I am trying to formulate a query that will allow me to find all records from a single column with 3 hyphens. An example of a record would be like XXXX-RP-XXXAS1-P.
I need to be able to sort through 1000s of records with either 2 or 3 hyphens.
You can REPLACE the hyphens in the string with an empty string and compute the difference of the length of original string and the replaced string to check for the number of hyphens.
select *
from yourtable
where len(column_name)-len(replace(column_name,'-',''))=3
and substring(column_name,9,1) not like '%[0-9]%'
If your records have 2 or 3 hyphens, then just do:
where col like '%-%-%-%'
This will get 3 or more hyphens. For exactly 3:
where col like '%-%-%-%' and col not like '%-%-%-%-%'
try this,
declare #t table(col1 varchar(50))
insert into #t values ('A-B'),('A-B-C-D-E'),('A-B-C-D')
select * from
(SELECT *
,(len(col1) - len(replace(col1, '-', ''))
/ len('-')) col2
FROM #T)t4
where col2=3
Related
So I have results that begins with 2 letters followed by 3 numbers, for example:
ID_Sample
AB001
BC003
AB100
BC400
How can I do a query that ignores the letters and just looks up the numbers to do a filter? For example:
WHERE ID_Sample >= 100
I tried using a "Replace" to get rid of known letters, but I figured there might be a better way. For example:
Select
Replace(id_sample,'AB','')
Choosing the 3 numerals on the right would work too.
For your sample data, you can just start at the third character and convert to a number:
where try_convert(int, stuff(ID_Sample, 1, 2, '')) > 100
Or, if you know that the number is 3 characters:
where try_convert(int, right(ID_Sample, 3)) > 100
+1 for Gordon's answer. This is a fun problem that you can solve using TRANSLATE if you're using SQL 2017+.
First, in case you've never used it, Per BOL TRANSLATE:
Returns the string provided as a first argument after some characters
specified in the second argument are translated into a destination set
of characters specified in the third argument.2
This:
SELECT TRANSLATE('123AABBCC!!!','ABC','XYZ');
Returns: 123XXYYZZ!!!
Here's the solution using TRANSLATE:
-- Sample Data
DECLARE #t TABLE (ID_Sample CHAR(6))
INSERT #t (ID_Sample) VALUES ('AB001'),('BC003'),('AB100'),('BC400'),('CC555');
-- Solution
SELECT
ID_Sample = t.ID_Sample,
ID_Sample_Int = s.NewString
FROM #t AS t
CROSS JOIN (VALUES('ABCDEFGHIJKLMNOPQRSTUVWXYZ', REPLICATE(0,26))) AS f(S1,S2)
CROSS APPLY (VALUES(TRY_CAST(TRANSLATE(t.ID_Sample,f.S1,f.S2) AS INT))) AS s(NewString)
WHERE s.NewString >= 100;
Without the WHERE clause filter you get:
ID_Sample ID_Sample_Int
--------- -------------
AB001 1
BC003 3
AB100 100
BC400 400
CC555 555
... the WHERE clause filters out the first two rows.
Check these methods- Unit test also done!
Declare #Table as table(ID_Sample varchar(20))
set nocount on
Insert into #Table (ID_Sample)
Values('AB001'),('BC003'),('AB100'),('BC400')
--substring_method
select * from #Table
where try_cast(substring(ID_Sample,3,3) as int) >100
--right_method
select * from #Table
where try_cast(right(ID_Sample,3) as int) >100
--stuff_method
select * from #Table
where try_cast(stuff(ID_Sample,1,2,'') as int) >100
--replace_method
select * from #Table
where try_cast(replace(ID_Sample,left(ID_Sample,2),'') as int) >100
How can I select records where in the column Value the 5th character is letter A?
For example the following records:
ID Value
-------------------------
1 1234A5636A6363
2 1234A4343B6363
3 1234B5353A6363
if I run
select * from table
where Value like '%A%'
this will return all records
but all I want is the first 2 where the 5th character is A, regardless if there are more A characters in the text or not
select *
from your_table
where substring(Value, 5, 1) = 'A'
The LIKE operator, in addition to %, which matches any number of any character, can use _, which matches any one single character. You may try:
SELECT *
FROM yourTable
WHERE Value LIKE '____A%'; -- 4 underscores here
use like below by using _(underscore)
LIKE '____A%'
SQL Server
select *
from YourTableName
where CHARINDEX('A', ColumnName) = 5
Note:- This finds where string 'A' starts at position 5
AND specify Your ColumnName
Hi I have one doubt in sql server .
how to get first position to right side specific character position.
table : empfiles
filename:
ab_re_uk_u_20101001
ax_by_us_19991001
abc_20181002
I want output like below:
filename
ab_re_uk_u
ax_by_us
abc
I tried like below :
select SUBSTRING(filename,1,CHARINDEX('2',filename) - 1) as filename from empfiles
above query is not given expected result please tell me how to write query to achive this task in sql server .
