Here are my tables CREATE Statements:
CREATE TABLE `User` (
`UserId` int NOT NULL AUTO_INCREMENT,
`CompanyId` int NOT NULL,
`FirstName` varchar(50) NOT NULL,
`LastName` varchar(50) NOT NULL,
`UserName` varchar(100) NOT NULL,
PRIMARY KEY (`UserId`),
KEY `FK_Company_User_idx` (`CompanyId`),
CONSTRAINT `FK_Company_User` FOREIGN KEY (`CompanyId`) REFERENCES `Company` (`CompanyId`)
) ENGINE=InnoDB AUTO_INCREMENT=256 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_0900_ai_ci;
CREATE TABLE `Notification` (
`NotificationId` int NOT NULL AUTO_INCREMENT,
`Email` varchar(100) DEFAULT NULL,
`Message` text,
`Subject` varchar(100) DEFAULT NULL,
`UserId` int NOT NULL,
PRIMARY KEY (`NotificationId`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_0900_ai_ci;
I am trying to create a FK between UserId in Notification and UserId in User but the dropdown for selecting the field is not populating.
Any help is appreciated.
Related
I have problem with creating FK. Am checking and checking and i cant see what is wrong.
I have product, sites and product_site_ref tables.
When i try to add FK to table product_site_ref i cant choose referenced column. Check screenshot.
Relation table
CREATE TABLE `product_site_ref` (
`id` bigint(20) NOT NULL,
`product_id` bigint(20) unsigned NOT NULL,
`site_id` bigint(20) DEFAULT NULL,
`created_at` timestamp NULL DEFAULT NULL,
`updated_at` timestamp NULL DEFAULT NULL,
`deleted_at` timestamp NULL DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `product_index` (`product_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;
Product table
CREATE TABLE `product` (
`id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
`code` varchar(125) COLLATE utf8mb4_unicode_ci DEFAULT NULL,
`name` varchar(225) COLLATE utf8mb4_unicode_ci DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=1 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;
Site table
CREATE TABLE `site` (
`id` bigint(20) NOT NULL,
`name` varchar(245) DEFAULT NULL,
`url` varchar(45) DEFAULT NULL,
`created_at` timestamp NULL DEFAULT NULL,
`updated_at` timestamp NULL DEFAULT NULL,
`deleted_at` timestamp NULL DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
Note: table proizvod on screenshot is product. I just translate for your goys to better understand. All is the some
These are tables before referances
CREATE TABLE olap.time (
idtime SERIAL NOT NULL PRIMARY KEY,
year integer,
month integer,
week integer,
day integer
);
CREATE TABLE olap.addressees (
idaddressee integer PRIMARY KEY NOT NULL,
name varchar(40) NOT NULL,
zip char(6) NOT NULL,
address varchar(60) NOT NULL
);
CREATE TABLE olap.customers (
idcustomer varchar(10) SERIAL PRIMARY KEY autoincrement,
name varchar(40) NOT NULL,
city varchar(40) NOT NULL,
zip char(6) NOT NULL,
address varchar(40) NOT NULL,
email varchar(40),
phone varchar(16) NOT NULL,
regon char(9)
);
After creating this tables I want to create this table
CREATE TABLE olap.fact(
idtime integer NOT NULL,
idaddressee integer NOT NULL,
idcustomer varchar(10) NOT NULL,
idfact integer NOT NULL,
price numeric(7,2),
PRIMARY KEY (idtime, idaddressee, idcustomer),
FOREIGN KEY (idaddressee) REFERENCES olap.addressees(idaddressee),
FOREIGN KEY (idcustomer REFERENCES olap.customers(idcustomer),
FOREIGN KEY (idtime) REFERENCES time(idtime)
));
But I get error as
"ERROR: syntax error at or near "REFERENCES"
LINE 9: FOREIGN KEY (idcustomer REFERENCES olap.customers(idcustom..."
