There are a few problems with your tables. I'll try to address the foreign keys first, since you question asked about them :)
But before that, we should realize that the two sets of tables (the first three you created and the second set, which you created after dropping the first set) are the same. Of course, the definition of Table3 in your second attempt has syntax and logical errors, but the basic idea is:
CREATE TABLE table3 (
"ID" bigint NOT NULL DEFAULT '0',
"DataID" bigint DEFAULT NULL,
"Address" numeric(20) DEFAULT NULL,
"Data" bigint DEFAULT NULL,
PRIMARY KEY ("ID"),
FOREIGN KEY ("DataID") REFERENCES Table1("DataID") on delete cascade on update cascade,
FOREIGN KEY ("Address") REFERENCES Table2("Address") on delete cascade on update cascade
);
This definition tell PostgreSQL roughly the following: "Create a table with four columns, one will be the primary key (PK), the others can be NULL. If a new row is inserted, check DataID and Address: if they contain a non-NULL value (say 27856), then check Table1 for DataIDΛ™and Table2 for Address. If there is no such value in those tables, then return an error." This last point which you've seen first:
ERROR: insert or update on table "Table3" violates foreign key constraint
"Table3_DataID_fkey" DETAIL: Key (DataID)=(27856) is not present in table "Table1".
So simple: if there is no row in Table1 where DataID = 27856, then you can't insert that row into Table3.
If you need that row, you should first insert a row into Table1 with DataID = 27856, and only then try to insert into Table3. If this seems to you not what you want, please describe in a few sentences what you want to achieve, and we can help with a good design.
And now about the other problems.
You define your PKs as
CREATE all_your_tables (
first_column NOT NULL DEFAULT '0',
[...]
PRIMARY KEY ("ID"),
A primary key means that all the items in it are different from each other, that is, the values are UNIQUE. If you give a static DEFAULT (like '0') to a UNIQUE column, you will experience bad surprises all the time. This is what you got in your third error message.
Furthermore, '0' means a text string, but not a number (bigint or numeric in your case). Use simply 0 instead (or don't use it at all, as I written above).
And a last point (I may be wrong here): in Table2, your Address field is set to numeric(20). At the same time, it is the PK of the table. The column name and the data type suggests that this address can change in the future. If this is true, than it is a very bad choice for a PK. Think about the following scenario: you have an address '1234567890454', which has a child in Table3 like
ID DataID Address Data
123 3216547 1234567890454 654897564134569
Now that address happens to change to something other. How do you make your child row in Table3 follow its parent to the new address? (There are solutions for this, but can cause much confusion.) If this is your case, add an ID column to your table, which will not contain any information from the real world, it will simply serve as an identification value (that is, ID) for an address.
Best Answer
The create table script 2 is correct.
InnoDB requires indexes on foreign keys and referenced keys so that foreign key checks can be fast and not require a table scan. In the referencing table, there must be an index where the foreign key columns are listed as the first columns in the same order. Such an index is created on the referencing table automatically if it does not exist. (This is in contrast to some older versions, in which indexes had to be created explicitly or the creation of foreign key constraints would fail.) index_name, if given, is used as described previously.