I have a data set that looks something like this:
A6177PE
A85506
A51SAIO
A7918F
A810004
A11483ON
A5579B
A89903
A104F
A9982
A8574
A8700F
And I need to find all the ENDings where they are non-numeric. In this example, that means PE, AIO, F, ON, B and F.
In pseudocode, I'm imagining I need something like
SELECT DISTINCT X FROM
(SELECT SUBSTR(COL,[SOME_CLEVER_LOGIC]) AS X FROM TABLE);
Any ideas? Can I solve this without learning regexp?
EDIT: To clarify, my data set is a lot larger than this example. Also, I'm only interested in the part of the string AFTER the numeric part. If the string is "A6177PE" I want "PE".
Disclaimer: I don't know Oracle SQL. But, I think something like this should work:
SELECT DISTINCT X FROM
(SELECT SUBSTR(COL,REGEXP_INSTR(COL, "[[:ALPHA:]]+$")) AS X FROM TABLE);
REGEXP_INSTR(COL, "[[:ALPHA:]]+$") should return the position of the first of the characters at the end of the field.
For readability, I'd recommend using the REGEXP_SUBSTR function (If there are no performance issues of course, as this is definitely slower than the accepted solution).
...also similar to REGEXP_INSTR, but instead of returning the position of the substring, it returns the substring itself
SELECT DISTINCT SUBSTR(MY_COLUMN,REGEXP_SUBSTR("[a-zA-Z]+$")) FROM MY_TABLE;
(:alpha: is supported also, as #Audun wrote )
Also useful: Oracle Regexp Support (beginning page)
For example
SELECT SUBSTR(col,INSTR(TRANSLATE(col,'A0123456789','A..........'),'.',-1)+1)
FROM table;
Related
I have got a table with a column that is type of VARCHAR2(255 BYTE). I would like to select only these rows that have numbers as a value, so I discard any other values as for example "lala","1z". I just want to have pure numbers from 1 to ..... 999999999 (just digital numbers in other words) :P
Could you tell me how to make it?
if you're using Oracle 12c r2 or later then use the built-in validate_conversion() function:
select *
from your_table
where validate_conversion(cast(your_column as number)) = 0
validate_conversion() returns 0 when the proposed conversion would succeed and 1 when it wouldn't. It also supports date and timestamp conversions. Find out more.
Something like this is the usual option. You could use regexp, but it's usually a bit slower.
select column1
from tableA
where translate(column1, '1234567890', '') is null;
Here's the regexp version kfinity referred to. The regex matches a line consisting of 1 or more digits.
select column1
from tableA
where regexp_like(column1, '^\d+$');
You don't want zero to start a number. So it seems like regular expressions are the way to go:
where regexp_like(column1, '^[1-9][0-9]*$');
I have a table with about 200 million records. One of the columns is defined as varchar(100) and it's included in a full text index. Most of the values are numeric. Only few are not numeric.
The problem is that it's not working well. For example if a row contains the value '123456789' and i look for '567', it's not returning this row. It will only return rows where the value is exactly '567'.
What am I doing wrong?
sql server 2012.
Thanks.
Full text search doesn't support leading wildcards
In my setup, these return the same
SELECT *
FROM [dbo].[somelogtable]
where CONTAINS (logmessage, N'28400')
SELECT *
FROM [dbo].[somelogtable]
where CONTAINS (logmessage, N'"2840*"')
This gives zero rows
SELECT *
FROM [dbo].[somelogtable]
where CONTAINS (logmessage, N'"*840*"')
You'll have to use LIKE or some fancy trigram approach
The problem is probably that you are using a wrong tool since Full-text queries perform linguistic searches and it seems like you want to use simple "like" condition.
If you want to get a solution to your needs then you can post DDL+DML+'desired result'
You can do this:
....your_query.... LIKE '567%' ;
This will return all the rows that have a number 567 in the beginning, end or in between somewhere.
99% You're missing % after and before the string you search in the LIKE clause.
es:
SELECT * FROM t WHERE att LIKE '66'
is the same as as using WHERE att = '66'
if you write:
SELECT * FROM t WHERE att LIKE '%66%'
will return you all the lines containing 2 'sixes' one after other
I'm trying to retrieve all columns that start with any non alpha characters in SQlite but can't seem to get it working. I've currently got this code, but it returns every row:
SELECT * FROM TestTable WHERE TestNames NOT LIKE '[A-z]%'
Is there a way to retrieve all rows where the first character of TestNames are not part of the alphabet?
Are you going first character only?
select * from TestTable WHERE substr(TestNames,1) NOT LIKE '%[^a-zA-Z]%'
The substr function (can also be called as left() in some SQL languages) will help isolate the first char in the string for you.
edit:
Maybe substr(TestNames,1,1) in sqllite, I don't have a ready instance to test the syntax there on.
Added:
select * from TestTable WHERE Upper(substr(TestNames,1,1)) NOT in ('A','B','C','D','E',....)
Doesn't seem optimal, but functionally will work. Unsure what char commands there are to do a range of letters in SQLlite.
