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.
Related
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 was practicing SQL injection skill, and I found that I could put = and LIKE in a single statement.
However, I'm not sure what does this mean and why it works?
SELECT 1 FROM users WHERE name='' LIKE '%'
So, what does that mean when I put = and LIKE in a statement, and when would I write something like this?
I am guessing that you are using MySQL, because this is syntactically correct in MySQL. It treats boolean types as numbers (which will be converted to integers and strings).
So, your code should be parsed as:
WHERE (name = '') LIKE '%'
This is because = and LIKE have the same precedence, and when operators have the same precedence, they are evaluated left-to-right (as explained in the documentation).
This, in turn evaluates to one of these three possibilities:
WHERE 1 LIKE '%' -- when name = ''
WHERE 0 LIKE '%' -- otherwise when name is not null
WHERE NULL like '%'
The first two will always evaluate to true. The third would discard any row where name is null.
(in MySQL and other popular DBMS) The LIKE operator is used to search for a specified pattern in a column. It admits "%" as a wildcard that represents zero, one, or multiple characters.
Your query always passes because the string '' meets this wildcard (zero characters). Incidentally, almost anything will. Some DBMS will react differently to such a query though.
I am working with Sybase SQL and want to exclude all entries that look like this:
(NOT PRESENT)
So I tried using:
SELECT col FROM table WHERE col NOT LIKE '(%)'
Do you guys know what is happening? I think I need to escap ( somehow, but I do not know how. The following returns an error:
SELECT col FROM table WHERE col NOT LIKE '\(%\)' ESCAPE '\'
Kind Regards
Try this :
SELECT col FROM table WHERE col NOT LIKE ('(%)')
You might find this helpful
Sybase Event Stream Processor 5.0 CCL Programmers Guide - String Functions
like()
Scalar. Determines whether a given string matches a specified pattern string.
Syntax
like ( string, pattern )
Parameters
string A string.
pattern A pattern of characters, as a string. Can contain wildcards.
Usage
Determines whether a string matches a pattern string. The function returns 1 if the string matches the pattern, and 0 otherwise. The pattern argument can contain wildcards: '_' matches a single arbitrary character, and '%' matches 0 or more arbitrary characters. The function takes in two strings as its arguments, and returns an integer.
Note: In SQL, the infix notation can also be used: sourceString like patternString.
Example
like ('MSFT', 'M%T') returns 1.
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.
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