I am trying to split a field by delimiter in LookML. This field either follows the format of:
Managers (AE)
Managers (AE - MM)
I was able to split to first case using this
sql: case
when rlike (${user_role_name}, '^.*[\\(\\)].*$') then split_part(${user_role_name}, ' ', -1)
However, I haven't been able to get the 2nd case to do the same. It's in a case statement so I am going to add another when statement, but am not able to figure out the regex for parentheses that contains spaces.
Thanks in advance for the help!
By "split" the string, I think you mean you want to extract the part in parentheses, right?
I would do this using a regex substring method. You didn't mention what warehouse you're using, and the syntax will vary a little, but on snowflake that would look like:
regexp_substr(${user_role_name}, '\\([^)]*\\)')
So, for example, with the inputs you gave:
select regexp_substr('Managers (AE)', '\\([^)]*\\)')
union all
select regexp_substr('Managers (AE - MM)', '\\([^)]*\\)')
result
(AE)
(AE - MM)
Related
CASE
WHEN <in_data> LIKE '[0-9][0-9][0-9][0-9][0-9][0-9]' THEN SUBSTR(<in_data>,1,3)
ELSE '000'
END
We're doing a migration project from Sybase to Teradata, and having a problem figuring this one out :) I'm still new to Teradata.
I would like to ask the equivalent TD code for this -
LIKE '[0-9][0-9][0-9][0-9][0-9][0-9]' to Teradata
Basically, it just checks whether the digits are numeric value.
Can someone give me a hint on this
You can also use REGEXP_SUBSTR to directly extract the three digits:
COALESCE(REGEXP_SUBSTR(in_data,'^[0-9]{3}(?=[0-9]{3}$)'), '000')
This looks for the first three digits and then does a lookahead for three following digits without adding them to the overall match.
^ indicates the begin of the string, '$' the end, so there are no other characters before or after the six digits. (?=...) is a so-called "lookahead", i.e. those three digits are checked, but ignored.
If there's no match the regex returns NULL which is changed to '000'.
You need to use regexp instead of like, since [0-9][0-9][0-9][0-9][0-9][0-9] is a regular expression.
To do an exact match, you need to add anchors. ie, to match the string which contains an exact 6 digit chars.
regexp '^[0-9]{6}$'
or
regexp '^[[:digit:]]{6}$'
I have strings like 'keepme:cutme' or 'string-without-separator' which should become respectively 'keepme' and 'string-without-separator'. Can this be done in PostgreSQL? I tried:
select substring('first:last' from '.+:')
But this leaves the : in and won't work if there is no : in the string.
Use split_part():
SELECT split_part('first:last', ':', 1) AS first_part
Returns the whole string if the delimiter is not there. And it's simple to get the 2nd or 3rd part etc.
Substantially faster than functions using regular expression matching. And since we have a fixed delimiter we don't need the magic of regular expressions.
Related:
Split comma separated column data into additional columns
regexp_replace() may be overload for what you need, but it also gives the additional benefit of regex. For instance, if strings use multiple delimiters.
Example use:
select regexp_replace( 'first:last', E':.*', '');
SQL Select to pick everything after the last occurrence of a character
select right('first:last', charindex(':', reverse('first:last')) - 1)
How can i query a column with Names of people to get only the names those contain exactly 2 “a” ?
I am familiar with % symbol that's used with LIKE but that finds all names even with 1 a , when i write %a , but i need to find only those have exactly 2 characters.
Please explain - Thanks in advance
Table Name: "People"
Column Names: "Names, Age, Gender"
Assuming you're asking for two a characters search for a string with two a's but not with three.
select *
from people
where names like '%a%a%'
and name not like '%a%a%a%'
Use '_a'. '_' is a single character wildcard where '%' matches 0 or more characters.
If you need more advanced matches, use regular expressions, using REGEXP_LIKE. See Using Regular Expressions With Oracle Database.
And of course you can use other tricks as well. For instance, you can compare the length of the string with the length of the same string but with 'a's removed from it. If the difference is 2 then the string contained two 'a's. But as you can see things get ugly real soon, since length returns 'null' when a string is empty, so you have to make an exception for that, if you want to check for names that are exactly 'aa'.
select * from People
where
length(Names) - 2 = nvl(length(replace(Names, 'a', '')), 0)
Another solution is to replace everything that is not an a with nothing and check if the resulting String is exactly two characters long:
select names
from people
where length(regexp_replace(names, '[^a]', '')) = 2;
This can also be extended to deal with uppercase As:
select names
from people
where length(regexp_replace(names, '[^aA]', '')) = 2;
SQLFiddle example: http://sqlfiddle.com/#!4/09bc6
select * from People where names like '__'; also ll work
I require a select query that adds a space to the data based on the placement of the capital letters i.e. 'HelpMe' using this query would be displayed as 'Help Me' . Note i cannot use a stored function to do this the it must be done in the query itself. The Data is of variable length and query must be in SQL. Any Help will be appreciated.
