I have email addresses like user1@gmail.com
, user2@ymail.com user3@hotmail.com
... etc.
I want a Mysql SELECT
tha开发者_运维百科t will trim user names and .com and return output as
gmail
,ymail
,hotmail
, etc.
Assuming that the domain is a single word domain like gmail.com, yahoo.com, use
select (SUBSTRING_INDEX(SUBSTR(email, INSTR(email, '@') + 1),'.',1))
The inner SUBSTR
gets the right part of the email address after @
and the outer SUBSTRING_INDEX
will cut off the result at the first period.
otherwise if domain is expected to contain multiple words like mail.yahoo.com
, etc, use:
select (SUBSTR(email, INSTR(email, '@') + 1, LENGTH(email) - (INSTR(email, '@') + 1) - LENGTH(SUBSTRING_INDEX(email,'.',-1))))
LENGTH(email) - (INSTR(email, '@') + 1) - LENGTH(SUBSTRING_INDEX(email,'.',-1))
will get the length of the domain minus the TLD (.com, .biz etc. part)
by using SUBSTRING_INDEX
with a negative count which will calculate from right to left.
I prefer:
select right(email_address, length(email_address)-INSTR(email_address, '@')) ...
so you don't have to guess how many sub-domains your user's email domain has.
For PostgreSQL:
split_part(email, '@', 2) AS domain
Full query:
SELECT email, split_part(email, '@', 2) AS domain
FROM users;
Ref: http://www.postgresql.org/docs/current/static/functions-string.html
Credit to https://stackoverflow.com/a/19230892/1048433
Using SUBSTRING_INDEX for "splitting" at '@' and '.' does the trick. See documentation at http://dev.mysql.com/doc/refman/5.1/de/string-functions.html#idm47531853671216.
SELECT SUBSTRING_INDEX(SUBSTRING_INDEX(email, '@', -1), '.', 1);
Example:
SELECT SUBSTRING_INDEX(SUBSTRING_INDEX("foo@bar.buz", '@', -1), '.', 1);
will give you "bar".
Here is what happens:
* Split "foo@bar.buz" at '@'. --> ["foo", "bar.buz"]
* Pick first element from right (index -1). --> "bar.buz"
* Split "bar.buz" at '.' --> ["bar", "buz"]
* Pick first element (index 1) --> "bar"
Result: "bar"
If you also need to get rid of subdomains, use:
SELECT SUBSTRING_INDEX(SUBSTRING_INDEX(SUBSTRING_INDEX(email, '@', -1), '.', -2), '.', 1);
Example:
SELECT SUBSTRING_INDEX(SUBSTRING_INDEX(SUBSTRING_INDEX("foo@1.2.3.bar.buz", '@', -1), '.', -2), '.', 1);
will give you "bar".
If you want to know the most used domain names from email addresses you have (can be usefull), you can do :
select (SUBSTRING_INDEX(SUBSTR(email, INSTR(email, '@') + 1),'.',1)) as a,count(*) as c
FROM User
group by a
order by c desc;
Result :
Try this, removes the @ from the domain and just leaves the domain, example: domain.com
select SUBSTR(SUBSTR(email_field, INSTR(email_field, '@'), INSTR(email_field, '.')), 2) as domain
DECLARE @Email Varchar(20)
SET @Email='abc@gmail.com'
SELECT SUBSTRING(@Email, CHARINDEX('@',@Email)+1,LEN(@Email))
Try this:
select SUBSTR(field_name, INSTR(field_name, '@'), INSTR(field_name, '.'))
select (SUBSTRING_INDEX(SUBSTR(email, INSTR(email, '@') + 1),'.',1) from tableName)
Some sql statements require the table name specified where the email column belongs to.
My suggestion would be (for mysql):
SELECT
LOWER(email) AS email,
SUBSTRING_INDEX(email, '@', + 1) AS account,
REPLACE(SUBSTRING_INDEX(email, '@', -1), CONCAT('.',SUBSTRING_INDEX(email, '.', -1)),'') -- 2nd part of mail - tld.
AS domain,
CONCAT('.',SUBSTRING_INDEX(email, '.', -1)) AS tld
FROM
...
ORDER BY domain, email ASC;
DECLARE @Email Varchar(20)
SET @Email='abc@gmail.com'
SELECT RIGHT(@Email,LEN(@Email)-CHARINDEX('@',@Email))
select SUBSTR(email, instr(email, '@') +1, instr(email,'.') -3)
exact answer is coming using this query
For MSSQL
declare @test as varchar(15) = 'foo@bar.buz'
SELECT SUBSTRING(@test, CHARINDEX('@', @test) + 1, LEN(@test)) AS ExtractString;
SELECT SUBSTR(NAME,INSTR(NAME,'@')+1) FROM ORACLE;
Oracle is my table.Don't be confuse.
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