Create Foreign Keys between two Schemata

Hello,
I use Oracle 10g and I trying to create a ForeignKey constraint between two tables in different schemata.
This is my DDL-Script
SQL> CREATE TABLE PROJECT.LOCATION (
ID INTEGER NOT NULL,
MAIN INTEGER,
KURZ VARCHAR(40),
NAME VARCHAR(40),
STRASSE VARCHAR(40),
ORT VARCHAR(40),
TELEFON VARCHAR(40),
FAX VARCHAR(40),
EMAIL VARCHAR(40),
PLZ VARCHAR(40),
CONSTRAINT PK_LOCATION PRIMARY KEY (ID)
Tablespace PROJECT;
CREATE TABLE Diary.Diary (
ID INTEGER NOT NULL,
LOCATION_ID INTEGER NOT NULL,
CONSTRAINT PK_Diary PRIMARY KEY (ID)
Tablespace Diary;
ALTER TABLE Diary.Diary
ADD CONSTRAINT FK_Diary_Has_LOCATION
FOREIGN KEY (LOCATION_ID) REFERENCES PROJECT.LOCATION (ID);
This is the Message that gives my SQLplus :
SQL> ALTER TABLE Diary.Diary
2 ADD CONSTRAINT FK_Diary_Has_LOCATION
3 FOREIGN KEY (LOCATION_ID) REFERENCES PROJECT.LOCATION (ID);
FOREIGN KEY (LOCATION_ID) REFERENCES PROJECT.LOCATION (ID)
FEHLER in Zeile 3:
ORA-00942: Tabelle oder View nicht vorhanden
All Grants (select,alter,references) are given to the User for the tables.
Whats then Problem?

You miss some priviledge:
SQL> create user project identified by project;
Utente creato.
SQL> grant connect, resource to project;
Concessione riuscita.
SQL> create user diary identified by diary;
Utente creato.
SQL> grant connect, resource to diary;
Concessione riuscita.
SQL> CREATE TABLE PROJECT.LOCATION (
  2  ID INTEGER NOT NULL,
  3  MAIN INTEGER,
  4  KURZ VARCHAR(40),
  5  NAME VARCHAR(40),
  6  STRASSE VARCHAR(40),
  7  ORT VARCHAR(40),
  8  TELEFON VARCHAR(40),
  9  FAX VARCHAR(40),
10  EMAIL VARCHAR(40),
11  PLZ VARCHAR(40),
12  CONSTRAINT PK_LOCATION PRIMARY KEY (ID)
13  )
14  ;
Tabella creata.
SQL> CREATE TABLE Diary.Diary (
  2  ID INTEGER NOT NULL,
  3  LOCATION_ID INTEGER NOT NULL,
  4  CONSTRAINT PK_Diary PRIMARY KEY (ID)
  5  )
  6  ;
Tabella creata.
SQL> ALTER TABLE Diary.Diary
  2  ADD CONSTRAINT FK_Diary_Has_LOCATION
  3  FOREIGN KEY (LOCATION_ID) REFERENCES PROJECT.LOCATION (ID);
FOREIGN KEY (LOCATION_ID) REFERENCES PROJECT.LOCATION (ID)
ERRORE alla riga 3:
ORA-00942: tabella o vista inesistente
-- DIARY CAN'T SEE PROJECT.LOCATION
SQL> grant select on project.location to diary;
Concessione riuscita.
SQL> ALTER TABLE Diary.Diary
  2  ADD CONSTRAINT FK_Diary_Has_LOCATION
  3  FOREIGN KEY (LOCATION_ID) REFERENCES PROJECT.LOCATION (ID);
FOREIGN KEY (LOCATION_ID) REFERENCES PROJECT.LOCATION (ID)
ERRORE alla riga 3:
ORA-01031: privilegi insufficienti
-- DIARY CAN SEE PROJECT.LOCATION BUT CAN'T REFERENCE IT
SQL> grant references on project.location to diary;
Concessione riuscita.
SQL> ALTER TABLE Diary.Diary
  2  ADD CONSTRAINT FK_Diary_Has_LOCATION
  3  FOREIGN KEY (LOCATION_ID) REFERENCES PROJECT.LOCATION (ID);
Tabella modificata.
-- NOW IT'S ALL OK!!Max
[My Italian Oracle blog|http://oracleitalia.wordpress.com/2009/12/18/table-elimination-oppure-join-elimination-lottimizzatore-si-libera-della-zavorra/]

