I need to escape special character from the pattern and skip last four digits:
08_27_19-13_00_34
Output:082719
Can anyone guide me how to use regexp_extract for this in impala?
Also can anyone please suggest links of article so that I can thoroughly understand regular expression working.
You could just use a combination of REPLACE and LEFT here:
SELECT
LEFT(REPLACE(col, '_', ''), 6) AS first_six
FROM yourTable;
The strategy here is to just remove the underscores, then take the first remaining six characters, which should be first six digits. You could also consider adding logic which checks if the col value would even match your pattern, but that would be up to you.
Related
I have a column that's string and I would like to use a WHERE statement for all strings that begin with the "sea_"
Any ideas on how to achieve that will be appreciated.
Thanks in advance
It's not clear what exactly your where clause should do. If necessary, you should please add more information. Anyway, in general, you can do such a select:
SELECT column FROM table WHERE column LIKE 'sea_%'
The % means that zero, one or more characters can follow.
If you need further conditions like an exact number of trailing characters etc., you please must specify your question.
If you like to ignore Upper/Lower case , it can help you
SELECT * FROM table WHERE LOWER(column) LIKE 'sea_%'
Be aware that underscore _ is itself a placeholder for any single character which must be escaped in order to use it literally.
The criteria you should use to ensure an underscore is also present is
where column like 'sea[_]%';
You can also define your own explicit escape character, such as
where column like 'sea#_%' escape '#';
I am using Firebird 2.5 and I have a field (called identifier) with mixed letters, numbers and special characters. I would like to use regex to extract only the numbers in a new column. I have tried something like below, but it is not working.
Any idea how I can achieve this using regex without using stored procedures or execute block
SELECT ORDER_ID,
ORDER_DATE,
SUBSTRING(IDENTIFIER FROM 1 TO 10) SIMILAR TO '^[0-9]{10}$' --- DESIRED EXTRACTION COLUMN
FROM ORDERS
Example of data
IDENTIFIER DESIRED OUTPUT
ANDRE 02869567995 02869567995
02869567995 MARIA 02869567995
028.695.67.995 02869567995
028695679-95 02869567995
You cannot do this in Firebird 2.5, at least not without help from a UDF, or a (selectable) stored procedure. I'm not aware of third-party UDFs providing regular expressions, so you might have to write this yourself.
In Firebird 3.0, you could also use a UDR or stored function to achieve this. Unfortunately, using the regular expression functionality available in Firebird alone will not be enough to solve this.
NOTE: The rest of the answer is based on the assumption to extract digits if the first 10 characters of string are digits. With the updated question, this assumption is no longer valid.
That said, if your need is exactly as shown in your question, that is only extract the first 10 characters from a string if they are all digits, then you could use:
case
when IDENTIFIER similar to '[[:DIGIT:]]{10}%'
then substring(IDENTIFIER from 1 for 10)
end
(as an aside, the positional SUBSTRING syntax is from <start> for <length>, not from <start> to <end>)
In Firebird 3.0 and higher, you can use SUBSTRING(... SIMILAR ...) with a SQL regular expression pattern. Assuming you want to extract 10 digits from the start of a string, you can do:
substring(IDENTIFIER similar '#"[[:DIGIT:]]{10}#"%' escape '#')
The #" delimits the pattern to extract (where # is a custom escape character as specified in the ESCAPE clause). The remainder of the pattern must match the rest of the string, hence the use of % here (in other cases, you may need to specify a pattern before the first #" as well.
See this dbfiddle for an example.
It is not possible in any version of Firebird.
I have data like "11223311" and I want all the multiple occurrence to be replaced by single occurrence i.e. the above should turn into '123'. I am working in SAP HANA.
But by using below logic I am getting
'1231' from '11223311'.
SELECT REPLACE_REGEXPR('(.)\1+' IN '11223331' WITH '\1' OCCURRENCE ALL) FROM DUMMY;
Your regular expression only replaces multiple consecutive occurrences of characters; that's what the \1+ directly after it's matching (.) is doing.
You can use look-ahead to remove all characters that also occur somewhere after that match. Note that this keeps the last occurrence, not the first:
SELECT REPLACE_REGEXPR('(.)(?=.*\1)' IN '11223331' WITH '' OCCURRENCE ALL) FROM DUMMY
This returns: 231
If you want to keep the first occurrence, I don't see a possibility just with one regex (I could be wrong though). Using a look-behind in the same way does not work because it would need to be variable-length, which is not supported in HANA and most other implementations. Often \K is recommended as alternative, but something like (.).*\K\1 wouldn't work with replace all, because all characters before \K are still consumed in replace. If you could run the same regex in a loop, it could work but then why not use a non-regex loop (like a user-defined HANA function) in the first place.
