I created triggers after creating a database in oracle apex. When testing in SQL code everything was fine, they worked. In the application I have the Professors table and when I change the value of the column "name", in the Courses table the value of the column "professor_name" should change. However when I change the value of the column "name" in the Professors table it throws me the error.
Here is the trigger definition:
CREATE OR REPLACE TRIGGER profesori_upd
AFTER UPDATE OF ime on profesori
FOR EACH ROW
BEGIN
UPDATE kursevi set ime_profesora = :new.ime
where pr_jmbg= :new.jmbg_pr;
END;
Here is the table Proffesors:
CREATE TABLE PROFESORI( jmbg_pr VARCHAR2(13) PRIMARY KEY,
ime VARCHAR2(20) NOT NULL,
prezime VARCHAR2(20) NOT NULL,
id_spreme NUMBER(5) NOT NULL,
CONSTRAINT profesor_sprema_fk FOREIGN KEY (id_spreme) REFERENCES
STRUCNE_SPREME(id)
);
Here is the table Courses:
CREATE TABLE KURSEVI(
opis VARCHAR2(40),
id_jezik NUMBER(10) NOT NULL,
id_nivo NUMBER(10) NOT NULL,
pr_jmbg VARCHAR2(13) NOT NULL,
CONSTRAINT kurs_jezik_fk FOREIGN KEY (id_jezik) REFERENCES JEZICI(id_j),
CONSTRAINT kurs_nivo_fk FOREIGN KEY (id_nivo) REFERENCES NIVOI_KURSEVA(id_n),
CONSTRAINT kurs_profesor_fk FOREIGN KEY (pr_jmbg) REFERENCES PROFESORI(jmbg_pr),
CONSTRAINT kurs_pk PRIMARY KEY(id_jezik,id_nivo,pr_jmbg)
);
After denormalization I added columns ime_profesora and prezime_profesora to the table Courses:
ALTER TABLE KURSEVI
ADD ime_profesora VARCHAR2(20);
ALTER TABLE KURSEVI
ADD prezime_profesora VARCHAR2(20);
UPDATE KURSEVI SET ime_profesora=( SELECT ime FROM PROFESORI WHERE profesori.jmbg_pr=kursevi.pr_jmbg);
UPDATE KURSEVI SET prezime_profesora=( SELECT prezime FROM PROFESORI WHERE profesori.jmbg_pr=kursevi.pr_jmbg);
ALTER TABLE KURSEVI
MODIFY ime_profesora varchar2(20) not null;
ALTER TABLE KURSEVI
MODIFY prezime_profesora varchar2(20) not null;
Error occurs after I try to change the value of the column name in the application, it does not say what the error is.
Related
I've created a MySQL Model with a few tables, some of them with fk's to another table. I usually export the SQL from MySQL Model to my database using the "Forward Engineer SQL CREATE Script" inside File -> Export -> Forward Engineer SQL CREATE Script. The problem here is that when I generate the creation script, all my fk's become unique. I didn't check UQ option in MySQL Model but it creates a script with unique fk's anyway, so, I need to change the SQL file generated and remove all the unwanted uniques. Anyone has a clue why this is happening?
Generated script:
CREATE TABLE IF NOT EXISTS `u514786799_detranleiloes`.`Lotes` (
`createdAt` DATE NOT NULL,
`updatedAt` DATE NOT NULL,
`id` INT UNIQUE NOT NULL AUTO_INCREMENT,
`LeiloesId` INT UNIQUE NOT NULL,
`conservado` TINYINT NULL,
`numero` INT NOT NULL,
`CRDsId` INT UNIQUE NULL,
PRIMARY KEY (`id`),
INDEX `fk_Lotes_Leiloes_idx` (`LeiloesId` ASC),
INDEX `fk_Lotes_CRDs1_idx` (`CRDsId` ASC),
CONSTRAINT `fk_Lotes_Leiloes`
FOREIGN KEY (`LeiloesId`)
REFERENCES `u514786799_detranleiloes`.`Leiloes` (`id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION,
CONSTRAINT `fk_Lotes_CRDs1`
FOREIGN KEY (`CRDsId`)
REFERENCES `u514786799_detranleiloes`.`CRDs` (`id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB
AUTO_INCREMENT = 1;
If a Cloud Spanner table is created with nullable columns, is it possible to add a NOT NULL constraint on a column without recreating the table?
You can add a NOT NULL constraint to a non-key column. You must first ensure that all rows actually do have values for the column. Spanner will scan the data to verify before fully applying the NOT NULL constraint. More information about how to alter tables is here and here.