Similar Messages

  • SSAS 2008 Linking two cubes on the foreign key between two fact tables

    Hi, all -- 
    I have two cubes:
    Cube 1 has Fact1 (F1, "Events") and 3 dimensions (D1, D2, D3)
    Cube 2 has Fact2 (F2, "Sales") and 3 dimensions (D4, D1, D6)
    As you can see, two cubes reuse D1 as their common dimension.  In addition, F2 foreign keys into F1, e.g. F1 is "Events", and F2 is "Sales".  Every "sale" is an "event", but not every "event" is
    a "sale".
    The question is, how to I link the two cubes and their two respective fact tables on the foreign key?
    Thanks, Austin.

    Hi Austin,
    According to your description, you want to retrieve data from two different cubes, right? In Analysis Services, to query multiple cubes from a single MDX statement you can use the LOOKUPCUBE function (you can't specify multiple cubes in your FROM statement).
    The LOOKUPCUBE function will only work on cubes that utilize the same source database as the cube on which the MDX statement is running. For the detail information about it, please refer to the link below to see the blog.
    Retrieving Data From Multiple Cubes in an MDX Query Using the Lookupcube Function
    If I have anything misunderstood, please point it out.
    Regards,
    Charlie Liao
    TechNet Community Support

  • Confusion On Creating Foreign Key

    I'm confused about creating a foreign key in my table since two tables in the schema will hold identical info:
    EMP_ID
    EMP_NAME
    EMP_MANAGER
    DEPT_ID
    DEPT_NAME
    DEPT_MANAGER
    Now my 1st confusion is if I'm creating a foreign key, do I point EMP_MANAGER to DEPT_ID or DEPT_MANAGER? How can I configure the EMPLOYEES.EMP_MANAGER to pull the values from DEPARTMENT.DEPT_MANAGER? Would I need to point it to DEPT_ID for any reason or would it be a direct link between the two _MANAGER columns?                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                   

    >
    the two column names have to be exact
    >
    No! The data in the columns needs to be exact. It doesn't matter what the column names are.
    The column names you provided appear to show that YOUR emp table is denormalized. It has an EMP_MANAGER column which presumably contains the NAME of the manager, rather than the DEPT_ID of the manager. If it actually contains the NAME and the DEPT_MANAGER column of the other table has names you could create a foreign key between them.
    But that means you are duplicating the manager name in both tables. If the EMP_MANAGER column actually contains the DEPT_ID of the employees manager then create a foreign key between EMP_MANAGER and DEPT_ID.

  • How to create foreign key automatically?

    I am writing to seek help, in regards creating foreign key automatically, when I insert data into my ''price'' table.  I have 2 tables, one called prices and names.  the relationship between them, if that one name can have many prices and one price
    can have many names (many-to-many). Hence, i have junction table called  "Name_Prices", as shown below in the sample database schema:
    Names
    name_id [pk]
    name
    type
    UploadDate
    Prices
    Price_id [pk]
    name_id [fk]
    price
    uploadDate
    Name_Prices
    name_id REFERENCE names (name_id)
    price_id REFERENCE prices (price_id)
    PRIMARY KEY (name_id, price_id)
    The  price's data input comes in as CSV file everyday. (please see the example below) :
    name name_type price UploadDate
    ALBA MBS 93.5 17/10/2014
    ALESC Trup 58 17/10/2014
    ALESC Trup 52 17/10/2014
    My desire goal/output is to be able to create a functionality, where I can insert the price's data into the database (''prices''), it will automatically insert foreign key in the price
    table (from the names table), and if there is a new price's name, then the database will create a new name id for it, in the name's table, transferring the name, its type, from the CSV input data. 
    In order to achieve this task, where would I start implementing this logic?(in SQL server or application-side) what steps does this involve and is this task achievable, all in sql server side (i.e. store procedure, functions etc..). 
    Apology in advance, if the question is not clear to understand, i happy to follow up with further questions, if required.  
    Any help would be very appreciated. Many thanks

