This one matches column_name like 'CharEndsHere%'
and
This one doesn't column_name like 'CharEndsHere'
I know that like operator will consider even the trailing spaces, so I just copied the exact column value (with trailing spaces) and pasted it.
Something like column_name like 'CharEndsHere ' yet it doesn't match -- why?.
I haven't used '=' operator since the columns type is ntext
Is there something I am missing here or shouldn't I use like operator in this way?
Edited : column_name like 'CharEndsHere__' (__ denoted the spaces) 'CharEndsHere ' is the exact value in that cell, using like in this way valid or no?
Edit :
This is the code I tried,
SELECT *
FROM [DBName].[dbo].[TableName]
WHERE [DBName].[dbo].[TableName].Address1 LIKE rtrim('4379 Susquehanna Trail S ')
I have also tried without using rtrim, yet the same result
Edit: According to Blindy's answer,
If a comparison in a query is to return all rows with the string LIKE 'abc' (abc
without a space), all rows that start with abc and have zero or more trailing
blanks are returned.
But in my case, I have queried, Like 'abc' and there is a cell containing 'abc '(with trailing spaces) which is not returned. That's my actual problem
This is a case of reading the documentation, it's very explicitly stated here: http://msdn.microsoft.com/en-us/library/ms179859.aspx
When you perform string comparisons by using LIKE, all characters in the pattern string are significant. This includes leading or trailing spaces. If a comparison in a query is to return all rows with a string LIKE 'abc ' (abc followed by a single space), a row in which the value of that column is abc (abc without a space) is not returned. However, trailing blanks, in the expression to which the pattern is matched, are ignored. If a comparison in a query is to return all rows with the string LIKE 'abc' (abc without a space), all rows that start with abc and have zero or more trailing blanks are returned.
Edit: According to your comments, you seem to be looking for a way to use like while ignoring trailing spaces. Use something like this: field like rtrim('abc '). It will still use indexes because rtrim() is a scalar operand and it's evaluated before the lookup phase.
Related
I have a db table in oracle where I have a column defined as a number.
The columns contains numbers like:
MyColumn
12540000000
78590000000
I want to find the records by searching MyColumn=12540000000 as well as MyColumn=1254 (without trailing zeros).
What could I try? TO_CHAR and a cutting logic or is there something more simple?
rtrim(MyColumn, '0') = '1254'
Note that on the right I enclosed the string within quotes (so it is really seen as a string, not a number). Apparently you are treating these as strings, right? Even if MyColumn is a number, it will be implicitly converted to a string before applying rtrim.
I am using Snowflake SQL. I would like to remove characters from a string after a special character ~. How can I do that?
here is the whole scenario. Let me explain. I do have a string like 'CK#123456~fndkjfgdjkg'. Now, i want only the number after #.And not anything after ~. This is number length varies for that field value. It might be 1 or 5 or 3. And i want to add the condition in where class where this number is equal to check_num from other table after joining. I am trying REGEXP_SUBSTR(A.SRC_TXT, '(?<=CK#)(.+?\b)') = C.CHK_NUM in the where condition. I am getting the error as 'No repititive argument after ?'
You can use a regex for this
-- To remove just the character after a ~
select regexp_replace('fo~o bar','~.', '');
-- returns 'fo bar'
--If you want to keep the ~
select regexp_replace('fo~o bar','~.', '~');
-- returns 'fo~ bar'
--If you want to remove everything after the ~
select regexp_replace('fo~o bar','~.*', '');
--returns 'fo'
If you need to remove other specific character sets after a ~, you can probably do this with a slightly more complicated regex, but I'd need examples of your desired input/output to help with that.
EDIT for updated question
This regex replace should get what you need.
select regexp_replace('CK#123456~fndkjfgdjkg','CK#(\\d*)~.*', '\\1');
-- returns 123456
(\\d*) gets ANY number of digits in a row, and the \\1 causes it to replace the match with what was in the first set of parenthesis, which is your list of digits. the CK# and ~.* are there to make sure the whole string gets matched and replaced.
If the CK# can vary as well, you can use .*? like this.
select regexp_replace('ABCD123HI#123456~fndkjfgdjkg','.*?#(\\d*)~.*', '\\1')
-- returns 123456
I'd probably do something like the following, easy enough but not as cool as RegEx type of functions.
set my_string='fooo~12345';
set search_for_me = '~';
SELECT SUBSTR($my_string, 1, DECODE(position($search_for_me, $my_string), 0, length($my_string), position($search_for_me, $my_string)));
I hope this helps...Rich
It looks like lookahead and lookbehinds are not supported in REGEXP functions, they seem to work in the PATTERN clause of a LIST command. Snowflake documentation makes no mention either way of lookahead or lookbehinds.
In your example:
It seems that the query engine is looking for that repeating argument, where you are attempting a lookbehind
You have not specified what you wanted extracted. You have two capture groups, but in this scenario everything would be returned
Since you are looking to remove everything after ~ you have a delimiter, why not use it in your REGEXP_SUBSTR function?