However, you can not add such a constraint to a key column. That kind of change would require rewriting all the data in the table, because the nullness of the key affects how the data is encoded. The only option for making that change is to create a new table that's set up the way you want, make code changes to support using both tables temporarily, gradually move the data from the old table to the new table, and eventually change the code to use only the new table and drop the old table. If you further then wanted the original table name, you'd have to do the whole thing again.
Unfortenately there is not way to add not null column
The way to do it:
1 add nullable column
ALTER TABLE table1 ADD COLUMN column1 STRING(255)
UPDATE table1.column1, SET NOT NULL VALUE to the column (if the table is not empty).
UPDATE TABLE table1 SET column1 = "<GENERATED DATA>"
Add constraint
ALTER TABLE table1 ADD COLUMN column1 STRING(255) NOT NULL
Thanks.
Creating a non-nullable column in Spanner on an existing table is typically a three step process:
# add new column to table
ALTER TABLE <table_name> ADD COLUMN <column_name> <value_type>;
# create default values
UPDATE <table_name> SET <column_name>=<default_value> WHERE TRUE;
# add constraint
ALTER TABLE <table_name> ALTER COLUMN <column_name> <value_type> NOT NULL;
I am using this link.
I have connected my cpp file with Eclipse to my Database with 3 tables (two simple tables
Person and Item
and a third one PersonItem that connects them). In the third table I use one simple primary and then two foreign keys like that:
CREATE TABLE PersonsItems(PersonsItemsId int not null auto_increment primary key,
Person_Id int not null,
Item_id int not null,
constraint fk_Person_id foreign key (Person_Id) references Person(PersonId),
constraint fk_Item_id foreign key (Item_id) references Items(ItemId));
So, then with embedded sql in c I want a Person to have multiple items.
My code:
mysql_query(connection, \
"INSERT INTO PersonsItems(PersonsItemsId, Person_Id, Item_id) VALUES (1,1,5), (1,1,8);");
printf("%ld PersonsItems Row(s) Updated!\n", (long) mysql_affected_rows(connection));
//SELECT newly inserted record.
mysql_query(connection, \
"SELECT Order_id FROM PersonsItems");
//Resource struct with rows of returned data.
resource = mysql_use_result(connection);
// Fetch multiple results
while((result = mysql_fetch_row(resource))) {
printf("%s %s\n",result[0], result[1]);
}
My result is
-1 PersonsItems Row(s) Updated!
5
but with VALUES (1,1,5), (1,1,8);
I would like that to be
-1 PersonsItems Row(s) Updated!
5 8
Can somone tell me why is this not happening?
Kind regards.
I suspect this is because your first insert is failing with the following error:
Duplicate entry '1' for key 'PRIMARY'
Because you are trying to insert 1 twice into the PersonsItemsId which is the primary key so has to be unique (it is also auto_increment so there is no need to specify a value at all);
This is why rows affected is -1, and why in this line:
printf("%s %s\n",result[0], result[1]);
you are only seeing 5 because the first statement failed after the values (1,1,5) had already been inserted, so there is still one row of data in the table.
I think to get the behaviour you are expecting you need to use the ON DUPLICATE KEY UPDATE syntax:
INSERT INTO PersonsItems(PersonsItemsId, Person_Id, order_id)
VALUES (1,1,5), (1,1,8)
ON DUPLICATE KEY UPDATE Person_id = VALUES(person_Id), Order_ID = VALUES(Order_ID);
Example on SQL Fiddle
Or do not specify the value for personsItemsID and let auto_increment do its thing:
INSERT INTO PersonsItems( Person_Id, order_id)
VALUES (1,5), (1,8);
Example on SQL Fiddle
I think you have a typo or mistake in your two queries.
You are inserting "PersonsItemsId, Person_Id, Item_id"
INSERT INTO PersonsItems(PersonsItemsId, Person_Id, Item_id) VALUES (1,1,5), (1,1,8)
and then your select statement selects "Order_id".
SELECT Order_id FROM PersonsItems
In order to achieve 5, 8 as you request, your second query needs to be:
SELECT Item_id FROM PersonsItems
Edit to add:
Your primary key is autoincrement so you don't need to pass it to your insert statement (in fact it will error as you pass 1 twice).
You only need to insert your other columns:
INSERT INTO PersonsItems(Person_Id, Item_id) VALUES (1,5), (1,8)
I've been trying to add a foreign key to my table using heidisql and I keep getting the error 1452.