    As noted above, I modified the design:
    Products
    Product_id [pk]
    Product
    type
    UploadDate
    Prices
    Price_id [pk]
    price
    uploadDate
    Product_Prices
    Product_id REFERENCE Products (Product_id)
    price_id REFERENCE prices (price_id)
    PRIMARY KEY (Product_id, price_id)
    Note
    Kalman Toth Database & OLAP Architect
    SQL Server 2014 Database Design
    New Book / Kindle: Beginner Database Design & SQL Programming Using Microsoft SQL Server 2014

  • Create view link between two view objects (from programmatic data source)

    Hi Experts,
    Can we create a link between two view objects (they are created from programmatic datasource ; not from either entity or sql query). If yes how to create the link; ( i mean the like attributes?)
    I would also like to drag and drop that in my page so that i can see as top master form and the below child table. Assume in my program i will be only have one master object and many child objects.
    Any hits or idea pls.
    -t

    Easiest way to do this is to add additional transient attributes to your master view object, and then include those additional transient attributes in the list of source attributes for your view link. This way, you can get BC4J to automatically refer to their values with no additional code on your part.

  • Unable to create foreign key: InvalidArgument=Value of '0' is not valid for 'index'. Parameter name: index

    I am running an SQL(CE) script to create a DB. All script commands succeed, but the DB get "broken" after creating the last costaint: after running the script, viewing table properties of Table2 and clicking on "Manage relations" gives the following error: Unable to create foreign key: InvalidArgument=Value of '0' is not valid for 'index'. Parameter name: index. Wondering what does that refer to...
    Here it is the script. Please note that no error is thrown by running the following queries (even from code that passing the queries by hand, one-by-one to sql server management studio).
    CREATE TABLE [table1] (
    [id_rubrica] numeric(18,0) NOT NULL
    , [id_campo] numeric(18,0) NOT NULL
    , [nome] nvarchar(100) NOT NULL
    GO
    ALTER TABLE [table1] ADD PRIMARY KEY ([id_rubrica],[id_campo]);
    GO
    CREATE UNIQUE INDEX [UQ__m_campi] ON [table1] ([id_campo] Asc);
    GO
    CREATE TABLE [table2] (
    [id_campo] numeric(18,0) NOT NULL
    , [valore] nvarchar(4000) NOT NULL
    GO
    ALTER TABLE [table2] ADD PRIMARY KEY ([id_campo],[valore]);
    GO
    ALTER TABLE [table2] ADD CONSTRAINT [campo_valoriFissi] FOREIGN KEY ([id_campo]) REFERENCES [table1]([id_campo]);
    GO
    Sid (MCP - http://www.sugata.eu)

    I know this is kind of old post, but did this realy solved your problem?
    I'm getting this same error message after adding a FK constraint via UI on VS2008 Server Explorer.
    I can add the constraint with no errors, but the constraint is not created on the DataSet wizard (strongly typed datasets on Win CE 6) and when I click "Manage Relations" on the "Table Properties" this error pop out:
    "InvalidArgument=Value or '0' is not valid for 'index'.
    Parameter name: index"
    Even after vreating my table with the relation in SQL the same occurs:
    CREATE TABLE pedidosRastreios (
        idPedidoRastreio INT NOT NULL IDENTITY PRIMARY KEY,
        idPedido INT NOT NULL CONSTRAINT FK_pedidosRastreios_pedidos REFERENCES pedidos(idPedido) ON DELETE CASCADE,
        codigo NVARCHAR(20) NOT NULL

  • Is it possible to create foreign key from composite key in other table.

