I'm trying to query the first letter of a last name in MySQL. I want to skip the first name and query the a certain letter in the last name. Thanks
In SKU_data, which SKU has a buyer whose last name begins with 'M'?
*/
select *
from sku_data
where buyer ;
In your where clause, search on WHERE buyer.LastName LIKE 'M%'. This will return all results that start with M.
You need to use a combination of the substring method and the substring_index method in your sql query.
Your select query should look something like this
SELECT SUBSTRING(last_name,1,1)
I'm assuming that your name field has both first and last name in it, instead of separate fields for first and last name. Use the substring_index method to figure out what last_name should be:
SUBSTRING_INDEX(full_name,' ',-1)
Combining this SQL, we have:
SELECT SUBSTRING(SUBSTRING_INDEX(buyer,' ',-1),1,1) as first_letter
Now you can use the first_letter field in a where clause to get your desired result:
SELECT *, SUBSTRING(SUBSTRING_INDEX(buyer,' ',-1),1,1) as first_letter
from sku_data
where first_letter = 'M' ;
Assuming that you have a field like say "full_name" which has first name and last name in the same column.
Lets say the full_name in table employee has a value "JEFF MOSTI"
You can simply use the following statement to get what you need
select * from employee where full_name regexp ' M';
I am assuming you are looking for the last name to start with 'M' and that there is a space (' ') between the first and last name.
Related
I have a teachers table that looks like this:
teacherid
teacherfname
teacherlname
salary
1
Alexander
Bennett
55.30
I would like to return any record that contains a given string in the teacherfname, teacherlname and salary columns.
What I have right now (this returns exact match only):
SELECT * FROM `teachers` WHERE 'Alexander' IN (teacherfname, teacherlname, salary)
What I would like to do is something like this (this would not return anything):
SELECT * FROM `teachers` WHERE '%Alex%' IN (teacherfname, teacherlname, salary)
What do I need to make the query work? Thank you.
I would assume that the value %Alex% won't ever match the salary column. If you want to search for any rows where the first name or last name include "Alex" I would use simple pattern matching, and force all comparisons to use the same letter case.
For example:
SELECT *
FROM `teachers`
WHERE lower(teacherfname) like '%alex%'
or lower(teacherlname) like '%alex%'
I want SQL to show / order the results for the column name first then show results for the description column last.
Current SQL query:
SELECT * FROM products WHERE (name LIKE '%$search_query%' OR description LIKE '%$search_query%')
I tried adding order by name, description [ASC|DESC] on the end but that didn't work.
It's for optimizing the search results. If a certain word is found in description it should go last if a certain word is also found in the name column.
You can use a CASE statement in an ORDER BY to prioritize name. In the example below all results where name is matched will come first because the CASE statement will evaluate to 1 whereas all other results will evaluate to 2.
I'm not sure by your problem description what exactly you want the behavior to be, but you can certainly use this technique to create more refined cases to prioritize your results.
SELECT *
FROM products
WHERE (name LIKE '%$search_query%' OR description LIKE '%$search_query%')
ORDER BY CASE WHEN name LIKE '%$search_query%' THEN 1 ELSE 2 END
If you want the names first, the simplest order by is:
order by (name like '%$search_query%') desc
MySQL treats booleans as numbers in a numeric context, with "1" for true and "0" for false.
While this is undocumented, when results sets combined by a UNION ALL and not sorted afterwards, they stay in the order returned, as UNION ALL just adds new results to the bottom of the result set. This should work for you:
SELECT * FROM products
WHERE name LIKE '%$search_query%'
UNION ALL
SELECT * FROM products
WHERE (description LIKE '%$search_query%' AND name NOT LIKE '%$search_query%')
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%'
SELECT *
FROM customers
WHERE Firstname LIKE 'George'
The problem is that i have more than 1 rows in the table with tha name Geoge and the result of the query shows only one row
You will want to include the wildcard % character to include the rows the have George present in the name:
SELECT *
FROM customers
WHERE Firstname LIKE '%George%';
If George will always appear at the beginning, then you can include the wildcard on the end:
SELECT *
FROM customers
WHERE Firstname LIKE 'George%';
you need to add a wildcard character % to match any value that contains george
SELECT *
FROM customers
WHERE Firstname LIKE '%George%'
MySQL LIKE Operator
the statement
WHERE Firstname LIKE 'George'
is equivalent with
WHERE Firstname = 'George'
that is why you are only getting one record which firstname is george.
UPDATE 1
SQLFiddle Demo
try
LOWER(Firstname) LIKE '%george%'
handles partial values and avoids case sensietivity issues.
I'm writing the following statement to select names that starts with a range of characters.
select name from db.table where name like '[c-e]%';
The statement does not return anything to me. But when I change it to:
select name from db.table where name like 'c%';
It returns records.
What is the problem ?
instead of LIKE use REGEXP
select name from db.table where name REGEXP '[c-e].+';
See MySQL Pattern Matching
#juergen d is right but a little change required as you want to check all the name starting from in range [c-d] so
select name from db.table where name REGEXP '^[c-e].+';
"^" indicates that match when starting must match range [c-e]
As others have said, you could use REGEXP; but when you want a range of starting characters, you can also do:
SELECT name FROM db.table WHERE name >= 'c' AND name < 'f';
This could be faster, as it can take advantage of an index in the name field to avoid a full scan.
You should use REGEXP as:
SELECT name FROM db.table where name REGEXP '[c-e].+';