Thanks in advance
The idcustomer from olap.fact and idcustomer from olap.customers has different datatype SERIAL and Varchar(10),
I have corrected the datatypes and validated the code below
CREATE TABLE olap.time (
idtime SERIAL NOT NULL PRIMARY KEY,
year integer,
month integer,
week integer,
day integer
);
CREATE TABLE olap.addressees (
idaddressee integer PRIMARY KEY NOT NULL,
name varchar(40) NOT NULL,
zip char(6) NOT NULL,
address varchar(60) NOT NULL
);
CREATE TABLE olap.customers (
idcustomer varchar(10) PRIMARY KEY ,
name varchar(40) NOT NULL,
city varchar(40) NOT NULL,
zip char(6) NOT NULL,
address varchar(40) NOT NULL,
email varchar(40),
phone varchar(16) NOT NULL,
regon char(9)
);
CREATE TABLE olap.fact(
idtime integer NOT NULL,
idaddressee integer NOT NULL,
idcustomer varchar(10) NOT NULL,
idfact integer NOT NULL,
price numeric(7,2),
PRIMARY KEY (idtime, idaddressee, idcustomer),
FOREIGN KEY (idaddressee) REFERENCES olap.addressees(idaddressee),
FOREIGN KEY (idcustomer) REFERENCES olap.customers(idcustomer),
FOREIGN KEY (idtime) REFERENCES olap.time(idtime)
);
My question is similar to the one posted earlier in the Community. questions/62936399/error-sql-state-42703-while-trying-to-insert-data-into-my-table
In PostgreSQL, I'm trying to run CREATE Table Statement so that I can continue with inserting values. Although my CREATE statement fails so I can't get on with INSERT statement. The error message that keeps coming up ERROR: relation "eventrequest" does not exist
SQL state: 42P01
I have re-did the entire CREATE Statement twice although the error message does not change.
CREATE TABLE CUSTOMER
(CustNo VARCHAR(8) CONSTRAINT CustNoNotNull NOT NULL,
CustName VARCHAR(30) CONSTRAINT CustNameNotNull NOT NULL,
Address VARCHAR(50) CONSTRAINT AddressNotNull NOT NULL,
Internal CHAR(1) CONSTRAINT InternalNotNull NOT NULL,
Contact VARCHAR(35) CONSTRAINT ContractNotNull NOT NULL,
Phone VARCHAR(11) CONSTRAINT CPhoneNotNull NOT NULL,
City VARCHAR(30) CONSTRAINT CityNotNull NOT NULL,
State VARCHAR(2) CONSTRAINT StateNotNull NOT NULL,
Zip VARCHAR(10) CONSTRAINT ZipNotNull NOT NULL,
CONSTRAINT PK_CUSTOMER PRIMARY KEY (CustNo)
);
CREATE TABLE FACILITY
(FacNo VARCHAR(8) CONSTRAINT FacNoNotNull NOT NULL,
FacName VARCHAR(30) CONSTRAINT FacNameNotNull NOT NULL,
CONSTRAINT PK_FACILITY PRIMARY KEY (FacNo),
CONSTRAINT Unique_FacName UNIQUE(FacName)
);
CREATE TABLE LOCATION
(LocNo VARCHAR(8) CONSTRAINT LocNoNotNull NOT NULL,
FacNo VARCHAR(8),
LocName VARCHAR(30) CONSTRAINT LocNameNotNull NOT NULL,
CONSTRAINT PK_LOCATION PRIMARY KEY (LocNo),
CONSTRAINT FK_FACNO FOREIGN KEY (FacNo) REFERENCES FACILITY (FacNo)
);
CREATE TABLE EMPLOYEE
(
EmpNo CHAR(11) CONSTRAINT EmpNoNotNull NOT NULL,
EmpName VARCHAR(30) CONSTRAINT EmpNameNotNull NOT NULL,
Department VARCHAR(30) CONSTRAINT DepartmentNotNull NOT NULL,