    SQL> desc PRODUCT_CONFIG_OPTION;
    Name Null? Type
    CONFIG_ITEM_ID NOT NULL VARCHAR2(20) --composite primary key
    CONFIG_OPTION_ID NOT NULL VARCHAR2(20) --composite primary key
    CONFIG_OPTION_NAME VARCHAR2(100)
    DESCRIPTION VARCHAR2(255)
    SEQUENCE_NUM NUMBER(18)
    LAST_UPDATED_STAMP TIMESTAMP(6)
    LAST_UPDATED_TX_STAMP TIMESTAMP(6)
    CREATED_STAMP TIMESTAMP(6)
    CREATED_TX_STAMP TIMESTAMP(6)
    SQL> DESC PRODUCT_CONFIG_ITEM;
    Name Null? Type
    CONFIG_ITEM_ID NOT NULL VARCHAR2(20)
    CONFIG_ITEM_TYPE_ID VARCHAR2(20)
    CONFIG_ITEM_NAME VARCHAR2(100)
    DESCRIPTION VARCHAR2(255)
    LONG_DESCRIPTION CLOB
    IMAGE_URL VARCHAR2(255)
    LAST_UPDATED_STAMP TIMESTAMP(6)
    LAST_UPDATED_TX_STAMP TIMESTAMP(6)
    CREATED_STAMP TIMESTAMP(6)
    CREATED_TX_STAMP TIMESTAMP(6)
    SQL> desc product;
    Name Null? Type
    PRODUCT_ID NOT NULL VARCHAR2(20)
    PRODUCT_TYPE_ID VARCHAR2(20)
    PRIMARY_PRODUCT_CATEGORY_ID VARCHAR2(20)
    MANUFACTURER_PARTY_ID VARCHAR2(20)
    FACILITY_ID VARCHAR2(20)
    INTRODUCTION_DATE TIMESTAMP(6)
    SUPPORT_DISCONTINUATION_DATE TIMESTAMP(6)
    SALES_DISCONTINUATION_DATE TIMESTAMP(6)
    SALES_DISC_WHEN_NOT_AVAIL CHAR(1)
    INTERNAL_NAME VARCHAR2(255)
    BRAND_NAME VARCHAR2(100)
    COMMENTS VARCHAR2(255)
    =========
    CREATE TABLE PROD_CONFIG_PROD_CONFIG_OPTION (
    PRODUCT_ID VARCHAR2(20),
    CONFIG_ITEM_ID VARCHAR2(20),
    CONFIG_OPTION_ID VARCHAR2(20),
    PAGE_NUM_TO NUMBER(18),
    ALTERNATE_PAGE_NUM_TO1 NUMBER(18),
    ALTERNATE_PAGE_NUM_TO2 NUMBER(18),
    ALTERNATE_PAGE_NUM_TO3 NUMBER(18),
    LAST_UPDATED_STAMP TIMESTAMP(6),
    LAST_UPDATED_TX_STAMP TIMESTAMP(6),
    CREATED_STAMP TIMESTAMP(6),
    CREATED_TX_STAMP TIMESTAMP(6),
    CONSTRAINT PK_PROD_CAT_CONFIG_MOD PRIMARY KEY (PRODUCT_ID),
    CONSTRAINT FK_PRODUCT_ID FOREIGN KEY (PRODUCT_ID) REFERENCES PRODUCT(PRODUCT_ID),
    CONSTRAINT FK_CONFIG_ITEM_ID FOREIGN KEY (CONFIG_ITEM_ID) REFERENCES PRODUCT_CONFIG_ITEM(CONFIG_ITEM_ID),
    CONSTRAINT FK_CONFIG_OPTION_ID FOREIGN KEY (CONFIG_OPTION_ID) REFERENCES PRODUCT_CONFIG_OPTION(CONFIG_OPTION_ID) )
    TABLESPACE DATA_SMALL
    i try to create this table if i omit 3rd foreign key constraint then table successfully created.but including trd foreign key constraint it return error "ORA-02270: no matching unique or primary key for this column-list"
    i checked everything is it possible to create foreign key from composite key in other table.

    And
    CONSTRAINT FK_CONFIG_OPTION_ID FOREIGN KEY (CONFIG_ITEM_ID,CONFIG_OPTION_ID) REFERENCES PRODUCT_CONFIG_OPTION(CONFIG_ITEM_ID,CONFIG_OPTION_ID)
    ?

  • How to create radio button between two slection screen

    hello all.
    could you please guide me how to create radio button between two SELECTION-SCREEN  in screen painter.
    Thank you,
    srinivas

