Trigger Oracle operator LIKE to respect a string format - sql

I'm trying to learn how triggers work and how to write them properly with a given exercise. I'm using SQL Developper with the version 4.2 of Oracle.
I have a table PERSON with these fields : 'matricule', 'pmatricule' and 'typpers'.
'typpers' is a letter to show if the person is a student (E), a professor (P) or other (X). 'matricule' is a chain of 3 to 7 numbers and 'matricule' is a chain with 'p' followed by a chain of 3 to 6 numbers. Example : 'p145', 'p123456'.
I want my trigger to check if the format of 'matricule' or 'pmatricule' is respected when someone wants to insert a row in the table PERSON.
CREATE OR REPLACE TRIGGER new_matricule
BEFORE INSERT ON PERSONNE
FOR EACH ROW
DECLARE
new_pmatricule PERSONNE.pmatricule%type;
BEGIN
IF((:NEW.TYPPERS = 'P' OR :NEW.TYPPERS = 'X') AND :NEW.pmatricule NOT LIKE 'p%______') THEN
RAISE_APPLICATION_ERROR(-20100, 'pmatricule format is not good');
END IF;
END;
My trigger works if I'm trying to insert a 'pmatricule' format like '52p' but it also activates when I'm trying to insert such as 'p145'.
I assume that my syntax is not good enough but I'm not finding the proper syntax to do that.
Has anyone an idea ?

First, simplify the syntax for the first part of the expression using in. Then, if you want to count the number of characters, leave out the '%'. Something like this:
IF((:NEW.TYPPERS IN ('P', 'X') AND :NEW.pmatricule NOT LIKE 'p______') THEN
RAISE_APPLICATION_ERROR(-20100, 'pmatricule format is not good');
END IF;
This says that the "p" needs to be followed by six characters. Your version says "at least 6 characters".
If you want two characters followed by a p, it would be:
:NEW.pmatricule NOT LIKE '__p'
Any number followed by a "p" would be:
:NEW.pmatricule NOT LIKE '%p'
If you want more advanced pattern matching, you probably want to use regexp_like(). It is much more versatile.
EDIT:
Based on your comment, you want:
where regexp_like(:NEW.pmatricule, '^p[0-9]{3,6}$')
This is probably the easiest way to write the condition.
If you mean 3-6 characters, you can use the cumbersome:
where :NEW.pmatricule like 'p___' or
:NEW.pmatricule like 'p____' or
:NEW.pmatricule like 'p_____' or
:NEW.pmatricule like 'p______'

Related

T-SQL - How to pattern match for a list of values?

I'm trying to find the most efficient way to do some pattern validation in T-SQL and struggling with how to check against a list of values. This example works:
SELECT *
FROM SomeTable
WHERE Code LIKE '[0-9]JAN[0-9][0-9]'
OR Code LIKE '[0-9]FEB[0-9][0-9]'
OR Code LIKE '[0-9]MAR[0-9][0-9]'
OR Code LIKE '[0-9]APRIL[0-9][0-9]
but I am stuck on wondering if there is a syntax that will support a list of possible values within the single like statement, something like this (which does not work)
SELECT *
FROM SomeTable
WHERE Code LIKE '[0-9][JAN, FEB, MAR, APRIL][0-9][0-9]'
I know I can leverage charindex, patindex, etc., just wondering if there is a simpler supported syntax for a list of possible values or some way to nest an IN statement within the LIKE. thanks!
I think the closest you'll be able to get is with a table value constructor, like this:
SELECT *
FROM SomeTable st
INNER JOIN (VALUES
('[0-9]JAN[0-9][0-9]'),
('[0-9]FEB[0-9][0-9]'),
('[0-9]MAR[0-9][0-9]'),
('[0-9]APRIL[0-9][0-9]')) As p(Pattern) ON st.Code LIKE p.Pattern
This is still less typing and slightly more efficient than the OR option, if not as brief as we hoped for. If you knew the month was always three characters we could do a little better:
Code LIKE '[0-9]___[0-9][0-9]'
Unfortunately, I'm not aware of SQL Server pattern character for "0 or 1" characters. But maybe if you want ALL months we can use this much to reduce our match:
SELECT *
FROM SomeTable
WHERE (Code LIKE '[0-9]___[0-9][0-9]'
OR Code LIKE '[0-9]____[0-9][0-9]'
OR Code LIKE '[0-9]_____[0-9][0-9]')
You'll want to test this to check if the data might contain false positive matches, and of course the table-value constructor could use this strategy, too. Also, I really hope you're not storing dates in a varchar column, which is a broken schema design.
One final option you might have is building the pattern on the fly. Something like this:
Code LIKE '[0-9]' + 'JAN' + '[0-9][0-9]'
But how you find that middle portion is up to you.
The native TSQL string functions don't support anything like that.
But you can use a workaround (dbfiddle) such as
WHERE CASE WHEN Code LIKE '[0-9]%[^ ][0-9][0-9]' THEN SUBSTRING(Code, 2, LEN(Code) - 3) END
IN
( 'JAN', 'FEB', 'MAR', 'APRIL' )
So first of all check that the string starts with a digit and ends in a non-space character followed by two digits and then check the remainder of the string (not matched by the digit check) is one of the values you want.
The reason for including the SUBSTRING inside the CASE is so that is only evaluated on strings that pass the LIKE check to avoid possible "Invalid length parameter passed to the LEFT or SUBSTRING function." errors if it was to be evaluated on a shorter string.