After reading around I made sure all my tables were running on InnoDB as well as checking that they had the same datatype and the only way I can add my key is if I drop all my data which I don't intend to do since I have spent quite a few hours on this.
here is my table create code:
CREATE TABLE `data` (
`ID` INT(10) NOT NULL AUTO_INCREMENT,
#bunch of random other columns stripped out
`Ability_1` SMALLINT(5) UNSIGNED NOT NULL DEFAULT '0',
#more stripped tables
`Extra_Info` SET('1','2','3','Final','Legendary') NOT NULL DEFAULT '1' COLLATE 'utf8_unicode_ci',
PRIMARY KEY (`ID`),
UNIQUE INDEX `ID` (`ID`)
)
COLLATE='utf8_unicode_ci'
ENGINE=InnoDB
AUTO_INCREMENT=650;
here is table 2
CREATE TABLE `ability` (
`ability_ID` SMALLINT(5) UNSIGNED NOT NULL AUTO_INCREMENT,
#stripped columns
`Name_English` VARCHAR(12) NOT NULL COLLATE 'utf8_unicode_ci',
PRIMARY KEY (`ability_ID`),
UNIQUE INDEX `ability_ID` (`ability_ID`)
)
COLLATE='utf8_unicode_ci'
ENGINE=InnoDB
AUTO_INCREMENT=165;
Finally here is the create code along with the error.
ALTER TABLE `data`
ADD CONSTRAINT `Ability_1` FOREIGN KEY (`Ability_1`) REFERENCES `ability` (`ability_ID`) ON UPDATE CASCADE ON DELETE CASCADE;
/* SQL Error (1452): Cannot add or update a child row: a foreign key constraint fails (`check`.`#sql-ec0_2`, CONSTRAINT `Ability_1` FOREIGN KEY (`Ability_1`) REFERENCES `ability` (`ability_ID`) ON DELETE CASCADE ON UPDATE CASCADE) */
If there is anything else I can provide please let me know this is really bothering me. I'm also using 5.5.27 - MySQL Community Server (GPL) that came with xampp installer.
If you are using HeidiSQL it is pretty easy.
Just see the image, click on the +Add to add foreign keys.
I prefer GUI way of creating tables and its attribute because it saves time and reduces errors.
I found it. Sorry everyone. The problem was that I had 0 as a default value for my fields while my original table had no value for 0.
Here is how you can do it ;
Create your Primary keys. For me this was straight forward so I won't post how to do that here
To create your FOREIGN KEYS you need to change the table / engine type for each table from MyIASM to InnoDb. To do this Select the table on the right hand side then select the OPTIONS tab on the right hand side and change the engine from MyIASM to InnoDb for every table.
Really hate to use other people's time, but it seems the problem is just not going away.
I considered all recommendations at http://verysimple.com/2006/10/22/mysql-error-number-1005-cant-create-table-mydbsql-328_45frm-errno-150/ and at http://forums.mysql.com/read.php?22,19755,19755#msg-19755 but nothing.
hope that someone points to a stupid mistake.
here are the tables:
CREATE TABLE IF NOT EXISTS `shop`.`category` (
`id` INT(11) NOT NULL AUTO_INCREMENT ,
`category_id` INT(11) NOT NULL ,
`parent_id` INT(11) NULL DEFAULT '0' ,
`lang_id` INT(11) NOT NULL ,
...other columns...
PRIMARY KEY (`id`, `category_id`) )
ENGINE = InnoDB
DEFAULT CHARACTER SET = utf8
COLLATE = utf8_unicode_ci;
CREATE TABLE IF NOT EXISTS `shop`.`product_category` (
`category_id` INT(11) NOT NULL ,
`product_id` INT(11) NOT NULL ,
INDEX `fk_product_category_category1_zxc` (`category_id` ASC) ,
CONSTRAINT `fk_product_category_category1_zxc`
FOREIGN KEY (`category_id` )
REFERENCES `shop`.`category` (`category_id` )
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB
DEFAULT CHARACTER SET = utf8
COLLATE = utf8_unicode_ci;
Error Code: 1005. Can't create table 'shop.product_category' (errno: 150)
You need an index on category_id in the category table (I see it's part of the primary key, but since it's the second column in the index, it can not be used). The field you are referencing in a foreign key always should be indexed.
In my case the issue was more like what was described in the first article you've linked to.
So I just had to make sure that:
Referenced Column is an index,
both Referencing Column and Referenced Column share the same type and length, i.e. e.g. both are INT(10),
both share the same not null, unsigned, zerofill etc. configuration.
both tables are InnoDB!
Here's the query template where Referencing Column is referencing_id and Referenced Column is referenced_id:
ALTER TABLE `db`.`referencing`
ADD CONSTRAINT `my_fk_idx`
FOREIGN KEY (`referencing_id`)
REFERENCES `db`.`referenced`(`referenced_id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION;
Update 2016-03-13: Ran into this problem again, ended up finding my own answer. This time it didn't help though. Turns out the other table was still set to MyISAM, as soon as I changed it to InnoDB everything worked.