I'm trying to add IDs from a inserted column to another inserted column in a Transaction. I tried like this:
begin
let doorOne = INSERT INTO doors SET color = green
let doorTwo = INSERT INTO doors SET color = blue
let car = INSERT INTO Cars SET doors = [$doorOne , $doorTwo]
commit retry 100
return $car
I get:
Unhandled rejection OrientDB.RequestError: The field 'Cars.doors' has been declared as LINKSET but the value is not a record or a record-id
I also tried to update it afterwards in the Transaction, but this won't work either (i think because the car is not created yet, so nothing to update) and i dont want to do it in two different calls, if there is a way to do it in one Transaction.
This is because INSERT returns, by default, the number of inserted entries. Try adding this clause at the end of each INSERT statement: RETURN #rid
Related
I have a trigger that basically get the last number inserted (folio) based on two columns (service_type and prefix) and then increment one value. Everything works fine but in some situations when there a lot of insert statements at the same time it causes that the function use the same last value inserted and it duplicates the folio column
CREATE OR REPLACE FUNCTION public.insert_folio()
RETURNS trigger
LANGUAGE plpgsql
AS $function$
DECLARE incremental INTEGER;
SELECT max(f.folio) INTO incremental FROM "folios" f
WHERE f.prefix = NEW."prefix" AND service_type = NEW."service_type";
NEW."folio" = incremental + 1;
IF NEW."folio" IS NULL THEN NEW."folio" = 1;
END IF;
RETURN NEW;
END;
$function$
CREATE TRIGGER insert_folio BEFORE INSERT ON "folios" FOR EACH ROW EXECUTE PROCEDURE insert_folio();
sample:
folio
service_type
prefix
1
DOCUMENT
DC
1
IMAGE
IMG
1
IMAGE
O
2
IMAGE
O
2 (This should be 3)
IMAGE
O
Any ideas?
Thanks!
It is because you have concurrent transactions that both see the same data. The second transaction does not see the record inserted by the first transaction until the first commits.
To prevent this behaviour you will have to lock the whole table when inserting to prevent concurrent writes or use advisory locks to prevent concurrent insertions of the same service_type and prefix.
I am building a personal comic book database and am having an issue with one of my SQL Server triggers.
My main comic entry (tabbed) form has a combo box for cover prices.
When the user clicks submit and inserts a comic into the database (comic_books table) on the comics entry page, I have a trigger that adds the cover price the user entered to a separate table (comic_prices table.) if the entry does not exist. This is working just fine.
However, I have a second tab ('Edit Comic') where the user can update an already inserted comic which uses a simple update script.
The user is able to change or add a new cover price of said comic from this tab also.
The issue I am having is that when the user clicks the 'Update Comic' button from the 'Edit Comic' tab, this newly entered comic price is not being inserted in the comic_prices table if it does not exist. So it looks like my trigger is only firing on my insert script and not on my update script.
Again, I have the trigger to only insert if the entry does not exist in the cover_prices table, otherwise it does nothing.
Please see my 'AFTER UPDATE, INSERT' trigger for this below and please let me know if you need any more information!
I appreciate any pointers or critiques!
DECLARE #cover_price varchar(50)
select #cover_price = cover_price from comic_books
If exists (SELECT cover_price FROM comic_prices where cover_price = #cover_price )
Begin
Return
End
IF not EXISTS (SELECT cover_price FROM comic_prices where cover_price = #cover_price)
INSERT INTO comic_prices(cover_price)VALUES(#cover_price)
UPDATE 04/22/17
After two or so weeks, I finally figured it out! I ended up making two separate triggers, one for the insert and another for the update. I kept the insert the same, however, the update looks like this:
DECLARE #publisher_name varchar(50)
--declare #comic_id bigint
select #publisher_name = inserted.issue_publisher from inserted
select #comic_id = comic_id from comic_books
If exists (select publisher_name from comic_publishers where publisher_name = #publisher_name)
Begin
return
End
IF not EXISTS (select publisher_name from comic_publishers where publisher_name = #publisher_name)
INSERT INTO comic_publishers (publisher_name)VALUES(#publisher_name)
I needed to tell the trigger to look for the inserted publisher!