Postgres SQL regexp_replace replace all number

I need some help with the next. I have a field text in SQL, this record a list of times sepparates with '|'. For example
'14613|15474|3832|148|5236|5348|1055|524' Each value is a time in milliseconds. This field could any length, for example is perfect correct '3215|2654' or '4565' (only 1 value). I need get this field and replace all number with -1000 value.
So '14613|15474|3832|148|5236|5348|1055|524' will be '-1000|-1000|-1000|-1000|-1000|-1000|-1000|-1000'
Or '3215|2654' => '-1000|-1000' Or '4565' => '-1000'.
I try use regexp_replace(times_field,'[[:digit:]]','-1000','g') but it replace each digit, not the complete number, so in this example:
'3215|2654' than must be '-1000|-1000', i get:
'-1000-1000-1000-1000|-1000-1000-1000-1000', I try with other combinations and more options of regexp but i'm done.
Please need your help, thanks!!!.
We can try using REGEXP_REPLACE here:
UPDATE yourTable
SET times_field = REGEXP_REPLACE(times_field, '\y[0-9]+\y', '-1000', 'g');
If instead you don't really want to alter your data but rather just view your data this way, then use a select:
SELECT
times_field,
REGEXP_REPLACE(times_field, '\y[0-9]+\y', '-1000', 'g') AS times_field_replace
FROM yourTable;
Note that in either case we pass g as the fourtb parameter to REGEXP_REPLACE to do a global replacement of all pipe separated numbers.
[[:digit:]] - matches a digit [0-9]
+ Quantifier - matches between one and unlimited times, as many times as possible
your regexp must look like
regexp_replace(times_field,'[[:digit:]]+','-1000','g')

Replacing characters in the oracle database value range

I have a database in an oracle, I filled in the "Number" fields with numbers starting from "A14602727" to "A14603000" but it turned out that I accidentally typed the symbol A (in Ukrainian) instead of A (in English). And now I when find the command:
select number from test where number 'А14602727'
...find me nothing. Is it possible somehow with the help of the command to replace all numbers from "A14602727" (Ukrainian) to "A14602727" (English) ?
I will be grateful for your help!)
You can use following trick to convert any character to its base character using accent-insesitive binary sorting.
select your_col,
utl_raw.cast_to_varchar2(nlssort(your_col, 'nls_sort=binary_ai')) converted_col
from your_table
Use it in update statement accordingly.
Cheers!!
You could use regexp_replace():
update test set number = regexp_replace(number, '^Ä', 'A');
Regexp '^Ä' represents character 'Ä' at the beginning of the string. I used 'Ä' to represent the A in Ukrainian: replace this with the correct character that you want to replace.
This can also be done, probably more efficiently, with substr and like:
update test set number = 'A' || substr(number, 2) where number like 'Ä%';
Use the simple Oracle REPLACE function:
UPDATE test
SET number = REPLACE(number, 'A', 'A');
Replace is taking 3 parameters:
String in which you will search for a string to replace (The number column)
The string you are searchin gto replace (Ukrainian A)
The string you will replace it with (English A)
Here you have a DEMO example.
Also please note that the query in your question:
select number from test where number 'А14602727';
Is not valid. It should be something like this:
select number from test where number like 'А14602727';
or like this:
select number from test where number = 'А14602727';
So first do check that!
Cheers!