I used 'upper' to make it so you don't need to do lower case letters in the not in statement...kinda hope SQLlite knows what that is.
try
SELECT * FROM TestTable WHERE TestNames NOT LIKE '[^a-zA-Z]%'
SELECT * FROM NC_CRIT_ATTACH WHERE substring(FILENAME,1,1) NOT LIKE '[A-z]%';
SHOULD be a little faster as it is
A) First getting all of the data from the first column only, then scanning it.
B) Still a full-table scan unless you index this column.
I have stored values in my database that look like 5XXXXXX, where X can be any digit. In other words, I need to match incoming SQL query strings like 5349878.
Does anyone have an idea how to do it?
I have different cases like XXXX7XX for example, so it has to be generic. I don't care about representing the pattern in a different way inside the SQL Server.
I'm working with c# in .NET.
You can write queries like this in SQL Server:
--each [0-9] matches a single digit, this would match 5xx
SELECT * FROM YourTable WHERE SomeField LIKE '5[0-9][0-9]'
stored value in DB is: 5XXXXXX [where x can be any digit]
You don't mention data types - if numeric, you'll likely have to use CAST/CONVERT to change the data type to [n]varchar.
Use:
WHERE CHARINDEX(column, '5') = 1
AND CHARINDEX(column, '.') = 0 --to stop decimals if needed
AND ISNUMERIC(column) = 1
References:
CHARINDEX
ISNUMERIC
i have also different cases like XXXX7XX for example, so it has to be generic.
Use:
WHERE PATINDEX('%7%', column) = 5
AND CHARINDEX(column, '.') = 0 --to stop decimals if needed
AND ISNUMERIC(column) = 1
References:
PATINDEX
Regex Support
SQL Server 2000+ supports regex, but the catch is you have to create the UDF function in CLR before you have the ability. There are numerous articles providing example code if you google them. Once you have that in place, you can use:
5\d{6} for your first example
\d{4}7\d{2} for your second example
For more info on regular expressions, I highly recommend this website.
Try this
select * from mytable
where p1 not like '%[^0-9]%' and substring(p1,1,1)='5'
Of course, you'll need to adjust the substring value, but the rest should work...
In order to match a digit, you can use [0-9].
So you could use 5[0-9][0-9][0-9][0-9][0-9][0-9] and [0-9][0-9][0-9][0-9]7[0-9][0-9][0-9]. I do this a lot for zip codes.
SQL Wildcards are enough for this purpose. Follow this link: http://www.w3schools.com/SQL/sql_wildcards.asp
you need to use a query like this:
select * from mytable where msisdn like '%7%'
or
select * from mytable where msisdn like '56655%'
Given this linq query against an EF data context:
var customers = data.Customers.Where(c => c.EmailDomain.StartsWith(term))
You’d expect it to produce SQL like this, right?
SELECT {cols} FROM Customers WHERE EmailDomain LIKE #term+’%’
Well, actually, it does something like this:
SELECT {cols} FROM Customer WHERE ((CAST(CHARINDEX(#term, EmailDomain) AS int)) = 1)
Do you know why?
Also, replacing the Where selector to:
c => c.EmailDomain.Substring(0, term.Length) == term
it runs 10 times faster but still produces some pretty yucky SQL.
NOTE: Linq to SQL correctly translates StartsWith into Like {term}%, and nHibernate has a dedicated LikeExpression.
I don't know about MS SQL server but on SQL server compact LIKE 'foo%' is thousands time faster than CHARINDEX, if you have INDEX on seach column. And now I'm sitting and pulling my hair out how to force it use LIKE.
http://social.msdn.microsoft.com/Forums/en-US/adodotnetentityframework/thread/1b835b94-7259-4284-a2a6-3d5ebda76e4b
The reason is that CharIndex is a lot faster and cleaner for SQL to perform than LIKE. The reason is, that you can have some crazy "LIKE" clauses. Example:
SELECT * FROM Customer WHERE EmailDomain LIKE 'abc%de%sss%'
But, the "CHARINDEX" function (which is basically "IndexOf") ONLY handles finding the first instance of a set of characters... no wildcards are allowed.
So, there's your answer :)
EDIT: I just wanted to add that I encourage people to use CHARINDEX in their SQL queries for things that they didn't need "LIKE" for. It is important to note though that in SQL Server 2000... a "Text" field can use the LIKE method, but not CHARINDEX.
Performance seems to be about equal between LIKE and CHARINDEX, so that should not be the reason. See here or here for some discussion. Also the CAST is very weird because CHARINDEX returns an int.
charindex returns the location of the first term within the second term.
sql starts with 1 as the first location (0 = not found)
http://msdn.microsoft.com/en-us/library/ms186323.aspx
i don't know why it uses that syntax but that's how it works
I agree that it is no faster, I was retrieving tens of thousands of rows from our database with the letter i the name. I did find however that you need to use > rather than = ... so use
{cols} FROM Customer WHERE ((CAST(CHARINDEX(#term, EmailDomain) AS int)) > 0)
rather than
{cols} FROM Customer WHERE ((CAST(CHARINDEX(#term, EmailDomain) AS int)) = 1)
Here are my two tests ....
select * from members where surname like '%i%' --12 seconds
select * from sc4_persons where ((CAST(CHARINDEX('i', surname) AS int)) > 0) --12 seconds
select * from sc4_persons where ((CAST(CHARINDEX('i', surname) AS int)) = 1) --too few results