I want to update OrigOrderNbr and OrigOrderType (QT type) because when I create first both of column are Null value. But after S2 was created (QT converted to S2) the OrigOrderType and OrigOrderNbr (S2) take from QT reference. Instead of that, I want to update it to QT also.
http://i.stack.imgur.com/6ipFa.png
http://i.stack.imgur.com/E6qzT.png
CREATE TRIGGER tgg_SOOrder
ON dbo.SOOrder
FOR INSERT
AS
DECLARE #tOrigOrderType char(2),
#tOrigOrderNbr nvarchar(15)
SELECT #tOrigOrderType = i.OrderType,
#tOrigOrderNbr = i.OrderNbr
FROM inserted i
UPDATE dbo.SOOrder
SET OrigOrderType = #tOrigOrderType,
OrigOrderNbr = #tOrigOrderNbr
FROM inserted i
WHERE dbo.SOOrder.CompanyID='2'
and dbo.SOOrder.OrderType=i.OrigOrderType
and dbo.SOOrder.OrderNbr=i.OrigOrderNbr
GO
After I run that trigger, it showed the message 'Error #91: Another process has updated 'SOOrder' record. Your changes will be lost.'.
Per long string of comments, including some excellent suggestions in regards to proper trigger writing techniques by #marc_s and #Damien_The_Unbeliever, as well as my better understanding of your issue at this point, here's the re-worked trigger:
CREATE TRIGGER tgg_SOOrder
ON dbo.SOOrder
FOR INSERT
AS
--Update QT record with S2 record's order info
UPDATE SOOrder
SET OrigOrderType = 'S2'
, OrigOrderNbr = i.OrderNbr
FROM SOOrder dest
JOIN inserted i
ON dest.OrderNbr = i.OrigOrderNbr
WHERE dest.OrderType = 'QT'
AND i.OrderType = 'S2'
AND dest.CompanyID = 2 --Business logic constraint
AND dest.OrigOrderNbr IS NULL
AND dest.OrigOrderType IS NULL
Basically, the idea is to update any record of type "QT" once a matching record of type "S2" is created. Matching here means that OrigOrderNbr of S2 record is the same as OrderNbr of QT record. I kept your business logic constraint in regards to CompanyID being set to 2. Additionally, we only care to modify QT records that have OrigOrderNbr and OrigOrderType set to NULL.
This trigger does not rely on a single-row insert; it will work regardless of the number of rows inserted - which is far less likely to break down the line.
I have to implement a trigger which would:
7) Show the DDL for how a trigger can be used to transfer all rental copies from a store being deleted from the Store information table to the central store
8) Show how this trigger can be extended to make
sure that the central store is never deleted from the database
So far I have done this:
CREATE OR REPLACE TRIGGER stores BEFORE DELETE ON stores FOR
EACH ROW BEGIN IF DELETING WHERE cvr = 123456789 THEN
Raise_Application_Error ( num => -20050, msg => 'You can
not delete Main Store.'); END IF; IF DELETING THEN
UPDATE store_id=123456789 ON movies WHERE isActive = 0 END
IF; END;
so main store is with cvr which is written, but it gives me a compilation error. Any help? Thanks in advance.
You have two errors in your code.
there is no "DELETING WHERE" expression, you have to use two boolean exceptions like this:
IF DELETING AND :old.cvr = 123456789 THEN...
:old.cvr refers to value of cvr column in deleted record
Syntax of UPDATE clause is
UPDATE table_name
SET column_name1 = value1,
column_name1 = value2
WHERE where_clause;
in your case probably somethink like this:
UPDATE movies
set store_id = 123456789
WHERE store_id = :old.cvr
I guess, I don't know required functionality
I m trying to write if statement to give error message if user try to add existing ID number.When i try to enter existing id i get error .untill here it s ok but when i type another id no and fill the fields(name,adress etc) it doesnt go to database.
METHOD add_employee.
DATA: IT_EMP TYPE TABLE OF ZEMPLOYEE_20.
DATA:WA_EMP TYPE ZEMPLOYEE_20.
Data: l_count type i value '2'.
SELECT * FROM ZEMPLOYEE_20 INTO TABLE IT_EMP.
LOOP AT IT_EMP INTO WA_EMP.