Please try this
SELECT REPLACE_REGEXPR(concat(concat('[^','11223331'),']') IN '0123456789' WITH '' OCCURRENCE ALL)
FROM DUMMY;
In Amazon Redshift tables, I have a string column from which I need to extract numbers only out. For this currently I use
translate(stringfield, '0123456789'||stringfield, '0123456789')
I was trying out REPLACE function, but its not gonna be elegant.
Any thoughts with converting the string into ASCII first and then doing some operation to extract only number? Or any other alternatives.
It is hard here as Redshift do not support functions and is missing lot of traditional functions.
Edit:
Trying out the below, but it only returns 051-a92 where as I need 05192 as output. I am thinking of substring etc, but I only have regexp_substr available right now. How do I get rid of any characters in between
select REGEXP_SUBSTR('somestring-051-a92', '[0-9]+..[0-9]+', 1)
might be late but I was solving the same problem and finally came up with this
select REGEXP_replace('somestring-051-a92', '[a-z/-]', '')
alternatively, you can create a Python UDF now
Typically your inputs will conform to some sort of pattern that can be used to do the parsing using SUBSTRING() with CHARINDEX() { aka STRPOS(), POSITION() }.
E.g. find the first hyphen and the second hyphen and take the data between them.
If not (and assuming your character range is limited to ASCII) then your best bet would be to nest 26+ REPLACE() functions to remove all of the standard alpha characters (and any punctuation as well).
If you have multibyte characters in your data though then this is a non-starter.
Better method is to remove all the non-numeric values:
select REGEXP_replace('somestring-051-a92', '[^0-9]', '')
You can specify "any non digit" that includes non-printable, symbols, alpha, etc.
e.g., regexp_replace('brws--A*1','[\D]')
returns
"1"
I have this regular expression in my sql query
DECLARE #RETURN_VALUE VARCHAR(MAX)
IF #value LIKE '%[0-9]%[^A-Z]%[0-9]%'
BEGIN
SET #RETURN_VALUE = NULL
END
I am not sure, but whenever I have this in my row 12 TEST then it gives me the value of 12, but if I have three digit number then it filters out the three digit numbers.How can I modify the regular expression to return me the three digits numbers too.
any help will be appreciated.
SQL doesn't have regular expressions: it has SQL wildcard expressions. They are much simpler than regular expressions and long predate regular expressions. For instance, there is no way to specify alternation (a|b) or repetition ( a*, a+, a?, a{m,n} ) such as you might find in a regular expression.
The 'like expression' that you have
LIKE '%[0-9]%[^A-Z]%[0-9]%'
will match any string containing the following pattern anywhere in the string
zero or more of any character, followed by...
a single decimal digit, followed by...
zero or more of any character, followed by...
a single character other than A–Z (whether it's case sensitive or not depends on the collating sequence in use), followed by...
zero or of any character, followed by...
a single decimal digit, followed by...
zero or more of any character
One should note that the % is likely to match perhaps more than you might like.
Have you tried ([0-9]*). I believe that this will capture every digit for you. However, I am not as strong at regex. When I ran this through rubular, it worked, though :) BTW, rubular is a great way to test out regular expressions
You can easily create a SQL CLR function and use this in your queries. Visual Studio has a project template for this and makes deploying the functions a snap.
Here is more information from Microsoft about how to create the function and how to use it (for boolean matches and for data extraction).
First of all, note that this is not really a "regular expression", it's a SQL-specific form of wildcard matching. You are very limited in what you can accomplish with SQL wildcards. As one example, you cannot "optionally" match a specific character or character set.
Your expression, as you've written it, will match any value that contains two digits with at least one non-letter character in between them, meaning it will match:
111
1^1
1?7
1AAAAAAAAAAA?AAAAAAAAA1
-----------------------5-----------------3-------
And infinitely more items of a similar structure.
Oddly, one string that would not match this pattern is "12 TEST" because there is no character between the 1 and 2. The pattern also won't "give you" the value of 12 back because it's not a parsing expression, just a matching expression: it returns 1 (true) or 0 (false).
There is clearly something else going on in your application, possibly even an actual regular expression, but it has nothing to do with the SQL you've included here.