFAULT public.gen_random_uuid() NOT NULL
);
CREATE TABLE public."favorite" (
userId uuid NOT NULL
photoId uuid NOT NULL
);
Now, I would like to query photos with a computed field isFavorite as boolean where the value is set to true when the current user has favorited the photo.
So, I created this custom SQL function:
CREATE OR REPLACE FUNCTION public.isfavorite(photo photo, hasura_session json)
RETURNS boolean
LANGUAGE sql
STABLE
AS $function$
SELECT EXISTS (
SELECT *
FROM public.favorite
WHERE "userId" = (VALUES (hasura_session ->> 'x-hasura-role'))::uuid AND "photoId" = photo.uuid
)
$function$
I can create this function with SQL in Hasura, but when I set this function to a computed field in the photo table, Hasura display this error:
in table "photo": in computed field "isFavorite": function "isfavorite" is overloaded. Overloaded functions are not supported
Where I made a mistake? Can we build a custom function that return boolean? How do you build a favorite (or like) functionality?
Solved: There was two isFavorite functions in the database that cause overloading...
So now there is a isFavorite field in the photo schema, but I need te provide $args with hasura_session as argument.
How to provide hasura_session without the need to fill in arguments?
You will need to track your computed column passing the session variable.
https://hasura.io/docs/1.0/graphql/manual/api-reference/schema-metadata-api/computed-field.html
{
"type":"add_computed_field",
"args":{
"table":{
"name":"photo",
"schema":"public"
},
"name":"isfavorite",
"definition":{
"function":{
"name":"isfavorite",
"schema":"public"
},
"table_argument":"photo_row",
"session_argument": "hasura_session"
}
}
}
This was also added recently. Make sure your are on version v1.3 or later. I would also change the function to accept photo_row as the variable, instead of photo photo this might cause issues with PostgreSQL.
CREATE OR REPLACE FUNCTION public.isfavorite(photo_row photo, hasura_session json)
RETURNS boolean
LANGUAGE sql
STABLE
AS $function$
SELECT EXISTS (
SELECT *
FROM public.favorite
WHERE "userId" = (VALUES (hasura_session ->> 'x-hasura-role'))::uuid AND "photoId" = photo.uuid
)
$function$
I have a query in mysql works well, but when I go to postgresql does not update me, I want to know where is my error.
I leave my php file the query update does not work
<?php
require_once "Controllers/conexion.php";
session_start();
$resultado=pg_query("SELECT nextval('user_id_seq') as key");
$row=pg_fetch_array($resultado, 0);
$key=$row['key'];
try {
$resultado = pg_query($conexion,"select * from encuesta_respuesta where id_user = '".$_SESSION['user']."' and id_encuesta = '".$_POST['id_encuesta']."'");
while( $row = pg_fetch_assoc($resultado)){
$data = $row;
}
if ($data['estado']=='F') {
header("Location: Inicio.php");
}
foreach($_POST['pregunta'] as $id_pregunta=>$valor){
$query="insert into encuesta_respuesta_opcion values (".$key.",".$_POST['id_encuesta'].",".$id_pregunta.",".$valor.")";
$resultado = pg_query($conexion,$query);
}
$query="update encuesta_respuesta set estado='F' where id_user=".$_SESSION['user']." and id_encuesta = ".$_POST['id_encuesta'];
$resultado = pg_query($conexion,$query);
$resp['error']=false;
} catch (Exception $e) {
$resp['error']=true;
}
header("Location: Inicio.php");
?>
Directly try to update data in your database, check this query works or not. If it works, then you have to change your query building procedure in your application. For example:
postgres=# create table test (id_user VARCHAR (50) PRIMARY KEY, id_encuesta VARCHAR (50), estado VARCHAR (10));
postgres=# insert into test values ('anower','engg.','A');
postgres=# update test set estado='F' where id_user='anower' and id_encuesta='engg.';
The query should work the same in MySql and postgres.
If you are getting different results during updates then your survey tables arent the same.
Most liked id_user and id_encuesta are autoincrement fields. So they dont necesary have the same values.
Try using a Select to see if they have same survey information.
SELECT *
FROM survey
where id_user=".$_SESSION['user']."
and id_encuesta = ".$_POST['id_encuesta'];
I have a table on Postgres that auto generates UUIDs, when I dd Customer::all(); on Laravel I get an array with "cs_id" => "d0402be5-e1ba-4cb2-a80c-5340b406e2c3" which is fine. When I loop or select one record with the only the cs_id the data it retuns 0,2,5 for the three records currently on the table which is incorrect data.
EDIT:
CREATE TABLE customers
(
cs_id character varying(255) NOT NULL DEFAULT gen_random_uuid(),
CONSTRAINT cs_customers_pkey PRIMARY KEY (cs_id),
}
On laravel
$customerData = Customer::where('cs_id','d0402be5-e1ba-4cb2-a80c-5340b406e2c3')->first();
dd($customerData['cs_id']);
For some reason Eloquent messes up there.
just add a getter and use it whenever you need the cs_id
public function getGuid()
{
return $this->attributes['cs_id'];
}
To use uuids auto-generated by the database, define your model as follows:
class Customer extends Model
{
// rename the id column (optional)
protected $primaryKey = 'cs_id';
// tell Eloquent that your id is not an integer
protected $keyType = 'string';
// do NOT set $incrementing to false
}
Then you can use all Eloquent's methods as you would with classic ids:
$customerData = Customer::findOrFail('d0402be5-e1ba-4cb2-a80c-5340b406e2c3');
Use Customer::findOrFail('d0402be5-e1ba-4cb2-a80c-5340b406e2c3');
to get the record matching that pk.
I'm assuming on top you have use App\Customer;