    hi
    SEE THIS CODE
    REPORT  ZNNR_REPORT NO STANDARD PAGE HEADING MESSAGE-ID ZNNR LINE-SIZE 100 LINE-COUNT 65(4).
    ******DATA DECLARATIONS**********
    DATA : BEGIN OF IT_PLANT OCCURS 0,
            MATNR LIKE MARA-MATNR,
            WERKS LIKE MARC-WERKS,
            PSTAT LIKE MARC-PSTAT,
            EKGRP LIKE MARC-EKGRP,
           END OF IT_PLANT.
    DATA : BEGIN OF IT_PONO OCCURS 0,
            EBELN LIKE EKKO-EBELN,
            EBELP LIKE EKPO-EBELP,
            MATNR LIKE EKPO-MATNR,
            WERKS LIKE EKPO-WERKS,
            LGORT LIKE EKPO-LGORT,
           END OF IT_PONO.
    TABLES EKKO.
    ********END OF DATA DECLARATIONS*********
    ********SELECTION SCREEN DESIGN ***********
    SELECTION-SCREEN BEGIN OF BLOCK B1 WITH FRAME TITLE TEXT-001.
    PARAMETER : P_WERKS LIKE MARC-WERKS MODIF ID S1.
    SELECT-OPTIONS : S_EBELN FOR EKKO-EBELN NO INTERVALS MODIF ID S2.
    SELECTION-SCREEN END OF BLOCK B1.
    SELECTION-SCREEN BEGIN OF LINE.
    PARAMETERS : R3 RADIOBUTTON GROUP G2 DEFAULT 'X'.
    SELECTION-SCREEN COMMENT 5(20) TEXT-003 FOR FIELD R3.
    SELECTION-SCREEN END OF LINE.
    SELECTION-SCREEN BEGIN OF LINE.
    PARAMETERS : R4 RADIOBUTTON GROUP G2.
    SELECTION-SCREEN COMMENT 5(20) TEXT-004 FOR FIELD R4.
    SELECTION-SCREEN END OF LINE.
    SELECTION-SCREEN BEGIN OF BLOCK B2 WITH FRAME TITLE TEXT-004.
    SELECTION-SCREEN BEGIN OF LINE.
    PARAMETERS : R1 RADIOBUTTON GROUP G1 DEFAULT 'X'.
    SELECTION-SCREEN COMMENT 5(20) TEXT-002 FOR FIELD R1.
    SELECTION-SCREEN END OF LINE.
    SELECTION-SCREEN BEGIN OF LINE.
    PARAMETERS : R2 RADIOBUTTON GROUP G1.
    SELECTION-SCREEN COMMENT 5(20) TEXT-003 FOR FIELD R2.
    SELECTION-SCREEN END OF LINE.
    SELECTION-SCREEN END OF BLOCK B2.
    ******END OF SELECTION SCREEN DESIGN****************
    <b>rEWARD IF USEFULL</b>

  • Creating a relationship between two tables  through configuration manager

    I've just created two tables in the underlying db, each with a primary key and established a primary key - foreign key relationship. I've also created a relationship through stellent and it shows up under the 'Relations' Tab.
    However, when I go to Information Fields > Field Info > Click on my Field > 'Edit' > Enable Option List and Configure > Use the view of one of the tables and click on Dependant Field , I don't see any Relationships defined. What could I have done wrong here?
    Thanks.

    ok...solved...followed metalink note 445363.1 for clarification on the right approach to creating Dependant Choice Lists (DCL).

  • Syntax for creating foreign key across users in a database

    There are two user present A,B.They are granted all privileges.Now in USER A, there is a table PARENT whose primary key is PARENT_NO.In USER B I have created a table CHILD whose primary key is CHILD_NO.
    In the CHILD table of USER B, I want to create a foreign key relation to the PARENT table of USER A.For this I have created a column CHILD_PARENT_NO in the CHILD table.If anybody knows the syntax please post the syntax for creating the required foreign key relationship?

    Please post your code. Cut'n'paste from SQL*Plus like this...
    SQL> conn a/a
    Connected.
    SQL> desc t1
    Name                                      Null?    Type
    COL1                                               NUMBER
    COL2                                               VARCHAR2(10)
    SQL> grant references on t1 to b;
    Grant succeeded.
    SQL> conn b/b
    SQL> create synonym a_t for a.t1;
    Synonym created.
    SQL> alter table test add constraint fk foreign key (n) references a_t(col1);
    Table altered.
    SQL> Note that Oracle will translate the synonym anyway...
    SQL> select constraint_name, r_owner, r_constraint_name
      2  from  user_constraints
      3  where table_name = 'TEST'
      4  /
    CONSTRAINT_NAME R_OWNER R_CONSTRAINT_NAME
    FK              A       T1_PK
    SQL> By the way, this ...
    GRANT ALL PRIVILEGES TO B;... is a mindbendingly unsafe way of proceeding. In real life you would have given user B the power to utterly destroy your database. It's always easier to start with good habits than to break bad ones so please get used to granting only the minimum set of privileges necessary.
    Cheers, APC

