I was wondering if there is a way to select the remaining characters from the varchar field if the field length is greater than 5.
For example:
ID_Num
99984
99923GF
99100EFK
99341LM
99145RL4C
What I am trying to get:
ID_Num
GF
EFK
LM
RL4C
STUFF is great for things like this:
SELECT STUFF(ID_Num,1,5,'')
FROM YourTable
WHERE LEN(ID_Num) > 5;
STUFF is used to "replace" characters within a string. In this case, it replaces the next 5 characters from position 1 (which is the start of the string), with the string ''; thus removing them.
Using RIGHT and LEN you can acheive your expected result.
SELECT RIGHT(ID_Num, LEN(ID_Num) - 5) AS ID_Num
FROM TestTable
WHERE LEN(ID_Num) > 5;
or using SUBSTRING
SELECT SUBSTRING(ID_Num, 6, LEN(ID_Num)) AS ID_Num
FROM TestTable;
Demo with sample data:
DECLARE #TestTable TABLE (ID_Num VARCHAR (20));
INSERT INTO #TestTable (ID_Num) VALUES
('99984'),
('99923GF'),
('99100EFK'),
('99341LM'),
('991456RL4C');
SELECT RIGHT(ID_Num, LEN(ID_Num) - 5) AS ID_Num
FROM #TestTable
WHERE LEN(ID_Num) > 5;
Output:
ID_Num
-------
GF
EFK
LM
6RL4C
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
I need to satisfy a condition in a string that has "ns[0-9]:" where [0-9] can be any number even greater than 10.
Example:
DECLARE #test TABLE ( value VARCHAR(20))
INSERT INTO #test VALUES
( 'ns1:'),
( 'NOT OK'),
( 'ns7:'),
( 'ns8:'),
( 'ns9:'),
( 'ns10:'),
( 'ns11:' )
SELECT *, PATINDEX( '%ns[0-9]:%', value ) passes
FROM #test
This only works on 1 to 9, not on 10 and above. I can use [0-9][0-9] but then it only works on 10 and above. I don't want a wild card between the number and the colon either.
I only want the following format to return a 1 with patindex
ns1:, ns2:, ns10:, ns11:, etc.
I also need a non-function solution. For performance reasons I want to use the string like functionality
Thanks
You can use:
select (case when value like 'ns[0-9]%:' and
value not like 'ns[0-9]%[^0-9]%:'
then 1 else 0
end) as passes_flag
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)
I'm trying to get the max value from a text field. All but two of the values are numbers with a single decimal. However, two of the values have something like 8.2.10. How can I pull back just the integer value? The values can go higher than 9.n, so I need to convert this field into a number so that I can get the largest value returned. So all I want to get back is the 8 from the 8.2.1.
Select cast(VERSION as int) is bombing out because of those two values with a second . in them.
You may derive by using regexp_substr with \d pattern :
with tab as
(
select regexp_substr('8.2.1', '\d', 1, 1) from dual
union all
select regexp_substr('9.0.1', '\d', 1, 1) from dual
)
select * from tab;
For Oracle you must attend the value as string for retire only the part before the dot. Ex:
SELECT NVL( SUBSTR('8.2.1',0, INSTR('8.2.1','.')-1),'8.2.1') AS SR FROM DUAL;
Check than the value is repeated 3 times in the sentence, and if the value is zero or the value didn't have decimal part then it will return the value as was set.
I had to use T-SQL rather PL/SQL, but the idea is the same:
DECLARE #s VARCHAR(10);
SELECT #s='8.2.1';
SELECT CAST(LEFT(#s, CHARINDEX('.', #s) - 1) AS INT);
returns the integer 8 - note that it won't work if there are no dots because it takes the part of the string to the left of the first dot.
If my quick look at equivalent functions was correct, then in Oracle that would end up as:
SELECT CAST(SUBSTR(VERSION, 1, INSTR(VERSION, '.') - 1) AS INT)
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