I'm using Informix and I want to create a table like this:
CREATE TABLE increment_table
(
id SERIAL NOT NULL,
increment_column INTEGER IDENTITY (START WITH 2000, INCREMENT BY 1),
name VARCHAR(255) NOT NULL,
PRIMARY KEY (id)
);
But when I run the statement, the result is a syntax error after the INTEGER keyword. I suppose is because Informix does not support this keyword IDENTITY.
Can you help me in some ideas of how can I create an auto increment column, with this characteristics:
The column need to increment automatically in every insert.
The column isn't primary key.
Allows me set the initial number of start and the value of increment.
Can edit the value of the column.
The table already have a serial column.
Related
I have created one table in PL SQL Developer.
CREATE TABLE Patient_List
(
Patient_ID number NOT NULL,
Patient_Name varchar(50) NOT NULL,
Patient_Address varchar(100) NULL,
App_Date date NULL,
Descr varchar(50),
CONSTRAINT patient_pk PRIMARY KEY(Patient_ID)
);
I want to auto increment Patient_ID, I tried altering the table and modifying the Patient_ID column but it's showing an error "invalid ALTER TABLE option"
ALTER TABLE Patient_List
MODIFY Patient_ID NUMBER NOT NULL GENERATED ALWAYS AS IDENTITY;
Please help, Thanks in advance.
This is not possible.
Oracle 10g didn't even have identity columns, they were introduced in Oracle 12.1
But even with a current Oracle version, you can't convert a regular column to an identity column. You would need to add a new one.
Before identity columns, the usual way was to create a sequence and a trigger to populate the column.
See here: How to create id with AUTO_INCREMENT on Oracle?
If anybody wants to modify existing column as auto_increment use this three lines
alter table Product drop column test_id;
create sequence Product_test_id_seq INCREMENT BY 1 nocache;
alter table Product add test_id Number default Product_test_id_seq.nextval not null;
I am trying to create a simple SQLite database that will allow me to store email addresses and timestamps. I have created the table like this:
$sql =<<<EOF
CREATE TABLE ENTRIES
(ID INT PRIMARY KEY NOT NULL,
EMAIL EMAIL NOT NULL,
TIMESTAMP DATETIME DEFAULT CURRENT_TIMESTAMP);
EOF;
And I am trying to insert an email like this:
$sql =<<<EOF
INSERT INTO ENTRIES (EMAIL)
VALUES (test#test.com);
EOF;
I am getting an error
NOT NULL constraint failed: ENTRIES.ID
I am assuming this is to do with the ID and autoincrement? I have read the docs and it advises against using autoincrement. Where am I going wrong?
The docs say:
If a table contains a column of type INTEGER PRIMARY KEY, then that column becomes an alias for the ROWID.
And because it becomes an alias for the ROWID, it's not necessary to explicitly specify a value.
You have INT PRIMARY KEY, not INTEGER PRIMARY KEY. If you change it to INTEGER PRIMARY KEY, it works the way you expect.
I have like 11 columns in my database table and i am inserting data in 10 of them. i want to have a unique number like "1101 and so on" in the 11th column.
Any idea what should i do?? Thanks in advance.
SQL Server 2012 and above you can generate Sequence
Create SEQUENCE RandomSeq
start with 1001
increment by 1
Go
Insert into YourTable(Id,col1...)
Select NEXT VALUE FOR RandomSeq,col1....
or else you can use Identity
Identity(seed,increment)
You can start the seed from 1101 and increment the sequence by 1
Create table YourTable
(
id INT IDENTITY(1101,1),
Col varchar(10)
)
If you want to have that unique number in a different field then you can manipulate that field with primary key and insert that value.
If you want in primary key value, then open the table in design mode, go to 'Identity specification', set 'identity increment' and 'identity seed' as you want.
Alternatively you can use table script like,
CREATE TABLE Persons
(
ID int IDENTITY(12,1) PRIMARY KEY,
FName varchar(255) NOT NULL,
)
here the primary key will start seeding from 12 and seed value will be 1.
If you have your table definition already in place you can alter the column and add Computed column marked as persisted as:
ALTER TABLE tablename drop column column11;
ALTER TABLE tablename add column11 as '11'
+right('000000'+cast(ID as varchar(10)), 2) PERSISTED ;
--You can change the right operator value from 2 to any as per the requirements.
--Also replace ID with the identity column in your table.
create table inc
(
id int identity(1100,1),
somec char
)
In PostgreSql, one can define a sequence and use it as the primary key of a table. In HsqlDB, one can still accomplish creating an auto-increment identity column which doesn't link to any user defined sequence. Is it possible to use a user defined sequence as the generator of an auto-increment identity column in HsqlDB?