IF wa_emp-EMPLOYEE_ID eq pa_id.
l_count = l_count * '0'.
else.
l_count = l_count * '1'.
endif.
endloop.
If l_count eq '2'.
WA_EMP-EMPLOYEE_ID = C_ID.
WA_EMP-EMPLOYEE_NAME = C_NAME.
WA_EMP-EMPLOYEE_ADDRESS = C_ADD.
WA_EMP-EMPLOYEE_SALARY = C_SAL.
WA_EMP-EMPLOYEE_TYPE = C_TYPE.
APPEND wa_emp TO it_emp.
INSERT ZEMPLOYEE_20 FROM TABLE it_emp.
CALL FUNCTION 'POPUP_TO_DISPLAY_TEXT'
EXPORTING
TITEL = 'INFO'
TEXTLINE1 = 'Record Added Successfully.'.
elseif l_count eq '0'.
CALL FUNCTION 'POPUP_TO_DISPLAY_TEXT'
EXPORTING
TITEL = 'INFO'
TEXTLINE1 = 'Selected ID already in database.Please type another ID no.'.
ENDIF.
ENDMETHOD.
I'm not sure I'm getting your explanation. Why are you trying to re-insert all the existing entries back into the table? You're just trying to insert C_ID etc if it doesn't exist yet? Why do you need all the existing entries for that?
If so, throw out that select and the loop completely, you don't need it. You have a few options...
Just read the table with your single entry
SELECT SINGLE * FROM ztable INTO wa WITH KEY ID = C_ID etc.
IF SY-SUBRC = 0.
"this entry exists. popup!
ENDIF.
Use a modify statement
This will overwrite duplicate entries with new data (so non key fields may change this way), it wont fail. No need for a popup.
MODIFY ztable FROM wa.
Catch the SQL exceptions instead of making it dump
If the update fails because of an exception, you can always catch it and deal with exceptional situations.
TRY .
INSERT ztable FROM wa.
CATCH sapsql_array_insert_duprec.
"do your popup, the update failed because of duplicate records
ENDTRY.
I think there's a bug when appending in internal table 'IT_EMP' and inserting in 'ZEMPLOYEE_20' table.
Suppose you append the first time and then you insert. But when you append the second time you will have 2 records in 'IT_EMP' that are going to be inserted in 'ZEMPLOYEE_20'. That is because you don't refresh or clear the internal table and there you will have a runtime error.
According to SAP documentation on 'Inserting Lines into Tables ':
Inserting Several Lines
To insert several lines into a database table, use the following:
INSERT FROM TABLE [ACCEPTING DUPLICATE KEYS] . This
writes all lines of the internal table to the database table in
one single operation. The same rules apply to the line type of
as to the work area described above. If the system is able to
insert all of the lines from the internal table, SY-SUBRC is set to 0.
If one or more lines cannot be inserted because the database already
contains a line with the same primary key, a runtime error occurs.
Maybe the right direction here is trying to insert the work area directly but before you must check if record already exists using the primary key.
Check the SAP documentation on this issue clicking the link before.
On the other hand, once l_count is zero because of l_count = l_count * '0'. that value will never change to any other number making that you won't append or insert again.
why are you retrieving all entries from zemployee_20 ?
You can directly check wether the 'id' exists already or not by using select single. If exists, then show message, if not, add.
It is recommended to retrieve only one field when its needed and not the entire table with asterisc *
SELECT single employee_id FROM ZEMPLOYEE_20 where employee_id = p_id INTO v_id. ( or field in structure )
if sy-subrc = 0. "exists
"show message
else. "not existing id
"populate structure and then add record to Z table
endif.
Furthermore, l_count is not only unnecessary but also bad implemented.
You can directly use the insert query,if the sy-subrc is unsuccessful raise the error message.
WA_EMP-EMPLOYEE_ID = C_ID.
WA_EMP-EMPLOYEE_NAME = C_NAME.
WA_EMP-EMPLOYEE_ADDRESS = C_ADD.
WA_EMP-EMPLOYEE_SALARY = C_SAL.
WA_EMP-EMPLOYEE_TYPE = C_TYPE.
INSERT ZEMPLOYEE_20 FROM WA_EMP.
If sy-subrc <> 0.
Raise the Exception.
Endif.