Email VARCHAR(255) CONSTRAINT EmailNotNull NOT NULL,
Phone VARCHAR(30) CONSTRAINT PhoneNotNull NOT NULL,
CONSTRAINT PK_EMPLOYEE PRIMARY KEY (EmpNo)
);
CREATE TABLE EVENTPLAN
(
PlanNo VARCHAR(8) NOT NULL,
EventNo VARCHAR(8) NOT NULL,
workdate DATE NOT NULL,
notes VARCHAR(40),
activity VARCHAR(20) NOT NULL,
empno VARCHAR(8),
CONSTRAINT PK_PLANNO PRIMARY KEY (PlanNo),
CONSTRAINT FK_EVENTNO FOREIGN KEY (EventNo) REFERENCES EventRequest (EventNo)
);
CREATE TABLE EVENTREQUEST
(
EventNo VARCHAR(8) NOT NULL,
DateHeld DATE NOT NULL,
DateReq DATE NOT NULL,
FacNo VARCHAR(8) NOT NULL,
CustNo VARCHAR(8) NOT NULL,
DateAuth DATE,
Status VARCHAR(8) NOT NULL CHECK (Status IN ('Pending', 'Denied', 'Approved')),
EstCost DECIMAL(10, 2) NOT NULL,
EstAudience INT NOT NULL CHECK (EstAudience > 0),
BudNo VARCHAR(8),
CONSTRAINT PK_EVENTNO PRIMARY KEY (EventNo),
CONSTRAINT FK_FACILITYNOEVENTREQ FOREIGN KEY (FacNo) REFERENCES Facility (FacNo),
CONSTRAINT FK_CUSTOMERNO FOREIGN KEY (CustNo) REFERENCES Customer (CustNo)
);
CREATE TABLE EVENTPLANLINE
(
PlanNo CHAR(8) NOT NULL,
LineNo CHAR(8) NOT NULL,
LocNo CHAR(8) NOT NULL,
ResNo CHAR(8) NOT NULL,
TimeStart TIMESTAMP NOT NULL,
TimeEnd TIMESTAMP NOT NULL,
NumberFLD INTEGER NOT NULL,
CONSTRAINT PK_EVENTPLANLINE PRIMARY KEY (PlanNo, LineNo),
CONSTRAINT FK_EVENTPLAN FOREIGN KEY (PlanNo) REFERENCES EventPlan (PlanNo),
CONSTRAINT FK_LOCATION FOREIGN KEY (LocNo) REFERENCES Location (LocNo),
CONSTRAINT FK_RESOURCETBL FOREIGN KEY (ResNo) REFERENCES ResourceTbl (ResNo)
);
CREATE TABLE RESOURCETBL
(
ResNo CHAR(8) NOT NULL,
ResName VARCHAR(30) NOT NULL,
Rate DECIMAL(8, 2) NOT NULL,
CONSTRAINT PK_RESOURCETBL PRIMARY KEY (ResNo)
);
In postgresql I have a table which I need to add a new column. the original table ddl is belowing:
CREATE TABLE survey.survey_response (
id uuid NOT NULL DEFAULT uuid_generate_v4(),
survey_id uuid NOT NULL,
survey_question_id uuid NULL,
user_id varchar(256) NULL,
device_id varchar(256) NULL,
user_country varchar(100) NULL,
client_type varchar(100) NULL,
product_version varchar(100) NULL,
answer text NULL,
response_date timestamptz NOT NULL DEFAULT now(),
survey_category varchar(100) NULL,
tags varchar(250) NULL,
tracking_id uuid NULL,
CONSTRAINT survey_response_pkey PRIMARY KEY (id)
)
WITH (
OIDS=FALSE
) ;
Then I alter the table to add a new column:
alter table survey.survey_response add column system_tags varchar(30) ;
But after that I found my instert statement cannot make change to this new column, for all the original columns it works fine:
INSERT INTO survey.survey_response
(id, survey_id, user_id, tags, system_tags)
VALUES(uuid_generate_v4(), uuid_generate_v4(),'1123','dsfsd', 'dsfsd');
select * from survey.survey_response where user_id = '1123';
The "tags" columns contains inserted value, however, system_tags keeps null.
