I am trying to find records that has the following scenario.
ID | name | email
1 Robert robert#gmail.com
2 William bill#gmail.com
3 Michael michael#gmail.com
4 Micahel mike#gmail.com
Based on the above table, I want to find the records where the "name" is contained in the "email field", here record 1 and 3 should be the output and not 2 and 4. Is there any way I can do this comparison?
I tried reading about regex but couldn't find anything. If it's comparison of same value, it will be straightforward, but I am not having any clue for this one. I thought of LIKE but looks like this cannot have field names.
The exact syntax will depend on how you want to define the relationship.
Are you looking for the name anywhere in the email address? (This will be slow)
select id,name,email
from your_table
where email like concat('%',name,'%')
Just at the beginning of the email address?
select id,name,email
from your_table
where email like concat(name,'%')
Just before the # sign?
select id,name,email
from your_table
where email like concat(name,'#%')
You can use LIKE, you just have to use it in combination with CONCAT.
SELECT
ID,
name,
email
FROM
yourTable
WHERE
email LIKE CONCAT(name, '%');
The CONCAT will return a string which can be used to match against email via LIKE.
This should work
SELECT * FROM table WHERE email LIKE (CONCAT('%',name,'%'))
select * from your_table where lower(substring_index(email,'#',1))=lower(name)
Related
I have a field called EMAIL_ADDRESS. One of the records would be:
john#gmail.com, mike#gmail.com, joe#yahoo.com, george#yahoo.com, fred#gmail.com
I wan to remove all yahoo addresses in my SELECT query to get:
john#gmail.com, mike#gmail.com, fred#gmail.com
If I use
REPLACE(SM.SCORECARD_EMAIL_ADDRESS, 'joe#yahoo.com,', '')
this works.
If I want to remove ALL yahoo email addresses this doesn't work:
REPLACE(SM.SCORECARD_EMAIL_ADDRESS, '%#yahoo.com,', '')
because wildcards don't seem to work as it's looking for % in the string.
You should probably fix your table design and stop storing CSV lists of email addresses. Instead, get each email onto a separate record. As a short term fix, if you're running MySQL 8+, you may use REGEXP_REPLACE():
UPDATE yourTable
SET EMAIL_ADDRESS = REGEXP_REPLACE(
REGEXP_REPLACE(EMAIL_ADDRESS, '(, )?\\S+#yahoo\\.com,?', ','), '^,+|,+$', '')
WHERE EMAIL_ADDRESS LIKE '%#yahoo.com%';
If you don't need to udpate records but you want them only in the SELECT query you can use NOT LIKE operator
SELECT * FROM your_table WHERE email NOT LIKE '%yahoo.com'
So you get records that doesn’t match the like pattern
i want some correction here. i want to select all people with name fred in database
Here's my query:
SELECT * FROM tdble WHERE CONCAT(name) LIKE CONCAT('%', REPLACE('fred', '')'%')
What you are asking can be simply achieved by either using the "=" operator of the wildcard operator "like" statement.
If you wish to find all records that have an exact match to the name 'Fred' then you should model your query as so:
Select * From tdble Where Name = 'fred'
However, if you want to get all results where the names have 'fred' included in it somewhere use the wildcard operator.
Select * From tdble Where Name like '%fred%'
Also you can further model your query to know where exactly in which form you want 'fred' to appear. Example if you want 'Fred' to be as the last characters of your name string, for instance you wish to get names which ends with fred then model your query like this:
Select * From tdble Where Name like '%fred'
(you will get results like 'alfred', provided there is an alfred in your table)
However if you wish to get all names that begin with fred, model the query like this:
Select * From tdble Where Name like 'fred%'
(you will get results like 'fredinane', provided there is a fredinane in your table)
Cheers
If you want to fetch record with name 'fred', you can simply do Select * from TableName Where Name = 'fred'.
If you want to fetch records which their names' string contain 'fred', you have to use select * from TableName where Name like '%fred%'
I'm trying to check if a field in a specific table contains also number, in particular I have a record that have the field name which contains this value: Besëlidhja Lezhë vs. Tërbuni Pukë 1 - 1, so I'm trying to get also all the rows of that table that contains a number inside the field name. I tried:
SELECT * FROM `venue` where `name` like '%[0-9]%'
but this will return an empty result, any idea?
This should tell you if name contains any digit (not tested)
SELECT * FROM venue WHERE name REGEXP '[0-9]'
You can try using a Regular Expression that filters for names in your name column with numeric characters . For example:
SELECT * FROM DATA WHERE name REGEXP '[a-z]...[0-9]';
mySQL allows you to use regular expression as a filter !
This should select out for names like Tërbuni Pukë 1 - 1. If you want to practice regular expressions this is a great website to test whether you have the right regex. https://regex101.com/
Hope this helps !
I believe this should work:
SELECT *
FROM venue
WHERE name like '%0%' or name like '%1%' or name like '%2%' or name like '%3%'
and so on til you get to 9. I hope this helps
I did some searching and from one question already posted on stackexchange, the answer was that it was not possible, but I figured to ask. I did not know if it was possible to form a SELECT query to dynamically select which columns will be displayed in a mysql SELECT statement result. Example:
Say I have column names Person, ID, Phone Number, Alt Number for this table:
John | 79 | 800-499-0000 | 800-499-5555
I would like to form a SELECT statement so that it will only pull down columns where string '800-499' is somewhere in the field. Thus the result from MySQL ideally would be:
800-499-0000 | 800-499-5555
The only problem is that I do not think dynamically selecting columns is possible.
Any help or confirmation is appreciated.
You could try something like:
select * from
(select concat(case when col1 like '%800-499-0000%' then concat('col1: ',col1,';') end,
case when col2 like '%800-499-0000%' then concat('col2: ',col1,';') end,
...
case when coln like '%800-499-0000%' then concat('coln: ',coln,';') end)
as search_results
from my_table) sq
where search_results is not null
I want to find the data from table artists where name is start with letter a, b, c.
i.e.
My o/p should contain
Adam
Alan
Bob
Ben
Chris
Cameron
But not GlAin, doC, dolBie
In MySQL use the '^' to identify you want to check the first char of the string then define the array [] of letters you want to check for.
Try This
SELECT * FROM artists WHERE name REGEXP '^[abc]'
You can use like 'A%' expression, but if you want this query to run fast for large tables I'd recommend you to put number of first button into separate field with tiny int type.
Try this:
select * from artists where name like "A%" or name like "B%" or name like "C%"
One can also use RLIKE as below
SELECT * FROM artists WHERE name RLIKE '^[abc]';
Try this simple select:
select *
from artists
where name like "a%"
I would go for substr() functionality in MySql.
Basically, this function takes account of three parameters i.e.
substr(str,pos,len)
http://www.w3resource.com/mysql/string-functions/mysql-substr-function.php
SELECT * FROM artists
WHERE lower(substr(name,1,1)) in ('a','b','c');
try using CHARLIST as shown below:
select distinct name from artists where name RLIKE '^[abc]';
use distinct only if you want distinct values only.
To read about it Click here.