I have name column in my table containing list of business names.
The problem is there are some business names containing - , ' characters.
I am using this query to search the name
$names = Brand::where(name, 'LIKE', '%' . $request->search . '%')->get();
Example : a name from database daddy's barber shop
If I search this using this term daddys barber shop without ' character, it returns nothing.
How to ignore the ' character from database so it will match the daddy's barber shop ?
How to achieve this search?
You can whether go with #Rohit shah answer using whereRaw instead of where, but that is heavy processing especially when you'll have a lot of raws in your table, so everytime the replace charcters is processed.
What I would do if I had that case, is to create an other column in the tabale "name_alt" where i store the names without the unwanted charcters an make the searches on that column
SELECT
Column_name
FROM
Table_name
WHERE
REPLACE(REPLACE(REPLACE(name, '-',''), ',', ''), ''', '') LIKE '%search_string%'
This type of Query might serve your purpose.
Hope this helps you. Let me know in case of any query.
Related
I'm working on a query which I thought should be quite intuitive, but somehow I'm facing a bit of issues when implementing it. I guess what I'm trying to achieve is to match a string stored in MySQL DB without space and punctuation (other creative approaches are more than welcome). At the same time I would like the query to handle Unicode characters in diacritics insensitive fashion (so options like REGEXP are kinda out of luck). And the last condition is I'm on MySQL 5.5 with InnoDB engine, so full-text indexing is not supported (but I'm open to upgrade to 5.6/5.7 if it helps sorting this out).
Consider the scenario which the string Hello-World from John Doe is stored in DB. I would like to find it when given the search string HelloWorld or JohnDoe. To be more general, the string in DB can contain brackets, understores and any other punctuation (not limited to ASCII but can compromise for now), while the search string can be a combination of words with or without any separators in between. The closest I've gotten so far is to daisy chain the REPLACE function for a list of known punctuation, like below:
SELECT text FROM table WHERE REPLACE(REPLACE(text, '-', ''), ' ', '') LIKE '%JohnDoe%'
My questions are:
Is there a better way instead of using the daisy chain above?
If that's the only solution, how will the performance be impacted when I chain up hundred or more REPLACE functions?
Thanks in advance for your help.
I don't know how restrictive your searches must be, but you could try to strip out all non-alphanumeric characters from it, so that you end up with a string like "HelloWorldfromJohnDoe" that you match with instead.
Have a look at this answer: How to remove all non-alpha numeric characters from a string?
You might have to change it around a bit though to make it fir your purposes. I changed it from CHAR(32) to CHAR(255) to make sure I could get the column, but you might want to look into changing the function altogether to fit your data more precisely.
Then you something like this:
SELECT *
FROM testing
WHERE alphanum(test) LIKE CONCAT('%', alphanum('John Doe'), '%')
which should give you a hit.
Method 1
I would have another column on the schema containing an "hashed" version of the name, for example, let's say you have the user:
John Doe The Great
This name hashes to
johndoethegreat
The hash function is coded in such a way that all of the following strings:
John_Doe_THE_great
John Doe The GREAT
John.Doe.The.Great
johnDOE___theGreat
john Doe the great
___john____DOE____THE____great
hash to the same value
johndoethegreat
It's trivial to write such a function. This way you can get the user input, hash it and then compare it against the hash column in your database
Names like:
Jon Doe
John Doo
will not be found of course
Method 2
Use the FULLTEXT search feature built-in in MySQL, sort the results by score and pick the first non zero entry
http://blog.oneiroi.co.uk/mysql/php/mysql-full-text-search-with-percentage-scoring/
I am totally missing the point of your question. You appear to have the string:
Hello-World from John Doe
If you want to find this when the search string is JohnDoe or John Doe, then you only need to substitute spaces:
where replace(text, ' ') like concat('%', 'JohnDoe', '%')
If you want a string that contains both "John" and "Doe" in that order, then:
where replace(text, ' ') like concat('%', 'John%Doe', '%')
I fail to see why 100 nested replace()s would be needed.
I am working on a legacy system a client has. Phone numbers are stored in a multitude of ways. Ex:
514-879-9989
514.989.2289
5147899287
The client wants to be able to search the database by phone number.
How could this be achieved without normalizing the data stored in the database? Is this possible?
I am wondering if it is possible to have a query that looks like:
SELECT FROM table WHERE phonenumber LIKE %input%
but that takes into account only the numerical characters in the db?
$sql = "SELECT * FROM tab
WHERE replace(replace(phone, '.', ''), '-', '') like '%". $input ."%'"
Yes you can add more replace according to values in your table, as mentioned by #spencer7593 eg:
$sql = "SELECT * FROM tab
WHERE replace(replace(replace(replace(replace(replace(phone, '.', ''), '-', ''), '+', ''), '(', ''), ')', ''), ' ', '') like '%". $input ."%'"
but I would prefer to cleanup the data before the query.
The approach I would take with this (i.e. not having a "normalized" value with only digits available, and a restriction of not adding an additional column with the normalized value...)