I tested the above scenario in my local postgreSQL 9.6, any ideas about this strange behavior? Thanks a lot
-----------------update----------
I found this survey.survey_response table has been partitioning based on month, So my inserted record will also be displayed in survey.survey_response_y2017m12. but the new system_tags column is also NULL
CREATE TABLE survey.survey_response_y2017m12 (
id uuid NOT NULL DEFAULT uuid_generate_v4(),
survey_id uuid NOT NULL,
survey_question_id uuid NULL,
user_id varchar(256) NULL,
device_id varchar(256) NULL,
user_country varchar(100) NULL,
client_type varchar(100) NULL,
product_version varchar(100) NULL,
answer text NULL,
response_date timestamptz NOT NULL DEFAULT now(),
survey_category varchar(100) NULL,
tags varchar(250) NULL,
tracking_id uuid NULL,
system_tags varchar(30) NULL,
CONSTRAINT survey_response_y2017m12_response_date_check CHECK (((response_date >= '2017-12-01'::date) AND (response_date < '2018-01-01'::date)))
)
INHERITS (survey.survey_response)
WITH (
OIDS=FALSE
) ;
If I run the same scenario in a non-partition table then the insert works fine.
So do I need any special settings for alter table for partition table?
Old thread but you need to drop and create again the RULE to fix the issue.
SELECT M.msg_id, M.uid_fk, M.message, M.created,
U.fname, U.lname, M.uploads
FROM messages M, users_friends F, users U
WHERE M.uid_fk=F.friendID
and F.userID = '5'
and status='2'
Building a facebook-like wall and want to grab messages(updates) from friends.
The query above returns an empty set, even though I've made sure there are messages from user 5 in the table.
Schema:
CREATE TABLE IF NOT EXISTS `messages` (
`msg_id` int(11) NOT NULL AUTO_INCREMENT,
`message` varchar(200) CHARACTER SET utf8 DEFAULT NULL,
`uid_fk` int(11) DEFAULT NULL,
`ip` varchar(30) DEFAULT NULL,
`created` int(11) DEFAULT '1269249260',
`uploads` varchar(30) DEFAULT NULL,
PRIMARY KEY (`msg_id`),
KEY `uid_fk` (`uid_fk`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=263 ;
CREATE TABLE IF NOT EXISTS `users` (
`fname` varchar(15) NOT NULL,
`lname` varchar(15) NOT NULL,
`userID` int(11) NOT NULL AUTO_INCREMENT,
`username` varchar(23) NOT NULL,
`email` varchar(50) NOT NULL,
`password` varchar(32) NOT NULL,
`DOB` date DEFAULT NULL,
`sex` varchar(1) DEFAULT NULL,
`about` text NOT NULL,
`location` varchar(20) DEFAULT NULL,
`last_login` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
`user_level` int(11) NOT NULL DEFAULT '0',
`profile_image` varchar(200) NOT NULL,
`profile_image_small` varchar(200) NOT NULL,
PRIMARY KEY (`userID`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=19 ;
CREATE TABLE IF NOT EXISTS `users_friends` (
`userID` int(11) NOT NULL,
`friendID` int(11) NOT NULL,
`status` int(1) NOT NULL DEFAULT '0',
PRIMARY KEY (`userID`,`friendID`),
KEY `fk_users_has_friends_users1` (`userID`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
You are single quoting your INT values in the WHERE clause → '5' & '2'.
Also, try JOINs.
SELECT M.msg_id,
M.uid_fk,
M.message,
M.created,
U.fname,
U.lname,
M.uploads
FROM messages M
INNER JOIN users_friends F ON F.friendID = M.uid_fk
AND F.userID = 5
AND F.status = 2
INNER JOIN users U ON U.userID = F.friendID;