Sample sql in PostgreSql:
CREATE SEQUENCE seq_company_id START WITH 1;
CREATE TABLE company (
id bigint PRIMARY KEY DEFAULT nextval('seq_company_id'),
name varchar(128) NOT NULL CHECK (name <> '')
);
What's the equivalent in HsqlDB?
Thanks.
In version 2.0, there is no direct feature for this. You can define a BEFORE INSERT trigger on the table to do this:
CREATE TABLE company ( id bigint PRIMARY KEY, name varchar(128) NOT NULL CHECK (name <> '') );
CREATE TRIGGER trigg BEFORE INSERT
ON company REFERENCING NEW ROW AS newrow
FOR EACH ROW
SET newrow.id = NEXT VALUE FOR seq_company_id;
and insert without using any vlue for id
INSERT INTO company VALUES null, 'test'
Update for HSQLDB 2.1 and later: A feature has been added to support this.
CREATE SEQUENCE SEQU
CREATE TABLE company ( id bigint GENERATED BY DEFAULT AS SEQUENCE SEQU PRIMARY KEY, name varchar(128) NOT NULL CHECK (name <> '') );
See the Guide under CREATE TABLE http://hsqldb.org/doc/2.0/guide/databaseobjects-chapt.html#dbc_table_creation
In addition, 2.1 and later has a PostgreSQL compatibility mode in which it accepts the PostgreSQL CREATE TABLE statement that references the sequence in the DEFAULT clause and translates it to HSQLDB syntax.
I need to have one column as the primary key and another to auto increment an order number field. Is this possible?
EDIT: I think I'll just use a composite number as the order number. Thanks anyways.
CREATE TABLE [dbo].[Foo](
[FooId] [int] IDENTITY(1,1) NOT NULL,
[BarId] [int] IDENTITY(1,1) NOT NULL
)
returns
Msg 2744, Level 16, State 2, Line 1
Multiple identity columns specified for table 'Foo'. Only one identity column per table is allowed.
So, no, you can't have two identity columns. You can of course make the primary key not auto increment (identity).
Edit: msdn:CREATE TABLE (Transact-SQL) and CREATE TABLE (SQL Server 2000):
Only one identity column can be created per table.
You can use Sequence for second column with default value IF you use SQL Server 2012
--Create the Test schema
CREATE SCHEMA Test ;
GO
-- Create a sequence
CREATE SEQUENCE Test.SORT_ID_seq
START WITH 1
INCREMENT BY 1 ;
GO
-- Create a table
CREATE TABLE Test.Foo
(PK_ID int IDENTITY (1,1) PRIMARY KEY,
SORT_ID int not null DEFAULT (NEXT VALUE FOR Test.SORT_ID_seq));
GO
INSERT INTO Test.Foo VALUES ( DEFAULT )
INSERT INTO Test.Foo VALUES ( DEFAULT )
INSERT INTO Test.Foo VALUES ( DEFAULT )
SELECT * FROM Test.Foo
-- Cleanup
--DROP TABLE Test.Foo
--DROP SEQUENCE Test.SORT_ID_seq
--DROP SCHEMA Test
http://technet.microsoft.com/en-us/library/ff878058.aspx
Add one identity column and then add a computed column whose formula is the name of the identity column
Now both will increment at the same time
No it is not possible to have more than one identity column.
The Enterprise Manager does not even allow you to set > 1 column as identity. When a second column is made identity
Also note that ##identity returns the last identity value for the open connection which would be meaningless if more than one identity column was possible for a table.
create table #tblStudent
(
ID int primary key identity(1,1),
Number UNIQUEIDENTIFIER DEFAULT NEWID(),
Name nvarchar(50)
)
Two identity column is not possible but if you accept to use a unique identifier column then this code does the same job as well. And also you need an extra column - Name column- for inserting values.
Example usage:
insert into #tblStudent(Name) values('Ali')
select * from #tblStudent
Ps: NewID() function creates a unique value of type uniqueidentifier.
The primary key doesn't need to be an identity column.
You can't have two Identity columns.
You could get something close to what you want with a trigger...
in sql server it's not possible to have more than one column as identity.
I've just created a code that will allow you inserting two identities on the same table. let me share it with you in case it helps:
create trigger UpdateSecondTableIdentity
On TableName For INSERT
as
update TableName
set SecondIdentityColumn = 1000000+##IDENTITY
where ForstId = ##IDENTITY;
Thanks,
A workaround would be to create an INSERT Trigger that increments a counter.
So I have a table that has one identity col : applicationstatusid. its also the primary key.
I want to auto increment another col: applicationnumber
So this is the trigger I write.
create trigger [applicationstatus_insert] on [ApplicationStatus] after insert as
update [Applicationstatus]
set [Applicationstatus].applicationnumber =(applicationstatusid+ 4000000)
from [Applicationstatus]
inner join inserted on [applicationstatus].applicationstatusid = inserted.applicationstatusid