I would take the user input for the search, and add wild cards in strategic locations. For example, if the user provides search input of 3155551212), then I'd run a query that has a predicate equivalent to this:
phonenumber LIKE '%315%555%1212%'
But if I'm not guaranteed that the provided search digits will be a full three digit area code, a three digit exchange (central office) code, and a four digit line number, for a broader search, I'd add wild cards between all of the provided digits, e.g.
phonenumber LIKE '%3%1%5%5%5%5%1%2%1%2%'
This latter approach is less than ideal, because it could potentially provide more matches than aren't intended. Especially if the user is providing fewer than ten digits. For example, consider a phonenumber value:
'+1 (315) 555-7172 ext. 123'
As a demonstration:
SELECT '+1 (315) 555-7172 ext. 123' LIKE '%3%1%5%5%5%5%1%2%1%2%'
, '+1 (315) 555-7172 ext. 123' LIKE '%315%555%1212%'
There's no builtin string function in MySQL that will extract the digit characters from a string.
If you want a function that does that, e.g.
SELECT only_digits_from('+1 (315) 555-7172 ext. 123')
to return
13155557172123
You'd have to create a stored function that does that. I wouldn't attempt doing it inline in the SQL statement, that would require an atrociously long and ugly expression.
This is piece of code i frequently use to clean up the database columns. I have modified to to be fit for your purpose.
Update Table SET Column =
replace
(replace
(replace(column,
'-','',
'.',''),
' ','')
)
WHERE Column is Not Null
My user table has a column "name" which contains information like this:
Joe Lee
Angela White
I want to search for either first name or last name efficiently. First name is easy, I can do
SELECT * FROM user WHERE name LIKE "ABC%"
But for last name, if I do
SELECT * FROM user WHERE name LIKE "%ABC"
That would be extremely slow.
So I am thinking about counting the characters of the input, for example, "ABC" has 3 characters, and if I can search only the last three characters in name column, that would be great. So I want something like
SELECT * FROM user WHERE substring(name, end-3, end) LIKE "ABC%"
Is there anything in MySQL that can do this?
Thanks so much!
PS. I cannot do fulltext because our search engine doesn't support that.
The reason that
WHERE name LIKE '%ith'
is a slow way to look for 'John Smith' by last name is the same reason that
WHERE Right(name, InStr(name, ' ' )) LIKE 'smi%'
or any other expression on the column is slow. It defeats the use of the index for quick lookup and leaves the MySQL server doing a full table scan or full index scan.
If you were using Oracle (that is, if you worked for a formerly wealthy employer) you could use function indexes. As it is you have to add some extra columns or some other helping data to accelerate your search.
Your smartest move is to split your first and last names into separate columns. Several other people have pointed out good reasons for doing that.
If you can't do that you could try creating an extra column which contains the name string reversed, and create an index on that column. That column will have, for example, 'John Smith' stored as 'htimS nhoJ'. Then you can search as follows.
WHERE nameReversed LIKE CONCAT(REVERSE('ith'),'%')
This search will use the index and be decently fast. I've had good success with it.
You're close. In MySQL you should be able to use InStr(str, substr) and Right(str, index) to do the following:
SELECT * FROM user WHERE Right(name, InStr(name, " ")) LIKE "ABC%"
InStr(name, " ") returns the index of the Space character (you may have to play with the " " syntax). This index is then used in the Right() function to search for only the last name (basically; problems arise when you have multiple names, multiple spaces etc). LIKE "ABC%" would then search for a last name starting with ABC.
You cannot use a fixed index as names that are more than 3 or less than 3 characters long would not return properly as you suggest.
However, as Zane said, it's a much better practise to use seperate fields.
If it is a MyIsam table, you may use Free text search to do the same.
You can use the REGEXP operator:
SELECT * FROM user WHERE name REGEXP "ABC$"
http://dev.mysql.com/doc/refman/5.1/en/regexp.html
I have a table of artist names such as Lady Gaga, Jason Mraz, Death Cab For Cutie, etc.
I want to be able to search for artists with no spaces. For instance, search the artists table where name is jasonmraz.
Is there any way to do this with mysql? Or should I create a new column in my table to hold these types of names?
SELECT REPLACE(your_col, " ", "") ...
or
SELECT your_col
WHERE REPLACE(your_col, " ", "") = "somestringsansspaces"
You can use the REPLACE function
SELECT REPLACE('Lady Gaga', ' ', '');
Will return LadyGaga
Using REPLACE(), as suggested, will work.
Bear in mind, however, that this is more work for your server, so if you have a large table of names (which I doubt you do, but never mind) you might want to consider adding a field of the names with no spaces.
Is ir possible to sort a column alphabetically but ignoring certain words like e.g 'The'
e.g.
A normal query would return
string 1
string 3
string 4
the string 2
I would like to return
string 1
the string 2
string 3
string 4
Is this possible?
EDIT
Please note I am looking to replace multiple words like The, A, etc... Can this be done?
You can try
SELECT id, text FROM table ORDER BY TRIM(REPLACE(LOWER(text), 'the ', ''))
but note that it will be very slow for large datasets as it has to recompute the new string for every row.
IMO you're better off with a separate column with an index on it.
For multiple stopwords just keep nesting REPLACE calls. :)
This will replace all leading "The " as an example
SELECT *
FROM YourTable
ORDER BY REPLACE(Val,'The ', '')
Yes, it should be possible to use expressions with the ORDER-part:
SELECT * FROM yourTable ORDER BY REPLACE(yourField, "the ", "")
I have a music listing that is well over 75,000 records and I had encountered a similar situation. I wrote a PHP script that checked for all string that began with 'A ', 'An ' or 'The ' and truncated that part off the string. I also converted all uppercase letters to lowercase and stored that string in a new column. After setting an index on that column, I was done.
Obviously you display the initial column but sort by the newly-created indexed column. I get results in a second or so now.