SELECT middle part of a String if it exists. Postgresql

i've got a problem with transferring "real-World" data into my schema.
It's actually a "project" for my Database course and they gave us ab table with dog race results. This Table has a column which contains the name of the Dog (which itself consists of the actuall name and the name of the breeder) and informations about the Birthcountry, actual living Country and the birth year.
Example filed are "Lillycette [AU 2012]" or "Black Bear Lee [AU/AU 2013]" or "Lemon Ralph [IE/UK 1998]".
I've managed it to get out the first word and save it in the right column with split_part like this:
INSERT INTO tblHund (rufname)
SELECT
split_part(name, ' ', 1) AS rufname,
FROM tblimport;
tblimport is a table where I dumped the data from the csv file.
That works just as it should.
Accessing the second part of the Name with this fails because sometimes there isn't a second part and sometimes times there second part consists of two words.
And this is the where iam stuck right now.
I tried it with substring and regular expressions:
INSERT INTO tblZwinger (Name)
SELECT
substring(vatertier from E'[^ ]*\\ ( +)$')AS Name
FROM tblimport
WHERE substring(vatertier from E'[^ ]*\\ ( +)$') != '';
The above code is executed without errors but actually does nothing because the SELECT statement just give empty strings back.
It took me more then 3h to understand a bit of this regular Expressions but I still feel pretty stupid when I look at them.
Is there any other way of doing this. If so just give me a hint.
If not what is wrong with my expression above?
Thanks for your help.
You need to use atom ., which matches any single character inside capturing group:
E'[^ ]*\\ (.+)$'
SELECT
tblimport.*,
ti.parts[1] as f1,
ti.parts[2] as f2, -- It should be the "middle part"
ti.parts[3] as f3
FROM
tblimport,
regexp_matches(tblimport.vatertier, '([^\s]+)\s*(.*)\s+\[(.*)\]') as ti(parts)
WHERE
nullif(ti.parts[2], '') is not null
Something like above.

How to make to_number ignore non-numerical values

Column xy of type 'nvarchar2(40)' in table ABC.
Column consists mainly of numerical Strings
how can I make a
select to_number(trim(xy)) from ABC
query, that ignores non-numerical strings?
In general in relational databases, the order of evaluation is not defined, so it is possible that the select functions are called before the where clause filters the data. I know this is the case in SQL Server. Here is a post that suggests that the same can happen in Oracle.
The case statement, however, does cascade, so it is evaluated in order. For that reason, I prefer:
select (case when NOT regexp_like(xy,'[^[:digit:]]') then to_number(xy)
end)
from ABC;
This will return NULL for values that are not numbers.
You could use regexp_like to find out if it is a number (with/without plus/minus sign, decimal separator followed by at least one digit, thousand separators in the correct places if any) and use it like this:
SELECT TO_NUMBER( CASE WHEN regexp_like(xy,'.....') THEN xy ELSE NULL END )
FROM ABC;
However, as the built-in function TO_NUMBER is not able to deal with all numbers (it fails at least when a number contains thousand separators), I would suggest to write a PL/SQL function TO_NUMBER_OR_DEFAULT(numberstring, defaultnumber) to do what you want.
EDIT: You may want to read my answer on using regexp_like to determine if a string contains a number here: https://stackoverflow.com/a/21235443/2270762.
You can add WHERE
SELECT TO_NUMBER(TRIM(xy)) FROM ABC WHERE REGEXP_INSTR(email, '[A-Za-z]') = 0
The WHERE is ignoring columns with letters. See the documentation