I am trying to query a mysql table which contains strings of numbers
(i.e. '1,2,3,4,5').
How do I search to see if it has '1' but not '11' bearing in mind if it is '9,10' '9%' doesnt work??
Fixed!
(field like '10' OR field like '%,10,%' OR field like '%,10' OR field like '10,%')
You could try the function find_in_set
select find_in_set('1','1,2,3,11,12')
You need the function FIND_IN_SET. Btw, '9%' should work, if the column contains the values you specified, are you sure you're querying
SELECT * FROM table WHERE field LIKE '9%'?
Standard SQL can do it as well:
...
WHERE
',' + SetValue + ',' LIKE '%,1,%'
AND ',' + SetValue + ',' NOT LIKE '%,11,%'
This expression cannot make use of an index, therefore performance will degrade quickly as the table size rises.
For better performance your table should be properly normalized, e.g.
SetId SetValue
1 1
1 2
1 3
1 4
1 5
instead of
SetId SetValue
1 '1,2,3,4,5'
Related
I have a column which has inconsistent data. The column named ID and it can have values such as
0897546321
ABC,0876455321
ABC,XYZ,0873647773
ABC,
99756
test only
The SQL query should fetch only Ids which are of 10 digit in length, should begin with a 08 , should be not null and should not contain all characters. And for those values, which have both digits and characters such as ABC,XYZ,0873647773, it should only fetch the 0873647773 . In these kind of values, nothing is fixed, in place of ABC, XYZ , it can be anything and can be of any length.
The column Id is of varchar type.
My try: I tried the following query
select id
from table
where id is not null
and id not like '%[^0-9]%'
and id like '[08]%[0-9]'
and len(id)=10
I am still not sure how should I deal with values like ABC,XYZ,0873647773
P.S - I have no control over the database. I can't change its values.
SQL Server generally has poor support regular expressions, but in this case a judicious use of PATINDEX is viable:
SELECT SUBSTRING(id, PATINDEX('%,08[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9],%', ',' + id + ','), 10) AS number
FROM yourTable
WHERE ',' + id + ',' LIKE '%,08[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9],%';
Demo
If you normalise your data, and split the delimited data into parts, you can achieve this some what more easily:
SELECT SS.value
FROM dbo.YourTable YT
CROSS APPLY STRING_SPLIT(YT.YourColumn,',') SS
WHERE LEN(SS.value) = 10
AND SS.value NOT LIKE '%[^0-9]%';
If you're on an older version of SQL Server, you'll have to use an alternative String Splitter method (such as a XML splitter or user defined inline table-value function); there are plenty of examples on these already on Stack Overflow.
db<>fiddle
I'm using SQL Server and would like to check if todays day name is in a list of values in a single field/column.
An example of the column "start_days" contents is:
'Monday','Tuesday','Sunday'
'Thursday'
'Friday','Sunday'
'Tuesday','Sunday'
'Tuesday','Wednesday','Thursday','Friday'
The code I am trying to run on this is:
case
when datename(weekday,getdate()) in (start_days) then 1
else 0
end as today_flag
And the result is 0 for every row.
Am I doing something wrong here or is it just not possible to use a single field as a list of values in the statement?
As a starter: you should fix your data model and not store multiple values in a single column. Storing list of values in a database column basically defeats the purpopose of a relational database. Here is a related reading on that topic.
That said, here is one option using pattern matching:
case
when ',' + start_days + ',' like '%,' + datename(weekday,getdate()) + ',%' then 1
else 0
end as today_flag
If you really have single quotes around values within the list, then we need to include them in the match:
case
when ',' + start_days + ',' like '%,''' + datename(weekday,getdate()) + ''',%' then 1
else 0
end as today_flag
If the values always are weekdays, this can be simplified since there is no risk of overlapping values:
case
when start_days like '%''' + datename(weekday,getdate()) + '''%' then 1
else 0
end as today_flag
The right answer to the question is fixing your data modal. Storing multiple values like that will leads you to many issue and you're stuck on one right now.
Until that, you could use LIKE operaor to get the desired results as:
SELECT *, CASE WHEN
Value LIKE CONCAT('%', QUOTENAME(DATENAME(WEEKDAY,GETDATE()), ''''), '%')
THEN 1
ELSE 0
END
FROM
(
VALUES
('''Monday'',''Tuesday'',''Sunday'''),
('''Thursday'''),
('''Friday'',''Sunday'''),
('''Tuesday'',''Sunday'''),
('''Tuesday'',''Wednesday'',''Thursday'',''Friday''')
) T(Value)
Here is a db<>fiddle where you can see how it's working online.
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.
In SQL Server, I have a string column that contains numbers. Each entry I need is only one number so no parsing is needed. I need some way to find all rows that contain numbers from 400 to 450. Instead of doing:
...where my stringcolumn like '%400%' or stringcolumn like '%401%' or stringcolumn like '%402%' or ...
is there a better that can save on some typing?
There are also other values in these rows such as: '5335154', test4559#me.com', '555-555-5555'. Filtering those out will need to be taken into account.
...where stringcolumn like '4[0-4][0-9]' OR stringcolumn = '450'
You don't need the wildcard if you want to restrict to 3 digits.
Use regex to accomplish this.
...where stringcolumn like '4[0-4][0-9]' OR stringcolumn like '450'
one way
WHERE Column like '%4[0-4][09]%'
OR Column LIKE '%500%'
keep in mind that this will pick anything with the number in it, so 5000 will be returned as well
I would do the following:
select t.*
from (select t.*,
(case when charindex('4', col) > 0
then substrint(col, charindex('4', col), charindex('4', col) + 2)
end) as col4xx
from t
) t
where (case when isnumeric(col4xx) = 1
then (case when cast(col4xx as int) between 400 and 450 then 'true'
end)
end) = 'true'
I'm not a fan of having case statements in WHERE clauses. However, to ensure conversion to a number, this is needed (or the conversion could become a column in another subquery). Note that the following is not equivalent:
where col4xx between '400' and '450'
Since the string '44A' would match.
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.