Thanks
You need to use user defined function for this until MS give us support for regular expressions. Solution would be something like:
SELECT col1, dbo.RegExReplace(col1, '([A-Z])',' \1') FROM Table
Aldo this would produce leading space that you can remove with TRIM.
Replace regular expresion function:
http://connect.microsoft.com/SQLServer/feedback/details/378520
About dbo.RegexReplace you can read at:
TSQL Replace all non a-z/A-Z characters with an empty string
Assume if you are using Oracle RDBMS, you use the following,
REGEX_REPLACE
SELECT REGEXP_REPLACE('ILikeToWatchCSIMiami',
'([A-Z.])', ' \1')
AS RX_REPLACE
FROM dual
;
Managed to get this output: * SQLFIDDLE
But as you see it doesn't treat well on words such as CSI though.
Does anyone have a LIKE pattern that matches whole words only?
It needs to account for spaces, punctuation, and start/end of string as word boundaries.
I am not using SQL Full Text Search as that is not available. I don't think it would be necessary for a simple keyword search when LIKE should be able to do the trick. However if anyone has tested performance of Full Text Search against LIKE patterns, I would be interested to hear.
Edit:
I got it to this stage, but it does not match start/end of string as a word boundary.
where DealTitle like '%[^a-zA-Z]pit[^a-zA-Z]%'
I want this to match "pit" but not "spit" in a sentence or as a single word.
E.g. DealTitle might contain "a pit of despair" or "pit your wits" or "a pit" or "a pit." or "pit!" or just "pit".
Full text indexes is the answer.
The poor cousin alternative is
'.' + column + '.' LIKE '%[^a-z]pit[^a-z]%'
FYI unless you are using _CS collation, there is no need for a-zA-Z
you can just use below condition for whitespace delimiters:
(' '+YOUR_FIELD_NAME+' ') like '% doc %'
it works faster and better than other solutions. so in your case it works fine with "a pit of despair" or "pit your wits" or "a pit" or "a pit." or just "pit", but not works for "pit!".
I think the recommended patterns exclude words with do not have any character at the beginning or at the end. I would use the following additional criteria.
where DealTitle like '%[^a-z]pit[^a-z]%' OR
DealTitle like 'pit[^a-z]%' OR
DealTitle like '%[^a-z]pit'
I hope it helps you guys!
Surround your string with spaces and create a test column like this:
SELECT t.DealTitle
FROM yourtable t
CROSS APPLY (SELECT testDeal = ' ' + ISNULL(t.DealTitle,'') + ' ') fx1
WHERE fx1.testDeal LIKE '%[^a-z]pit[^a-z]%'
If you can use regexp operator in your SQL query..
For finding any combination of spaces, punctuation and start/end of string as word boundaries:
where DealTitle regexp '(^|[[:punct:]]|[[:space:]])pit([[:space:]]|[[:punct:]]|$)'
Another simple alternative:
WHERE DealTitle like '%[^a-z]pit[^a-z]%' OR
DealTitle like '[^a-z]pit[^a-z]%' OR
DealTitle like '%[^a-z]pit[^a-z]'
This is a good topic and I want to complement this to someone how needs to find some word in some string passing this as element of a query.
SELECT
ST.WORD, ND.TEXT_STRING
FROM
[ST_TABLE] ST
LEFT JOIN
[ND_TABLE] ND ON ND.TEXT_STRING LIKE '%[^a-z]' + ST.WORD + '[^a-z]%'
WHERE
ST.WORD = 'STACK_OVERFLOW' -- OPTIONAL
With this you can list all the incidences of the ST.WORD in the ND.TEXT_STRING and you can use the WHERE clausule to filter this using some word.
You could search for the entire string in SQL:
select * from YourTable where col1 like '%TheWord%'
Then you could filter the returned rows client site, adding the extra condition that it must be a whole word. For example, if it matches the regex:
\bTheWord\b
Another option is to use a CLR function, available in SQL Server 2005 and higher. That would allow you to search for the regex server-side. This MSDN artcile has the details of how to set up a dbo.RegexMatch function.
Try using charindex to find the match:
Select *
from table
where charindex( 'Whole word to be searched', columnname) > 0