I'm an access beginner.
I'm trying to write a query that finds all partial matches between a field in one table and a field in another.
I'll give a simplified example of what I'm looking for as a final result and maybe you can help me write the SQL for it.
Table 1: Strings
pre
ing
up
Table 2: Words
present
door
cup
making
kingdom
Desired Result Table:
pre present
up cup
ing kingdom
ing making
Here's what I have - it can give me a prompt where I type in what I'm looking for. (e.g. if I type in 'up', then I get cup from the word list). What I want is a way to automate the whole process for all of the strings in the first list.
SELECT *
FROM word
WHERE (((word) LIKE "*" & string & "*"));
Thanks,
I don't have Access in front of me, but I think you can do something similar to this (this is for SQL Server):
select stringname, wordname
from strings
join words on wordname like '%' + stringname + '%'
SQL Fiddle Demo
If I had to guess at the Access syntax, replace % with * and possibly change the + to &, but hopefully this will get you going in the right direction.
Related
My database is Oracle 11g.
I want to do a in query in sql. The query criteria is to matched some of the characters of a field:
description
CWLV321900017391;EFHU3832239
CWLV321900017491;ERHU3832239
CWLV321900017591;ERHU3832239
CWLV321900017691;ERHU3832239
My query is like this:
select * from product where description in ('CWLV321900017391', 'CWLV321900017491');
It returns no records in result.
I expect the result like below:
description
CWLV321900017391;EFHU3832239
CWLV321900017491;ERHU3832239
How to get it by SQL?
thanks.
You are using IN to search for a partial string match in the description table. This is not going to return the results, as IN will only match exact values.
Instead, one way to achieve this is to use a LIKE operator with %:
select *
from product
where (description LIKE 'CWLV321900017391%' OR
description LIKE 'CWLV321900017491%');
The % at the end indicates that anything can follow after the specified text.
This will return any description that starts with CWLV321900017391 or CWLV321900017491.
Incidentally, if your search term occurs anywhere in the description field, you will need to use a % at each end of the search term:
description LIKE '%CWLV321900017391%' OR description LIKE '%CWLV321900017491%'
There are various ways to solve it. Here is one:
select * from product
where instr (description, 'CWLV321900017391') > 0
or instr (description, 'CWLV321900017491') > 0;
If you know you're always searching for the start of the description you can use:
select * from product
where substr (description, 1, 16) in ('CWLV321900017391','CWLV321900017491')
Also, there's LIKE or REGEX_LIKE solution. It depends really on what strings you're searching for.
Of course none of these solutions is truly satisfactory, and for large volumes of data may exhibit suck-y performance. The problem is the starting data model violates First Normal Form by storing non-atomic values. Poor data models engender clunky SQL.
You can try below way -
select * from product
where substr(description,1,instr(description,';',1,1)-1) in ('CWLV321900017391', 'CWLV321900017491')
I am opposed to storing multiple values in a delimited string format. There are many better alternatives, but let me assume that you don't have a choice on the data model.
If you are stuck with strings in this format, you should take the delimiters into account. You can do this using like:
where ';' || description || ';' like '%;CWLV321900017391;%'
You can also do something similar with regexp_like() if you want to look for one of several values:
where regexp_like(';' || description || ';',
'(;CWLV321900017391;)|(;CWLV321900017391;)'
)
In REGEX you can do something like [a-c]+, which will match on
aaabbbccc
abcccaabc
cbccaa
b
aaaaaaaaa
In SQL LIKE it seems that one can either do the equivalent of ".*" which is "%", or [a-c]. Is it possible to use the +(at least one) quantifier in SQL to do [a-c]+?
EDIT: Just to clarify, the desired end-query would look something like
SELECT * FROM table WHERE column LIKE '[a-c]+'
which would then match on the list above, but would NOT match on e.g "xxxxxaxxxx"
As a general rule, SQL Server's LIKE patterns are much weaker than regular expressions. For your particular example, you can do:
where col not like '%[^a-c]%'
That is, the column contains no characters that are not a, b, or c.
