The teacher gave us a team assignment, and me and my teammate are quite struggling with it (especially since we need to use things like TRIGGERS and PROCEDURES, things we didn't see in class yet …).
We need to implement an arc-relationship, and we fail to understand how …
But before I tell you guys what I need to accomplish, I will give you part of the description of the task, so you guys can understand the situation a bit better …
We basically need to make an ERD for a VLSI CAD-system and we need to implement it. Now, we have our CELL entity, the attributes of which aren't really relevant … The only thing you guys need to know in order to help us is that it has a primary key, CELL_CODE, which is a VARCHAR.
Each CELL has many (I think at least four, I don't think you can have triangular CELLS, but doesn't matter anyways) SIDES. A SIDE can be logically identified by its CELL, and to make matters ridiculously difficult, each SIDE has to be numbered by its CELL, like so:
CELLS:
CELL_CODE
1
2
SIDES:
SEQUENCE_NUMBER CELL_CODE
1 1
2 1
3 1
1 2
2 2
3 2
Now, each SIDE has its CONNECTION_PINS. CONNECTION_PINS is also uniquely identified by SIDES, which are basically numbered in a similar manner:
CELLS:
CELL_CODE
1
2
SIDES:
SEQUENCE_NUMBER CELL_CODE
1 1
2 1
3 1
1 2
2 2
3 2
CONNECTION_PINS:
SEQUENCE_NUMBER SIE_SEQUENCE_NUMBER CELL_CODE
1 1 1
2 1 1
1 2 1
2 2 1
1 3 1
2 3 1
1 1 2
2 1 2
1 2 2
2 2 2
1 3 2
2 3 2
I tried to explain the numbering issue we have here: Data model - PRIMARY KEY numbering issue, but yeah, I didn't really explain it the way it should be explained ...
Now, we have one final entity, which is where the Arc comes in: CONNECTIONS. CONNECTIONS has 2 CONNECTION_PINS: one for START_FROMand one for END_OF. Now, logically seen the start pin can't be the end pin as well, for a given connection. And that's our struggle. Basically, this shouldn't be allowed:
CELLS:
CELL_CODE
1
2
SIDES:
SEQUENCE_NUMBER CELL_CODE
1 1
2 1
3 1
1 2
2 2
3 2
CONNECTION_PINS:
SEQUENCE_NUMBER SIE_SEQUENCE_NUMBER CELL_CODE
1 1 1
2 1 1
1 2 1
2 2 1
1 3 1
2 3 1
1 1 2
2 1 2
1 2 2
2 2 2
1 3 2
2 3 2
CONNECTIONS:
(you shouldn't be able to put this in …)
CPI_SEQNUM_START SIE_SEQNUM_START CELL_CODE_START CPI_SEQNUM_END SIE_SEQNUM_END CELL_CODE_END
1 1 1 1 1 1
Now, this is basically the ERD for this part:
ERD with barred relationships and the arc-relationship in question
and this is the physical model:
Physical model
I basically thought a simple CHECK might do (CHECK (CPI_SEQNUM_START <> CPI_SEQNUM_END AND CELL_CODE_START <> CELL_CODE_END AND SIE_SEQNUM_START <> SIE_SEQNUM_END) ), but that prevented us from inserting anything somehow … Any advice?
Your approach was correct to use a CHECK constraint. Your logic for the constraint was wrong though. You need an OR condition. Only one of the three fields needs to be different.
CPI_SEQNUM_START <> CPI_SEQNUM_END OR
CELL_CODE_START <> CELL_CODE_END OR
SIE_SEQNUM_START <> SIE_SEQNUM
... assuming all three fields are not nullable.
Related
I want to generate some mock data in mockaroo.
The format of the data should be like this
Claim ID Claim subid
1 1
1 2
1 3
2 1
2 2
2 3
3 1
So, basically the claimdid column can have multiple subids. Is this possible?
I'm trying to join several tables, where one of the tables is acting as a
key-value store, and then after the joins find the maximum value in a
column less than another column. As a simplified example, I have the following three tables:
Documents:
DocumentID
Filename
LatestRevision
1
D1001.SLDDRW
18
2
P5002.SLDPRT
10
Variables:
VariableID
VariableName
1
DateReleased
2
Change
3
Description
VariableValues:
DocumentID
VariableID
Revision
Value
1
2
1
Created
1
3
1
Drawing
1
2
3
Changed Dimension
1
1
4
2021-02-01
1
2
11
Corrected typos
1
1
16
2021-02-25
2
3
1
Generic part
2
3
5
Screw
2
2
4
2021-02-24
I can use the LEFT JOIN/IS NULL thing to get the latest version of
variables relatively easily (see http://sqlfiddle.com/#!7/5982d/3/0).