  • Programmatically create a relationship between two positions in HR

    Hi,
    I have a requirement to create relationships in HRP1001 between two given positions with a start and end date.
    I need to write an upload program to do this but want to avoid Batch Input if possible.
    Are there any relevant function modules that can do what transaction PP01 does?
    Many Thanks
    David

    Hi,
    Try using this code
    LOOP AT T_MAINTAIN INTO WA_MAINTAIN.
        WA_MAINTAIN-FCODE = 'INSE'.
        WA_MAINTAIN-PLVAR = '01'.
        WA_MAINTAIN-ISTAT = '1'.
    *Relate account to project
        IF WA_MAINTAIN-OTYPE = 'O' AND WA_MAINTAIN-SCLAS = 'O'.
          WA_MAINTAIN-RSIGN = 'B'.
          WA_MAINTAIN-RELAT = '002'.
    *Relate position to project
        ELSEIF WA_MAINTAIN-OTYPE = 'O' AND WA_MAINTAIN-SCLAS = 'S'.
          WA_MAINTAIN-RSIGN = 'B'.
          WA_MAINTAIN-RELAT = '003'.
    *Relate job to position
        ELSEIF WA_MAINTAIN-OTYPE = 'S' AND WA_MAINTAIN-SCLAS = 'C'.
          WA_MAINTAIN-RSIGN = 'B'.
          WA_MAINTAIN-RELAT = '007'.
    *Relate employee to position
        ELSEIF WA_MAINTAIN-OTYPE = 'S' AND WA_MAINTAIN-SCLAS = 'P'.
          WA_MAINTAIN-RSIGN = 'A'.
          WA_MAINTAIN-RELAT = '008'.
        ENDIF.
    *Relate position to position
        ELSEIF WA_MAINTAIN-OTYPE = 'S' AND WA_MAINTAIN-SCLAS = 'S'.
          WA_MAINTAIN-RSIGN = 'A'.
          WA_MAINTAIN-RELAT = '002'.
        ENDIF.
        WA_MAINTAIN-ENDDA = '99991231'.
    *FM to create relationship
        CALL FUNCTION 'RH_RELATION_MAINTAIN'
          EXPORTING
            ACT_FCODE           = WA_MAINTAIN-FCODE
            ACT_PLVAR           = WA_MAINTAIN-PLVAR
            ACT_OTYPE           = WA_MAINTAIN-OTYPE
            ACT_OBJID           = WA_MAINTAIN-OBJID
            ACT_ISTAT           = WA_MAINTAIN-ISTAT
            ACT_RSIGN           = WA_MAINTAIN-RSIGN
            ACT_RELAT           = WA_MAINTAIN-RELAT
            ACT_SCLAS           = WA_MAINTAIN-SCLAS
            ACT_SOBID           = WA_MAINTAIN-SOBID
            ACT_BEGDA           = WA_MAINTAIN-BEGDA
            ACT_ENDDA           = WA_MAINTAIN-ENDDA
            ACT_PROZT           = WA_MAINTAIN-PROZT
          EXCEPTIONS
            MAINTAINANCE_FAILED = 1
            OTHERS              = 2.
        IF SY-SUBRC <> 0.
         WRITE : WA_INPUT-OBJID.
         MESSAGE ID SY-MSGID TYPE SY-MSGTY NUMBER SY-MSGNO
                 WITH SY-MSGV1 SY-MSGV2 SY-MSGV3 SY-MSGV4.
        ENDIF.
    ENDLOOP.

  • Create a relationship between two existing tables

    When I create a new table the wizard exposes the ability to add a foreign key column. However when editing an existing table the only option I seem to have is to create a lookup table, which is a new table that is autogenerated. I find it hard to believe that it's impossible to create and manage relationships in APEX...but I've unable to find the access point if there is one.

