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How to update the password for username (of type Admin) with next condition: first 3 letters of his name, 2 last letters of the last name and 4 last digits of his phone
create table company
(
CODE_COMPANY char(30),
NAME_COMPANY varchar2(30) not null,
MAIL_COMPANY varchar2(30) null,
constraint PK_CODE_COMPANY primary key (CODE_COMPANY),
);
create table USERNAME
(
NAME_USERNAME varchar2(30),
USER_LOCATION number,
fNAME varchar2 (30) not null,
lNAME varchar2 (30) not null,
PHONE_USER char(13) null,
PASSWORD varchar2(30) not null,
USER_POSITION varchar2 (30),
check (USER_POSITION in('Admin', 'Superadmin', 'Technician', 'Student')),
constraint PK_NAME_USERNAME primary key (NAME_USERNAME),
constraint FK_USER_LOCATION foreign key (USER_LOCATION) references uLOCATION (LOCATION)
);
create table uLOCATION
(
LOCATION number,
CODE_COMPANY char(30),
NAME_LOCATION varchar2(30) not null,
FLOOR_LOCATION varchar2(10),
check (FLOOR_LOCATION in ('MAIN_FLOOR', '1ST FLOOR', '2ND FLOOR', '3RD FLOOR')),
constraint PK_LOCATION primary key (LOCATION),
constraint FK_CODE_COMPANY_L foreign key (CODE_COMPANY) references company (CODE_COMPANY),
);
Oracle concat function accepts only two parameters, so there must be two nested concats to join 3 strings together.
update username set password =
concat(
concat(
substr(fname,1,3),
substr(lname, length(lname) - 1)
),
substr(phone_user, length(phone_user) - 3)
)
where user_position = 'Admin';
These links explain how substr() works with length()
https://www.techonthenet.com/oracle/functions/substr.php
https://www.w3resource.com/oracle/character-functions/oracle-length-function.php
Related
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CREATE TABLE DRIVER
(
Driver_Licence VARCHAR2(8) NOT NULL,
SSN VARCHAR2(7),
First_Name VARCHAR2(7) NOT NULL,
Last_Name VARCHAR2(7),
Birth_Date DATE DEFAULT '01/01/1900',
Hire_Date DATE,
State_Name CHAR(2),
CONSTRAINT Driver_PK PRIMARY KEY (Driver_Licence),
CONSTRAINT Driver_UC UNIQUE (SSN),
CONSTRAINT Driver_CHK CHECK (Hire_Date > Birth_Date)
);
/*
Hire Date corresponds to the date the employee was first hired
*/
Trying to create a table in SQL but keep getting the error in the title
ORA-01401: inserted value too large for column
I think you misplaced error
ORA-01401: inserted value too large for column for creating table, that error is for inserting statement.
For creating the table, you need to convert to date for default value of column "Birth_Date". I try to run this script on my env and running well.
CREATE TABLE DRIVER
(
Driver_Licence VARCHAR2(8) NOT NULL,
SSN VARCHAR2(7),
First_Name VARCHAR2(7) NOT NULL,
Last_Name VARCHAR2(7),
Birth_Date DATE DEFAULT TO_DATE('01/01/1900', 'DD/MM/YYYY'),
Hire_Date DATE,
State_Name CHAR(2),
CONSTRAINT Driver_PK PRIMARY KEY (Driver_Licence),
CONSTRAINT Driver_UC UNIQUE (SSN),
CONSTRAINT Driver_CHK CHECK (Hire_Date > Birth_Date)
);
Hope this can help you.
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Here is my code. I can't figure out why I can't add the constraint.
Create table customer(
UserId Integer
)
CREATE TABLE Date(
DateID INTEGER,
User1ID INTEGER NOT NULL,
User2ID INTEGER NOT NULL,
Date CHAR(20) NOT NULL,
GeoLocation CHAR(20) NOT NULL,
BookingFee INTEGER NOT NULL,
CustomerRepresentative CHAR(20) NOT NULL,
Comments CHAR(200),
User1Ratings CHAR(20),
User2Ratings CHAR(20),
Primary Key (DateID),
Check ( User1Ratings IN (‘Excellent’, ‘VeryGood’, ‘Good’, ‘Fair’, ‘Poor’) ),
Check ( User2Ratings IN (‘Excellent’, ‘VeryGood’, ‘Good’, ‘Fair’, ‘Poor’) ),
FOREIGN KEY (User1ID) REFERENCES customer(UserID)
)
The most likely explanation for the behavior is that UserID column is not defined as the PRIMARY KEY or a UNIQUE KEY in the customer table
One fix for this would be to re-write the create for the customer table
For SQL Server:
CREATE TABLE customer
( UserId Integer NOT NULL
, CONSTRAINT PK_customer PRIMARY KEY CLUSTERED (UserId)
);
For MySQL:
CREATE TABLE customer
( UserId INTEGER NOT NULL PRIMARY KEY
);
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create table empl
(
empid number primary key,
empname varchar(25),
email varchar(25) not null unique,
doj date not null,
sal number not null check (sal > 0),
deptid number FOREIGN KEY REFERENCES dept(deptid)
);
Remove the FOREIGN KEY:
create table empl (
empid number primary key,
empname varchar(25),
email varchar(25) not null unique,
doj date not null,
sal number not null check (sal>0),
deptid number REFERENCES dept(deptid)
);
SQL Fiddle is here.
