I'm trying to list all users beginning with a letter, e.g. D
Would the following be the right method of doing this.
Select concat(firstname, '',lastname) from users where concat(lastname) = "D*"
SELECT concat(firstname, '',lastname) FROM users WHERE lastname LIKE "D%"
If you want to use wildcards, you need the LIKE operator. Also, in your where clause you only have one column (lastname), so you don't need concat.
i would try:
select * from users where lastname like 'D%';
For getting list starting with eg: "D":
SELECT firstname FROM users WHERE LEFT(firstname,1)= 'D';
Related
I m trying to query a database with about 2000 entries. I want to select the entries in which the names may contain any one of the vowel.
I tried using the following query, but it gives me those entries that contain all the given characters in that order.
select * from myTable where name like '%a%e%i%';
How do I modify the above query to select those entries with names that may contain at least anyone of the vowels.
Try this for SQL Server:
SELECT * FROM myTable WHERE name LIKE '%[AEIOU]%';
I hope this helps you...
SELECT * FROM myTable WHERE name REGEXP 'a|e';
or.....
SELECT * FROM myTable WHERE name REGEXP 'a|e|i';
In SQL Server, you would do:
where name like '%[aeiou]%';
In MySQL, you would do something similar with a regular expression.
Use OR like this.
This will work for both SQL Server and MySql.
select * from myTable where name like '%a%' OR name like '%e%' OR name like '%i%';
Use LIKE and OR.
Query
select * from myTable
where name like '%a%'
or name like '%e%'
or name like '%i%'
or name like '%o%'
or name like '%u%'
I have a select statement like this:
Select * from A where name like 'a%' or name like 'b%' or name like 'j%' or name like ... etc
Is it possible to store a%, b%, j% in a table somewhere so I can more easily manage this list and convert my query to something like:
Select * from A where name like (Select * from StringPatternToMatch)
Try this:
SELECT * FROM A
JOIN StringPatternToMatch patt ON A.name LIKE '%' + patt.pattern + '%';
Replace patt.pattern with the name of the column in your StringPatternToMatch
You can do a regexp search instead.
select *
from A where name regexp '^[abjf]';
It's easier query to maintain than a ton of or'd likes.
demo here
'^[abjf]' means match the start of the string (^), followed by any of the characters in the list ([abjf]). It doesn't care what comes after that.
Just add more letters to the list if you find names starting with them.
This is my users table:
http://ezinfotec.com/Capture.PNG
I need to select all rows those are not contain 2 in except column. How to write a query for this using php & Mysql.
The result i expect for this query is only return last row only.
Thank you.
Don't store comma separated values in your table, it's very bad practice, nevertheless you can use FIND_IN_SET
SELECT
*
FROM
users
WHERE
NOT FIND_IN_SET('2', except)
Try this:
SELECT *
FROM users
WHERE CONCAT(',', except, ',') NOT LIKE '%,2,%'
this should work for you
SELECT *
FROM table
WHERE table.except NOT LIKE '%,2%'
OR table.except NOT LIKE '%2,%';
Say I want to search for a user, 'Richard Best'. Is it possible to compare the full name is concatenated first name and last name? I do not have a full name field.
select * from users where last_name + ' ' + first_name like '%richa%'
I am using Mysql
These are equivalent:
select * from users where concat(last_name,' ',first_name) like '%richa%'
select * from users where concat_ws(' ',last_name,first_name) like '%richa%'
This might also work:
select * from users where last_name like '%richa%' or first_name like '%richa%'
Take a look at this thread.
Using mysql concat() in WHERE clause?
select * from users where (first_name + ' ' + last_name) like '%richa%'
In laravel Eloquent you can use whereRaw("concat(first_name,' ',last_name) LIKE %$search%")
I have this MySQL query.
I have database fields with this contents
sports,shopping,pool,pc,games
shopping,pool,pc,games
sports,pub,swimming, pool, pc, games
Why does this like query does not work?
I need the fields with either sports or pub or both?
SELECT * FROM table WHERE interests LIKE ('%sports%', '%pub%')
Faster way of doing this:
WHERE interests LIKE '%sports%' OR interests LIKE '%pub%'
is this:
WHERE interests REGEXP 'sports|pub'
Found this solution here: http://forums.mysql.com/read.php?10,392332,392950#msg-392950
More about REGEXP here: http://www.tutorialspoint.com/mysql/mysql-regexps.htm
The (a,b,c) list only works with in. For like, you have to use or:
WHERE interests LIKE '%sports%' OR interests LIKE '%pub%'
Why not you try REGEXP. Try it like this:
SELECT * FROM table WHERE interests REGEXP 'sports|pub'
You can also use REGEXP's synonym RLIKE as well.
For example:
SELECT *
FROM TABLE_NAME
WHERE COLNAME RLIKE 'REGEX1|REGEX2|REGEX3'
Don't forget to use parenthesis if you use this function after an AND parameter
Like this:
WHERE id=123 and(interests LIKE '%sports%' OR interests LIKE '%pub%')
Or if you need to match only the beginning of words:
WHERE interests LIKE 'sports%' OR interests LIKE 'pub%'
you can use the regexp caret matches:
WHERE interests REGEXP '^sports|^pub'
https://www.regular-expressions.info/anchors.html
Your query should be SELECT * FROM `table` WHERE find_in_set(interests, "sports,pub")>0
What I understand is that you store the interests in one field of your table, which is a misconception. You should definitively have an "interest" table.
Like #Alexis Dufrenoy proposed, the query could be:
SELECT * FROM `table` WHERE find_in_set('sports', interests)>0 OR find_in_set('pub', interests)>0
More information in the manual.
More work examples:
SELECT COUNT(email) as count FROM table1 t1
JOIN (
SELECT company_domains as emailext FROM table2 WHERE company = 'DELL'
) t2
ON t1.email LIKE CONCAT('%', emailext) WHERE t1.event='PC Global Conference';
Task was count participants at an event(s) with filter if email extension equal to multiple company domains.