    Hello,
    I'm not sure what version of APEX you are using as you don't say, but in my 2.2.0.00.32 installation, when I select a table using the Object Browser, select Constraints from the choices running along the top of the page, then click the Create button/link, the Foreign Key option is available in the Constraint Type drop-down list.
    Hope that helps a bit,
    Mark

  • Create line extension between two SPA-3102

    I`m having problems to create a line extension between two SPA-3102
    I have one SPA-3102 connected to an analog PBX system with IP 192.168.0.201, and the other SPA-3102 with analog phone and IP 192.168.0.200
    I succesfully setup them to make a call from the first to the second
    But I couldn`t setup them to make a call from the second (192.168.0.200) and give me the dialtone of the PBX connected to the first SPA-3102 (192.168.0.201).
    I could setup a hot line on the second SPA-3102 (192.168.0.200) and call to 192.168.0.201, but it doesn`t take the line to hear the pstn dialtone.
    I saw many answers about this problem, but no one resolve the problem, i have the latest firmware. please, anyone could help me and if it`s possible to work please send me all the configuration needed.
    Thanks again

    Hi Jeremy,
    I have a similar problem, I have one PSTN line (say Line1) with free minutes to mobiles, so its good for outgoing calls. The other line (say Line2) which i have is acually VoIP but it comes with its own hardware (magicJack if you have heard) so I can't use a SIP client and have to use the supplied Hw client, but it does give me an option to connect any normal phone to this magicJack (i suppose that would make it a fxs port). Now this magicJack is cheap for other people to call me.
    I want to find a solution so that all the calls I receive on Line2 get forwarded to my mobile number via Line1. And if I receive any calls on Line1 they should be treated normally (my home phone rings). Do you have some idea how I can achieve this with minimal spend? Thanx
    Atif

  • Create AlE Relationship between two company code F210 to FN21 (UHP To UPI)

    Hi,
    I want to know how to create ALE realtion ship between two company code ( F210 to FN21(Server name UHP to UPI )  and how to process it?
    Please reply ASAP.
    Regards,
    Achin

    Hi Sathish,
    Thanks for reply!..
    But My question I want to Add ALE relationship between two company code F210 to FN21 (UHP(Production server) To UPI(Third Party System)).
    Regards!
    Achin

  • Foreign Keys involving two different databases/schemas

    Hi -- I'm wondering if it's possible to set a foreign key that involves tables that live in different databases/schemas. And, if it is possible, is it a horrible design idea?
    Thanks,
    ~Christine

    cav0227, you have asked two different two part questions.
    1 - Is it possible to set up FK that span owners and is this good design?
    Yes, it is possilbe and there if the data is related then yes it is good design. We like to minimize the number of object owners in our system and use a table naming standard where the first two or three letters of the table name represent a system code: ap, ar, gl for accounts receivable, accounts payable, and general ledger. Other may prefer to have an AP owner, AR owner, and a GL owner.
    Ownership has no direct bearing on quality of design. If the tables are related then you should define the FK no matter who the owner is.
    2 - Is it possible to define FK to remote (distributed) tables and is the good design? The answer is no. FK cannot reference remote objects. To enforce referencial integrity accross databases via the database, rather than in the application, requires the use of table triggers.
    As far as good design that depends on why the data is remote and why the tables need to be linked. There are issues related to availability, recovery, and performance when data is distributed. Having such a relationship could be good design or bad design depending on if all the inter-related factors have been considered.
    IMHO -- Mark D Powell --

Maybe you are looking for

  • Kernel Panic right after start up on my PowerBook G4

    I get the standard kernel panic message quite frequently after I start up my PowerBook. I have used the Disk Utility (software), and the Hardware Utility to check the system hardware, and both say all is well. I would appreciate any ideas on how to f

  • Weblogic 10 doesn't start through admin console

    Hi, While starting a managed server on WebLogic 10.3 through the admin console I get the below error. However when starting the same managed server with a script I can successfully start it up. <Sep 30, 2010 10:31:52 AM EEST> <Emergency> <Management>

  • Every other row (background color)

    I want to keep it simple and in css.  How can I get this to work properly so every other row has a yellowish background color with no gaps between the words. This is what I have but its producing an odd result. <!DOCTYPE html PUBLIC "-//W3C//DTD XHTM

  • Please Help! Buying Songs!

    Hello! I had a question. Today I downloaded the song Just Stand Up! and it came with a digital booklet. I downloaded it on my ipod touch and it told me that when i connected to the computer that the digital booklet would download but it didnt. Can an

  • Unable to install iTunes - error message

    I keep trying to install iTunes with the same error all the time. "error writing to C:/program files/itunes/itunes.resources/pt.1proj/ipodsetup.nib/objects.xib" I have deleted the Temp files folder, uninstalled both iTunes and QuickTime and restarted