When defining a column, the foreign key relationship is established by the keyword references. You use foreign key when you want to introduce it as a constraint after the columns are defined. So you could also write:
create table empl (
empid number primary key,
empname varchar(25),
email varchar(25) not null unique,
doj date not null,
sal number not null check (sal>0),
deptid number,
foreign key (deptid) REFERENCES dept(deptid)
);
By the way, when using Oracle, one usually uses varchar2() instead of varchar().
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it generates an error,i cant understand whats wrong with the script..every suggestion would be helpful thnx
CREATE TABLE ar_abonent
(
ar_nr_klienti_primary_key int,
emri varchar(25),
mbiemri varchar(25)
);
create table ar_celular
(
NR_CEL INT (12),
ar_nr_klienti FOREIGN Key REFERENCES ar_abonent (ar_nr_klienti),int,
identifikues boolean,
data_aktivizimit date,
marka varchar(25)
constraint chk_celular check (NR_CEL IN ('35566%','35567%','35568%','35569%' AND IDENTIFIKUES='TRUE')
);
CREATE TABLE ar_abonent
(
ar_nr_klienti int NOT NULL primary key , --<-- This column needs to be a primary key
emri varchar(25),
mbiemri varchar(25)
);
create table ar_celular
(
NR_CEL INT ,
ar_nr_klienti int FOREIGN Key REFERENCES ar_abonent (ar_nr_klienti) ,
identifikues bit,
data_aktivizimit date,
marka varchar(25),
constraint chk_celular check (NR_CEL IN ('35566%','35567%','35568%','35569%') AND IDENTIFIKUES= 1)
);
I tried correcting it using SQL Fiddle (set to Oracle 11G R2) as I don't own any Oracle servers. This is how it should look to run without issues:
CREATE TABLE ar_abonent
(
ar_nr_klienti int NOT NULL PRIMARY KEY ,
emri VARCHAR2(25),
mbiemri VARCHAR2(25)
);
CREATE TABLE ar_celular
(
NR_CEL int,
ar_nr_klienti int,
identifikues number(1),
data_aktivizimit date,
marka VARCHAR2(25),
CONSTRAINT fk_column FOREIGN KEY (ar_nr_klienti) REFERENCES ar_abonent (ar_nr_klienti),
CONSTRAINT chk_celular CHECK (NR_CEL IN ('35566%','35567%','35568%','35569%') AND IDENTIFIKUES= 1)
);
Sample SQL Fiddle
Turns out Oracle doesn't have a BOOLEAN type (at table level, although it does exist in PL/SQL), and VARCHAR2 might be preferred to VARCHAR (unless nulls are a concern).
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I am running these two commands to create new tables, but the second create table command keeps giving me error.
CREATE TABLE TEAMSTADIUM(
stadium_name varchar2(40) not null,
stadium_max_capcity number(10) not null,
stadium_field_serface varchar2(40) not null,
stadium_year_built number(4) not null,
stadium_location varchar2(40) not null,
Primary KEY(stadium_name)
)
CREATE TABLE TEAMINFO(
team_name varchar2(40) not null,
team_owner varchar2(40) not null,
team_coach varchar2(40) not null,
team_created Date() not null,
PRIMARY KEY(team_name)
foreign key(stadium_name) references TEAMSTADIUM(stadium_name)
)
Your TEAMINFO table references TEAMSTADIUM.stadium_name, but has no such column of its own. Add it, and ensure it has exactly the same data type as the parent table:
CREATE TABLE TEAMINFO(
team_name varchar2(40) not null,
team_owner varchar2(40) not null,
team_coach varchar2(40) not null,
team_created Date not null,
-- Remove () ^^
-- This column must exist in both tables
stadium_name varchar2(40) not null,
PRIMARY KEY(team_name),
-- missing comma ^^
foreign key(stadium_name) references TEAMSTADIUM(stadium_name)
)
After applying the three modifications above, it will execute correctly: http://sqlfiddle.com/#!4/883a4
Try using Date not null; and put ; after each ) of end table
CREATE TABLE TEAMINFO(
stadium_name varchar2(40) not null,
team_name varchar2(40) not null,
team_owner varchar2(40) not null,
team_coach varchar2(40) not null,
team_created Date not null,
PRIMARY KEY(team_name),
foreign key(stadium_name) references TEAMSTADIUM(stadium_name)
);