Is there a way to truncate a field value in CloudWatch? - amazon-cloudwatch

Suppose I have a simple Insights query like so
fields #timestamp, #message
Is there a way to truncate the #message field. For example, say I only want to skip the 1st 50 characters.
I know I can use the parse function but is there a simpler way, an substring equivalent that I can use in the fields line perhaps

There is a substr function:
Returns a substring from the index specified by the number argument to the end of the string. If the function has a second number argument, it contains the length of the substring to be retrieved. For example, substr("xyZfooxyZ",3, 3) returns "foo".
https://docs.aws.amazon.com/AmazonCloudWatch/latest/logs/CWL_QuerySyntax.html

Related

PostgreSQL - Substring to nth occurence of character

I'm trying to run a query that returns the substring of a value up to the nth occurence of a particular character.
For example:
This is the value abc|123|xyz098|password|xxxxx and I'd like to get abc|123|xyz098 So everything up to the third '|'. The number of characters between the delimiters changes, so I can't use a fixed index.
My thinking is to use SUBSTRING() with POSITION(). But I'm not having any success using it it to get to the nth occurence. Does anyone have any solutions?
You can use regexp_matches:
select regexp_matches('abc|123|xyz098|password|xxxxx', '\w+\|\w+\|\w+(?=\|)')
For a string result, this:
select split_part('abc|123|xyz098|password|xxxxx', split_part('abc|123|xyz098|password|xxxxx','|',4),1);

extract all numbers from start of string?

I have a table which contains some bad data I am trying to clean up.
An example of the fields is below
36234735HAN876
2342JOE9823
554444PUT003
What I want to do is remove all the numeric characters before the first alphabetical character so it would look like the below:
HAN876
JOE9823
PUT003
What would be the best way to achieve this? I have used the below method but this can only be used to extract ALL numeric from the string, not the ones before the alphabetical characters
How to get the numeric part from a string using T-SQL?
You could achieve this using PATINDEX to locate the first position of an alphabetical character in the string, and then use SUBSTRING to only return the characters after that position:
CREATE TABLE #temp (val VARCHAR(50));
INSERT INTO #temp VALUES ('36234735HAN876'), ('2342JOE9823'), ('554444PUT003'), ('TEST1234');
SELECT val,
SUBSTRING(val, PATINDEX('%[A-Z]%', val), LEN(val)) AS output
FROM #temp;
DROP TABLE #temp;
Outputs:
val output
36234735HAN876 HAN876
2342JOE9823 JOE9823
554444PUT003 PUT003
TEST1234 TEST1234
Note that I have created a temporary table with a column named val. You should change this to work with whatever the actual column is called.
About case sensitivity: If you are using a non-case sensitive collation this will work without issue. If your collation is case sensitive then you may need to alter the pattern being matched to cater for upper- and lower-case letters.
Use PATINDEX to find the first non-numeric character (or first alpha character, depending on the logic) and STUFF to remove them:
SELECT STUFF(V.YourString,1,ISNULL(NULLIF(PATINDEX('%[^0-9]%',V.YourString),0)-1,0),'')
FROM (VALUES('36234735HAN876'),
('2342JOE9823'),
('554444PUT003'),
('ABC123'))V(YourString)
If the logic is the first alpha character, instead of the first non-numeric, then the pattern would be [A-z].
The NULLIF and ISNULL are in there for when/if the string starts with a alpha/non-numeric and thus doesn't cause STUFF to error due to the 3rd parameter being -1. The is demonstrated with the additional example I put into the sample data ('ABC123').

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')

How to retrieve specific character positions within rows of database column using REGEX in Oracle SQL?

What Oracle SQL query could return the second, third and fourth positions of characters contained within rows of a specific column using the REGEXP_SUBSTR method instead of using SUBSTR method like my example provided below?
SELECT SUBSTR(city,2,3) AS "2nd, 3rd, 4th"
FROM student.zipcode;`
One way that works for me (with test data) is:
SELECT REGEXP_SUBSTR(city, '\S{3}', 2) AS partial FROM student.zipcode;
Note that this is set to find three non-whitespace characters beginning at the second position of the string.
You could also use:
SELECT REGEXP_SUBSTR(city, '.{3}', 2) AS partial FROM student.zipcode;
which will instead match any three characters in the 2nd to 4th position.
However, I'm not sure what advantage this has over simply:
SELECT SUBSTR(city,2,3) AS partial FROM student.zipcode;
The REGEXP_INSTR function is not what you want, as it returns an index (position number) for the search item in the searched string. You can read about it here: http://www.techonthenet.com/oracle/functions/regexp_instr.php

How to remove part of the string in oracle

Input data:
abcdef_fhj_viji.dvc
Expected output:
fhj_viji.dvc
The part to be trimmed is not constant.
Use the REPLACE method
Select REPLACE('abcdef_fhj_viji.dvc','abcde','')
If you want this query for your table :
Select REPLACE(column,'abcde','') from myTable
For update :
UPDATE TABLE
SET column = REPLACE(column,'abcde','')
select substr('abcdef_fhj_viji.dvc',instr('abcdef_fhj_viji.dvc','_')+1) from dual
So, Its all depends on INSTR function, define from which position and which occurrence, you will get the index and pass that index to SUBSTR to get your string.
Since you didn't give a lot of information I'm gonna assume some.
Let's assume you want a prefix of some string to be deleted. A good way to do that is by using Regular Expressions. There's a function called regexp_replace, that can find a substring of a string, depending on a pattern, and replace it with a different string. In PL/SQL you could write yourself a function using regexp_replace, like this:
function deletePrefix(stringName in varchar2) return varchar2 is
begin
return regexp_replace(stringName, '^[a-zA-Z]+_', '');
end;
or just use this in plain sql like:
regexp_replace(stringName, '^[a-zA-Z]+_', '');
stringName being the string you want to process, and the ^[a-zA-Z]+_ part depending on what characters the prefix includes. Here I only included upper- and lowercase letters.