Try the following:
SELECT $1,REGEXP_SUBSTR($1,'\\w+#(.+?)~',1,1,'is',1)
FROM VALUES
('CK#123456~fndkjfgdjkg')
,('QH#128fklj924~fndkjfgdjkg')
;
This looks for:
One or more word characters
Followed by #
Capturing one or more characters upto and not including ~
Returns the characters within the capture group
You can change the .+? to \\d+? to make sure the pattern is only digits. Backslashes must be escaped with a backslash.
The descriptions for each argument of the function can be found here:
https://docs.snowflake.net/manuals/sql-reference/functions/regexp_substr.html
You could check this!!
select substr('CK#123456~fndkjfgdjkg',4,6) from dual;
OUTPUT
123456
https://docs.snowflake.net/manuals/sql-reference/functions/substr.html
I need to find rows where the phone number field contains unexpected characters.
Most of the values in this field look like:
123456-7890
This is expected. However, we are also seeing character values in this field such as * and #.
I want to find all rows where these unexpected character values exist.
Expected:
Numbers are expected
Hyphen with numbers is expected (hyphen alone is not)
NULL is expected
Empty is expected
Tried this:
WHERE phone_num is not like ' %[0-9,-,' ' ]%
Still getting rows where phone has numbers.
from https://regexr.com/3c53v address you can edit regex to match your needs.
I am going to use example regex for this purpose
select * from Table1
Where NOT REGEXP_LIKE(PhoneNumberColumn, '^[+]*[(]{0,1}[0-9]{1,4}[)]{0,1}[-\s\./0-9]*$')
You can use translate()
...
WHERE translate(Phone_Number,'a1234567890-', 'a') is NOT NULL
This will strip out all valid characters leaving behind the invalid ones. If all the characters are valid, the result would be NULL. This does not validate the format, for that you'd need to use REGEXP_LIKE or something similar.
You can use regexp_like().
...
WHERE regexp_like(phone_num, '[^ 0123456789-]|^-|-$')
[^ 0123456789-] matches any character that is not a space nor a digit nor a hyphen. ^- matches a hyphen at the beginning and -$ on the end of the string. The pipes are "ors" i.e. a|b matches if pattern a matches of if pattern b matches.
Oracle has REGEXP_LIKE for regex compares:
WHERE REGEXP_LIKE(phone_num,'[^0-9''\-]')
If you're unfamiliar with regular expressions, there are plenty of good sites to help you build them. I like this one
I have a column that has values that look like the following:
17_data...
18_data...
1801151...data
The data isn't the cleanest in this columns, so I am trying to use a LEFT function to identify the rows that have the 2017 year followed by an underscore LEFT(column, 3) = '17[_]' This doesn't return a single column. So to troubleshoot, I added this WHERE clause to the SELECT statement to see what was getting returned, and I got the value 175 where the actual first three characters are "17_".
Why is this, and how can I structure my WHERE clause to pick up those rows?
When you tried adding 'where' with a rule of LEFT(column, 3) = '17[_]', it was doomed to fail. Operator '=' performs exact comparison: both sides must be equal. That is, it would look for rows whose first 3 characters (left,3) are equal to 17[_], that is, 5 characters, one, seven, bracket, underscore, bracket. Text of 3 characters will not exactly-match 5 characters, ever.
You should have written simply:
WHERE LEFT(column, 3) = '17_'
I guess that you've got the idea for adding a bracket from reading about LIKE patterns. LIKE operator allows you to look for strings contained at start/end/middle of the data.
WHERE column LIKE 'mom%' - starts with mom
WHERE column LIKE '%dad' - ends with dad
and so on. LIKE supports '%' meaning "and then text of any length", and also "_" meaning "and then just one character". This forms a problem: when you want to say "starts with _mom", you cannot write
WHERE column LIKE '_mom%'
because it would also match 9mom, Bmom, and so on, due to _ meaning 'any single character'. That's why in such cases, only in LIKE, you have to write the underscore in brackets:
WHERE column LIKE '[_]mom%' - starts with _mom
Knowing that, it's obvious that you could construct your 'starts with 17_' with LIKE as well:
SELECT column1, column2, ..., columnN
FROM sometable
WHERE column LIKE '17[_]%'
I need a like expression that will match a character whether or not it exists. It needs to match the following values:
..."value": "123456"...
..."value": "123456"...
"...value":"123456"...
This like statement will almost work: LIKE '%value":%"123456"%'
But there are values like this one that would also match, but I don't want returned:
..."value":"99999", "other":"123456"...
A regex expression to do what I'm looking to do is 'value": *?"123456"'. I need to do this in SQL Server 2008 and I don't believe there is good regex support in that version. How can I match using a like statement?
Remove the whitespace in your compare with REPLACE():
WHERE REPLACE(column,' ','') LIKE '%"value":"123456"%'
May need a double replace for tabs:
REPLACE(REPLACE(column,' ',''),' ','')
I don't think you can with the like operator. You could exclude ones you could match, like if you want to make sure it just doesn't contain other:
[field] LIKE '%value":%"123456"%` AND [field] NOT LIKE '%"other"%'
Otherwise I think you'd have to do some processing on the string. You could write a UDF to take the string and parse it to find the value for 'value' and compare based on that:
dbo.fn_GetValue([field], 'value') = '123456'
The function could find the index of '"' + #name + '"', find the next index of a quote, and the one after that, then get the string between those two quotes and return it.