I'm trying to create a SQL query which will supply values for auto completion for a text field. Everything is working however I can't seem to create an SQL query which is exact enough for the purposes I want. I am using MySQL.
If there is a space (or multiple spaces) in the search term, I only want the query to do a LIKE comparison on the part of the string after the last space.
For example, say I have two possible values in the database:
Bolt
Bolts Large
Currently if the user types 'Bolt' then a space, both values above are returned using this query -
SELECT name FROM items WHERE name LIKE 'SEARCH_TERM%'
What I want is that if the user types 'Bolt' then a space, then only Bolt is returned from the database.
Effectively meaning that only the last part of the search term after the space is compared using LIKE, the results should match exactly up until the last space.
I've also tried:
SELECT name FROM items WHERE name LIKE 'SEARCH_TERM[a-z]%'
But that actually returns no results using the above scenario.
Is what I'm after possible? I've also tried to explore using Full Text Search but have had no look with that. I believe full text search is enabled on the name field, however I have limited experience with this. The query below didn't work.
SELECT name FROM items WHERE MATCH(name) AGAINST('SEARCH_TERM')
Any advice or points would be very appreciated.
The query
SELECT name FROM items WHERE name LIKE 'Bolt %'
doesn't return any record, because both 'Bolt' and 'Bolts Large' don't match 'Bolt %'.
SELECT name FROM items WHERE name LIKE 'Bolt%'
returns both records, because both 'Bolt' and 'Bolts Large' match 'Bolt%'.
To look for 'Bolt' and not 'Bolts', you must add a space to both your search string and the column string:
SELECT name FROM items WHERE concat(name, ' ') LIKE 'Bolt %'
returns 'Bolt' but not 'Bolts Large'.
SELECT name FROM items WHERE REPLACE(name, ' ', '') LIKE 'SEARCH_TERM%'
You could also use CONCAT and TRIM, or just trim
SELECT name FROM items WHERE name LIKE TRIM('SEARCH_TERM')
or your choice
SELECT name FROM items WHERE name LIKE CONCAT(TRIM('SEARCH_TERM'), '%')
SELECT name FROM items WHERE name LIKE CONCAT('%',TRIM('SEARCH_TERM'))
SELECT name FROM items WHERE name LIKE CONCAT('%',TRIM('SEARCH_TERM'), '%')
Related
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'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'm trying to find all entries that contain a backslash anywhere, like so:
SELECT * FROM animals WHERE bodyType LIKE '%\\%'
I also tried:
SELECT * FROM animals WHERE bodyType LIKE '%\\\\%'
and
SELECT * FROM animals WHERE bodyType LIKE '%\\\\\\\\%'
Neither worked. Anyone know how to do this?
I am running the commands in MySQL Quick Admin v1.5.4
This question has not answer.
You try to match searching of content with backslash on names of columns.
Your (a bit general) query
SELECT * FROM tableName WHERE columnName LIKE '%\\%'
will give you any result if content of any column will contain backslash. In this case your query is correct. But you cannot match name of any column.
I looked into some books I have and all they say the same: this will select all records written in column of chosen name that are containing backslash. But columns have to be chosen exactly (they cannot be selected by name with using of SQL query).
I have a typeahead select box for finding users in a database. Frequently there are meny users with the same first name & different last names.
A person searching for a user would generally type in: "John doe".
The problem with this is that the typeahead will start getting all the John users but then return no results when the space is typed in.
"WHY" this is happening is obvious, user first names are trimmed & anything with a space will not match ~ let alone with another character following that, first & last names are stored in separate columns.
Here is the query as it stands:
SELECT Entities.* FROM `Entities`
WHERE `Entities`.`first_name` LIKE '%john doe%'
OR `Entities`.`last_name` LIKE '%john doe %'
How can I re-write this query so that the results would be:
all users whose first name is john
ignore any amount of spaces
AND where last name is doe
when a user types in"john doe" in the search control
(Keeping in mind that the search control is ONE text type field)
If you want the first name like john and the last name like doe, then try something like:
where concat(Entities`.`first_name`, ' ', `Entities`.`last_name`) like '%john% doe%'
If you want do do this based on a search string, then something like:
where concat(`Entities`.`first_name`, ' ', `Entities`.`last_name`) like
concat('%', replace('john doe', ' ', '%'), '%')
Or, use full text search.
I want to select all rows, where the column g_folder exists in the beginning of the search word.
For example: One element in the table c_groups is '/f1' (in column g_folder). So I want to have only the items, which are in the beginning of the search word:
It should look like: g_folder% is like '/f1/xx';
But I can't put this percent behind a variable. How can I do this?
My statement:
select count(*) from c_groups where g_folder is like '/f1/xx'
If this is SQL Server, you can concatenate your column like so:
SELECT
COUNT(*)
FROM
c_groups
WHERE
'/f1/xx' LIKE (g_folder + '%')