What I want is the latest version of variables that are less than or equal
to a revision which has a DateReleased, for example:
DocumentID
Filename
Variable
Value
VariableRev
DateReleased
ReleasedRev
1
D1001.SLDDRW
Change
Changed Dimension
3
2021-02-01
4
1
D1001.SLDDRW
Description
Drawing
1
2021-02-01
4
1
D1001.SLDDRW
Description
Drawing
1
2021-02-25
16
1
D1001.SLDDRW
Change
Corrected Typos
11
2021-02-25
16
2
P5002.SLDPRT
Description
Generic Part
1
2021-02-24
4
How do I do this?
I figured this out. Add another JOIN at the start to add in another version of the VariableValues table selecting only the DateReleased variables, then make sure that all the VariableValues Revisions selected are less than this date released. I think the LEFT JOIN has to be added after this table.
The example at http://sqlfiddle.com/#!9/bd6068/3/0 shows this better.
I have a requirement from client where I need to store a value against list of combination.
For example I have following LOBs and against each combination I need to store a value.
Auto
WC
Personal
I purposed multiple solutions he is not satisfied with anyone.
Solution 1: create single table, insert value against all possible combination(string) something like
LOB Value
Auto 1
WC 2
Personal 3
Auto,WC 4
Auto, personal 5
WC, Personal 6
Auto, WC, Personal 7
Solution 2: create lkp_lob, lob_group and lob_group_detail tables. Each group combination represent a group.
Lkp_lob
Lob_key Name
1 Auto
2 WC
3 Person
Lob_group (unique query constrain on lob_group_key and lob_key)
Lob_group_key Lob_key
1 1
2 2
3 3
4 1
4 2
5 1
5 3
6 2
6 3
7 1
7 2
7 3
Lob_group_detail
Lob_group_key Value
1 1
2 2
3 3
4 4
5 5
6 6
7 7
Any suggestion would be highly appreciated.
First of all I did not understood that terms you said.
But from database perspective it is always good to have multiple tables for each module. You will be facing less difficulties when doing CRUD. And will be more faster.
We have R(A,B,C)
1 2 3
1 3 2
1 2 2
3 2 1
3 2 3
Question : Which of the following multivalued dependencies does this instance of R not satisfy?
BC ↠ A
BC ↠ C
C ↠ A
✔ A ↠ B
Right answer is 4 but why i dont understand
please explain
It doesn't satisfy statement 4 because for every unique combination of A and B, you have to have matching rows with C.
To make Statement 4 valid, you'd need a 1 3 3 in your table (since you have a 1 2 3).
Good luck.
Hello I have doubt regarding how to create the table for the pairwise testing.
For example if I have three parameter which can each attain two different values. How do I create a table of input with all possible combination then? Would it look something like this?
| 1 2 3
-----------
1 | 1 1 1
2 | 1 2 2
3 | 1 1 2
4 | 1 2 1
Does each parameter corresponds to each column?
However since I have 3 parameter, which each can take 2 different value. The number of test cases should be 2^3 isn't it?
There's a good article with links to some useful tools here:
http://blog.josephwilk.net/ruby/pairwise-testing-with-cucumber.html
For the parameters: each column is a parameter, and each row is a possible combination. Here is the table:
| 1 2 3
-----------
1 | 1 1 1
2 | 2 1 1
3 | 1 2 1
4 | 1 1 2
5 | 2 2 1
6 | 2 1 2
7 | 1 2 2
8 | 2 2 2
so 2^3=8 possible combinations as you can see :)
For the values: each column is a value, and each row is a possible combination:
| 1 2
--------
1 | 1 1
2 | 2 1
3 | 1 2
4 | 2 2
They are 2^2=4 possible combinations. Hope it helps.
1) Please note that pair-wise testing is not about scanning exhaustively all possible combination of values of all parameters. Firstly, such a scanning would give you an enormous amount of test cases that almost no existing system could be able to run all of them.
Secondly, pair-wise testing for a software system is based on the hope that the two parameters having the highest number of possible values are the culprit for the highest percentage of faults of that system.
This is of course only a hope and almost no rigorous scientific research has existed so far to prove that.
2) What I often see in the documentations discussing pair wise testing, like this is that the list of all possible values (aka the pair-wise test table) is not constructed in a well thought way. This creates confusions.
In your case, all the parameters have the same number of possible values (2 values), therefore you could choose any two parameters of those three to build the table. What you could pay attention is the ordering of the combination: you iterate first the top-right parameter, then the next parameter to the left, and so on, ...
Say if you have two parameters p1 and p2, p1 has two possible values apple and orange; and p2 has two possible values red and blue, then your pair-wise test table would be:
index| p1 p2
------------------
1 | apple red
2 | apple blue
3 | orange red
4 | orange blue