I have a scenario which I am not able to do in 10.7 version. Basically, I have a data column in which I need to find the Reg Exp pattern inside the data which is in the form of notes/comments/description.
For example, Data in the column : The SSN number is 760-56-6289
In the above data 760-56-6289 is the actual SSN number which I need to find across all schemas/tables/columns for the defined reg exp pattern. And, we can have a pre or post text for actual SSN value.
Could you please let me know how to achieve this PostgreSQL 10.7?
Please let me know if you need more information for the same.
demo:db<>fiddle
SELECT
(regexp_matches(mycolumn, '^.*([\d]{3}-[\d]{2}-[\d]{4}).*$'))[1]
FROM mytable
The RegEx means:
Start of text: ^
arbitrary number of characters: .*
group of your number: (...)
3 digit characters: [\d]{3}
- character
2 digits: [\d]{2}
- character
4 digits: [\d]{4}
arbitrary number of characters: .*
end of text: $
regexp_matches() gives out all found groups as an array. So, there is only one group, the array contains only one value. This is your number which can be get with the index [1]
Related
I have this inputs:
John/Bean/4000-M100
John/4000-M100
John/4000
How can I get just the 4000 but note that the 4000 there will be change from time to time it can be 3000 or 2000 how can I treat that using regex pattern?
Here's my output so far, it statisfies John/400-M100 and John/4000 but the double slash doesnt suffice the match requirements in the regex I have:
REGEXP_REPLACE(REGEXP_SUBSTR(a.demand,'/(.*)-|/(.*)',1,1),'-|/','')
You can use this query to get the results you want:
select regexp_replace(data, '^.*/(\d{4})[^/]*$', '\1')
from test
The regex looks for a set of 4 digits following a / and then not followed by another / before the end of the line and replaces the entire content of the string with those 4 digits.
Demo on dbfiddle
This would also work, unless you need any digit followed by three zeros. See it in action here, for as long as it lives, http://sqlfiddle.com/#!4/23656/5
create table test_table
( data varchar2(200))
insert into test_table values('John/Bean/4000-M100')
insert into test_table values('John/4000-M100')
insert into test_table values('John/4000')
select a.*,
replace(REGEXP_SUBSTR(a.data,'/\d{4}'), '/', '')
from test_table a
The following will match any multiple of 1000 less than 10000 when its preceded by a slash:
\/[1-9]0{3}
To match any four-digit number preceded by a slash, not followed by another digit, such as 4031 in—
Sal_AS_180763852/4200009751_S5_154552/4031
—try:
\/\d{3}(?:(?:\d[^\d])|(?:\d$))
https://regex101.com/r/Am34WO/1
We pull in a list of phone numbers as part of a datafeed. They are all for North America based companies. I would like to remove any leading "1" or "+1" and any trailing information like "x100", " EXT400", etc. They are stored in MariaDB so I would like to do
UPDATE `CompanyPhone` SET `number`= REGEXP_SUBSTR(`number`,pattern)
to remove the unwanted stuff, I just need the REGEX to select the correct part of the phone number.
"1 (555) 555-5555 x100" -> "(555) 555-5555"
"+15555555555 EXT400" -> "5555555555"
" 555-555-5555" -> "555-555-5555" (remove leading space)
Basically, I need just the first 10 digits, ignoring the first digit if it is a 1, and the formatting currently in the first 10 digits ("()" or " " or "-") if it is possible to keep it.
If everything could be reformatted to (555) 555-5555 that would be a bonus but is not required. I could do this a 2nd query if needed.
You could use REGEXP_REPLACE for this. Assuming you are using MariaDB 10.0.5 or later, you can use PCRE regular expressions. For your sample expressions, this regexp will give you the desired results (demo on Regex101). It looks for 3 groups of numbers (3 digits, 3 digits and then 4 digits) possibly preceded by a 1, and with other non-digit characters (e.g. +, -) around them.
^(?:\D*)1?(?:\D*)(\d{3})(?:\D*)(\d{3})(?:\D*)(\d{4}).*$
So your UPDATE statement will become
UPDATE `CompanyPhone` SET `number`= REGEXP_REPLACE(`number`, '^(?:\\D*)1?(?:\\D*)(\\d{3})(?:\\D*)(\\d{3})(?:\\D*)(\\d{4}).*$', '(\\1) \\2-\\3')
I need to parse the following expression:
Fertilizer abc 7-15-15 5KG BOX 250 KG
in 3 fields:
The product description: Fertilizer abc 7-15-15
Size: 250
Size unit: KG
Do not know how to proceed. Please, any help and explanation?
Try this in the alteryx REGEX Tool with Parse selected as the Method:
([A-z ]* [\d-]{6,8}) ([A-Z\d]{2,6}) (.{1,5}?) (\d*) ([A-Z]*)
You can test it at Regexpal to see the breakdown of each group but essentially the first set of brackets will get you your product description (text and spaces until 6-8 characters made up of digits and dashes), the 2nd & 3rd parts will deal with the erroneous info that you don't want, the 4th group will be just digits and the 5th group will be any text afterwards.
Note that this will change dramatically if your data has digits where there is characters currently etc.
You can always break it up into even smaller groups and then concatenate back together as well.
SELECT name
FROM players
WHERE name ~ '(.*){8,}'
It is really simple but I cannot seem to get it.
I have a list with names and I have to filter out the ones with at least 8 characters... But I still get the full list.
What am I doing wrong?
Thanks! :)
A (.*){8,} regex means match any zero or more chars 8 or more times.
If you want to match any 8 or more chars, you would use .{8,}.
However, using character_lenth is more appropriate for this task:
char_length(string) or character_length(string) int Number of characters in string
CREATE TABLE table1
(s character varying)
;
INSERT INTO table1
(s)
VALUES
('abc'),
('abc45678'),
('abc45678910')
;
SELECT * from table1 WHERE character_length(s) >= 8;
See the online demo
I am trying to extract a specific text from an Outlook subject line. This is required to calculate turn around time for each order entered in SAP. I have a subject line as below
SO# 3032641559 FW: Attached new PO 4500958640- 13563 TYCO LJ
My final output should be like this: 3032641559
I have been able to do this in MS excel with the formulas like this
=IFERROR(INT(MID([#[Normalized_Subject]],SEARCH(30,[#[Normalized_Subject]]),10)),"Not Found")
in the above formula [#[Normalized_Subject]] is the name of column in which the SO number exists. I have asked to do this in oracle but I am very new to this. Your help on this would be greatly appreciated.
Note: in the above subject line the number 30 is common in every subject line.
The last parameter of REGEXP_SUBSTR() indicates the sub-expression you want to pick. In this case you can't just match 30 then some more numbers as the second set of digits might have a 30. So, it's safer to match the following, where x are more digits.
SO# 30xxxxxx
As a regular expression this becomes:
SO#\s30\d+
where \s indicates a space \d indicates a numeric character and the + that you want to match as many as there are. But, we can use the sub-expression substringing available; in order to do that you need to have sub-expressions; i.e. create groups where you want to split the string:
(SO#\s)(30\d+)
Put this in the function call and you have it:
regexp_substr(str, '(SO#\s)(30\d+)', 1, 1, 'i', 2)
SQL Fiddle