If last position has always numeric values then you can use patindex():
select *, substring(filename, 1, patindex('%[0-9]%', filename)-2) as NewFile
from empfiles e;
If you want to get characters after than _ to right sight of string then you can use combo to reverse() and substring()
select *,
reverse(substring(reverse(filename),charindex('_', reverse(filename))+1, len(filename)))
from empfiles e;
Another way is to use reverse in combination with STUFF.
create table f(filename nvarchar(100));
insert into f values
('ab_re_uk_u_20101001')
,('ax_by_us_19991001')
,('abc_20181002');
select
filename=reverse(stuff(reverse(filename),1,charindex('_',reverse(filename)),''))
from f
Try This
CREATE TABLE #DATA([FILENAME] NVARCHAR(100));
INSERT INTO #DATA VALUES
('ab_re_uk_u_20101001')
,('ax_by_us_19991001')
,('abc_20181002');
SELECT [filename],
SUBSTRING([filename],0,PATINDEX('%[0-9]%',[filename])-1) AS ExpectedResult
FROM #Data
Result
filename ExpectedResult
--------------------------------------
ab_re_uk_u_20101001 ab_re_uk_u
ax_by_us_19991001 ax_by_us
abc_20181002 abc
Well, obviously the last position value is a date, and the format is YYYYMMDD so its 8 characters, plus, added by underscore character, so that makes its 9 character.
Assumed by the above statement applied, the following logic of the query should work
SELECT SUBSTRING(ColumnText, 1, LEN(ColumnText) - 9)
Which means, only display characters from character position 1, to character position LEN - 9, which LEN is the length of characters, and 9 is the last 9 digit of number to be removed
Try with this ..
select [filename],SUBSTRING([filename],1,PATINDEX('%_[0-9]%',[filename])-1) from empfiles
Individual Select records
SELECT SUBSTRING('ab_re_uk_u_20101001',1,PATINDEX('%_[0-9]%','ab_re_uk_u_20101001')-1)
SELECT SUBSTRING('ax_by_us_19991001',1,PATINDEX('%_[0-9]%','ax_by_us_19991001')-1)
SELECT SUBSTRING('abc_20181002',1,PATINDEX('%_[0-9]%','abc_20181002')-1)
Is there a way to use Pattern Matching with SQL LIKE, to match a variable number of characters with an upper limit?
For example, I have one column which can have "correct values" of 2-10 numbers, anything more than 10 and less than 2 is incorrect.
If you really want to use like you can do:
where col like '__' or col_like '___' or . . .
or:
where col like '__%' and -- at least two characters
col not like '_________%' -- not 9 characters
The more typical method would be:
where len(col) between 2 and 10
If you want to ensure they are numbers:
where len(col) between 2 and 10 and
col not like '%[^0-9]%'
You could make a function to do this the long way of inspecting each character like below:
DECLARE #TempField varchar(max)= '1v3Abc567'
DECLARE #FieldLength int = LEN(#TempField)
DECLARE #CountNumeric int = 0
WHILE #FieldLength > 0
BEGIN
SELECT #CountNumeric = #CountNumeric + ISNUMERIC( SUBSTRING (#TempField,#FieldLength,1))
SET #FieldLength = #FieldLength - 1
END
SELECT #CountNumeric NumericCount
You can use Regex.
SELECT * FROM mytable WHERE mycol LIKE '%[0-9]{2,10}%';
should do it (that's off the top of my head, so double-check!
I need to create a SQL Query.
This query need to select from a table where a column contains regular expression.
For example, I have those values:
TABLE test (name)
XHRTCNW
DHRTRRR
XHRTCOP
CPHCTPC
CDDHRTF
PEOFOFD
I want to select all the data who have "HRT" after 1 char (value 1, 2 and 3 - Values who looks like "-HRT---") but not those who might have "HRT" after 1 char (value 5).
So I'm not sure how to do it because a simple
SELECT *
FROM test
WHERE name LIKE "%HRT%"
will return value 1, 2, 3 and 5.
Sorry if I'm not really clear with what I want/need.
You can also change the pattern. Instead of using % which means zero-or-more anything, you can use _ which means exactly one.
SELECT * FROM test WHERE name like '_HRT%';
You can use substring.
SELECT * FROM test WHERE substring(name from 2 for 3) = 'HRT'
Are the names always 7 letters? Do:
SELECT substring (2, 4, field) from sometable
That will just select the 2-4th characters and then you can use like "%HRT"