I want to compare and select a field from DB using Like keyword or any other technique.
My query is the following:
SELECT * FROM Test WHERE name LIKE '%xxxxxx_Ramakrishnan_zzzzz%';
but my fields only contain 'Ramakrishnan'
My Input string contain some extra character xxxxxx_Ramakrishnan_zzzzz
I want the SQL query for this. Can any one please help me?
You mean you want it the other way round? Like this?
Select * from Test where 'xxxxxx_Ramakrishnan_zzzzz' LIKE '%' + name + '%';
You can use the MySQL functions, LOCATE() precisely like,
SELECT * FROM WHERE LOCATE("Ramakrishnan",input) > 0
Are the xxxxxx and zzzzz bits always 6 and 5 characters? If so, then this is doable with a bit of string cutting.
with Test (id,name) as (
select 1, 'Ramakrishnan'
union
select 2, 'Coxy'
union
select 3, 'xxxxxx_Ramakrishnan_zzzzz'
)
Select * from Test where name like '%'+SUBSTRING('xxxxxx_Ramakrishnan_zzzzz', 8, CHARINDEX('_',SUBSTRING('xxxxxx_Ramakrishnan_zzzzz',8,100))-1)+'%'
Results in:
id name
1 Ramakrishnan
3 xxxxxx_Ramakrishnan_zzzzz
If they are variable lengths, then it will be a horrible construction of SUBSTRING,CHARINDEX, REVERSE and LEN functions.
Related
I need to find (exclude in fact) any results that contain '%' sign, wherever in a string field. That would mean ... WHERE string LIKE '%%%'. Googling about escaping gave me the following ideas. The first throws syntax error, the second returns rows but there are records actually contain '%'.
1st:
SELECT * FROM table
WHERE string NOT LIKE '%!%%' ESCAPE '!'
///tried with different escape characters
2nd:
SELECT * FROM table
WHERE string NOT LIKE '%[%]%'
Trying on GCP BigQuery.
Try:
SELECT *
FROM table
WHERE string NOT LIKE '%!%%' {ESCAPE '!'}
With curly braces as shown in microsoft sql server docs
Or also:
WITH indata(s) AS (
SELECT 'not excluded'
UNION ALL SELECT '%excluded'
UNION ALL SELECT 'Ex%cluded'
UNION ALL SELECT 'Excluded%'
)
SELECT * FROM indata WHERE INSTR(s,'%') = 0;
-- out s
-- out --------------
-- out not excluded
find (exclude in fact) any results that contain '%'
Consider below simple approach
select *
from your_table
where not regexp_contains(string , '%')
I have a query
select distinct(tad.ASP_NAME)
from TABLE_ASP_DETAILS tad
where tad.ASSIGNED_FE_LAST_NAME = 'asurekam2'
where ASSIGNED_FE_LAST_NAME will be equivalent to SureKAM2 and it should return SureKAM2 for the above query.
Similarly
select distinct(tad.ASP_NAME)
from TABLE_ASP_DETAILS tad
where tad.ASSIGNED_FE_LAST_NAME = 'ABT_Dallas1_TX'
should return ABT from table.
So basically I want contains like functionality in my input string and it should be able to search for something similar in tad.ASP_NAME. ASP Name would be a substring of the input string
ACS_ITALY_CATANIA,ACS_ITALY_BARI,ACS_xxxxx any input should be able to find ACS value in tad.ASP_NAME
ADNTELECOM_Sayedur_Rahman , ADNTELECOM_Reza_Bin_Mujib, ADNTELECOM_Reza_Bin_Mujib
should be able to find ADNTELECOM value in tad.ASP_NAME
This seems to do what you want:
where lower('asurekam2') like '%' || lower(tad.ASSIGNED_FE_LAST_NAME) || '%'
Would this do? You didn't provide test case so I improvised. You need lines 6-8.
SQL> with table_asp_details(asp_name, assigned_fe_last_name) as
2 (select 'ACS_ITALY_CATANIA', 'ACS_Dallas1_TX' from dual union all
3 select 'ACS_ITALY_BARI', 'ACS_Dallas1_TX' from dual
4 )
5 -- this is what you need
6 select distinct regexp_substr(asp_name, '[[:alpha:]]+') result
7 from table_asp_details tad
8 where tad.assigned_fe_last_name = 'ACS_Dallas1_TX';
RESULT
-----------------
ACS
SQL>
I think you are looking for something like this, but its hard to say with so few details:
select distinct(tad.ASP_NAME)
from TABLE_ASP_DETAILS tad
where regexp_like('asurekam2', tad.ASSIGNED_FE_LAST_NAME)
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"
I want to select names from a table where the 'name' column contains '%' anywhere in the value. For example, I want to retrieve the name 'Approval for 20 % discount for parts'.
SELECT NAME FROM TABLE WHERE NAME ... ?
You can use like with escape. The default is a backslash in some databases (but not in Oracle), so:
select name
from table
where name like '%\%%' ESCAPE '\'
This is standard, and works in most databases. The Oracle documentation is here.
Of course, you could also use instr():
where instr(name, '%') > 0
One way to do it is using replace with an empty string and checking to see if the difference in length of the original string and modified string is > 0.
select name
from table
where length(name) - length(replace(name,'%','')) > 0
Make life easy on yourselves and just use REGEXP_LIKE( )!
SQL> with tbl(name) as (
select 'ABC' from dual
union
select 'E%FS' from dual
)
select name
from tbl
where regexp_like(name, '%');
NAME
----
E%FS
SQL>
I read the documentation mentioned by Gordon. The relevent sentence is:
An underscore (_) in the pattern matches exactly one character (as opposed to one byte in a multibyte character set) in the value
Here was my test:
select c
from (
select 'a%be' c
from dual) d
where c like '_%'
The value a%be was returned.
While the suggestions of using instr() or length in the other two answers will lead to the correct answer, they will do so slowly. Filtering on function results simply take longer than filtering on fields.
How do I do a like search on a number column in SQL?
I want numbers which are like '0.0000%'.
I tried with
select * from emp where emp_id & '' like '123%'
select * from emp where CONVERT(varchar(20), emp_id) like '123%'
but in vain.
Please help me
Regardless of which DBMS you are using AND assuming you have a valid reason to do this, you have several ways to solve problems like these. I can think of three right now:
Convert the number to a string and use a LIKE operator on this:
select *
from emp
where to_char(emp_id) like '123%';
Use mathematical operators directly (like Andrey suggests), for example:
select *
from table
where num between 0 and 0.0001;
Construct a mathematical expression (actually, this is just another case of method 2), for example:
select *
from table
where abs(num - round(num, 5)) < 0.00001;
Use comparison operators (> and <):
select * from table where num > 0 and num < 0.00001
select 0.0001*1000 from dual if it is <1 means the 0.0001 has 3 or more zeros.
so i did like
select * from emp where emp_id*10000<1