You can use regex in SQL with combination of LIKE e.g :
SELECT * FROM Table WHERE Field LIKE '%[^a-z0-9 .]%'
This works in SQL
Or in your case
SELECT * FROM Table WHERE Field LIKE '%[^a-c]%'
I seems you want some data from database, That is you don't know exactly, You must show your column and the all character that you want in that filed.
i've got a problem with transferring "real-World" data into my schema.
It's actually a "project" for my Database course and they gave us ab table with dog race results. This Table has a column which contains the name of the Dog (which itself consists of the actuall name and the name of the breeder) and informations about the Birthcountry, actual living Country and the birth year.
Example filed are "Lillycette [AU 2012]" or "Black Bear Lee [AU/AU 2013]" or "Lemon Ralph [IE/UK 1998]".
I've managed it to get out the first word and save it in the right column with split_part like this:
INSERT INTO tblHund (rufname)
SELECT
split_part(name, ' ', 1) AS rufname,
FROM tblimport;
tblimport is a table where I dumped the data from the csv file.
That works just as it should.
Accessing the second part of the Name with this fails because sometimes there isn't a second part and sometimes times there second part consists of two words.
And this is the where iam stuck right now.
I tried it with substring and regular expressions:
INSERT INTO tblZwinger (Name)
SELECT
substring(vatertier from E'[^ ]*\\ ( +)$')AS Name
FROM tblimport
WHERE substring(vatertier from E'[^ ]*\\ ( +)$') != '';
The above code is executed without errors but actually does nothing because the SELECT statement just give empty strings back.
It took me more then 3h to understand a bit of this regular Expressions but I still feel pretty stupid when I look at them.
Is there any other way of doing this. If so just give me a hint.
If not what is wrong with my expression above?
Thanks for your help.
You need to use atom ., which matches any single character inside capturing group:
E'[^ ]*\\ (.+)$'
SELECT
tblimport.*,
ti.parts[1] as f1,
ti.parts[2] as f2, -- It should be the "middle part"
ti.parts[3] as f3
FROM
tblimport,
regexp_matches(tblimport.vatertier, '([^\s]+)\s*(.*)\s+\[(.*)\]') as ti(parts)
WHERE
nullif(ti.parts[2], '') is not null
Something like above.
I have column store_name (varchar). In that column I have entries like prime sport, best buy... with a space. But when user typed concatenated string like primesport without space I need to show result prime sport. how can I achieve this? Please help me
SELECT *
FROM TABLE
WHERE replace(store_name, ' ', '') LIKE '%'+#SEARCH+'%' OR STORE_NAME LIKE '%'+#SEARCH +'%'
Well, I don't have much idea, and even I am searching for it. But may be what I know works for you, You can achieve this by performing different type of string operations:
Mike can be Myke or Myce or Mikke or so on.
Cat an be Kat or katt or catt or so on.
For this you should write a function to generate number of possible strings and then form a SQL Query using all these, and query the database.
A similar kind of search in known as Soundex Search from Oracle and Soundex Search from Microsoft. Have a look of it. this may work.
And overall make use of functions like upper and lower.
Have you tried using replace()
You can replace the white space in the query then use like
SELECT * FROM table WHERE replace(store_name, ' ', '') LIKE '%primesport%'
It will work for entries like 'prime soft' querying with 'primesoft'
Or you can use regex.
I am working on a small relational database for school and having trouble with a simple query. I am supposed to find all records in a table that have a field with the word 'OMG' somewhere in the text.
I have tried a few other and I can't figure it out. I am getting an invalid relational operator error.
my attempts:
select * from comments where contains('omg');
select * from comments where text contains('omg');
select * from comments where about('omg');
and several more variations of the above. As you can see I am brand spanking new to SQL.
text is the name of the field in the comments table.
Thanks for any help.
Assuming that the column name is text:
select * from comments where text like '%omg%';
The percentage % is a wild-card which means that any text can come before/after omg
Pay attention, if you don't want the results to contain words in which omg` is a substring - you might want to consider adding whitespaces:
select * from comments where text like '% omg %';
Of course that it doesn't cover cases like:
omg is at the end of the sentence (followed by a ".")
omg has "!" right afterwards
etc
You may want to use the LIKE operator with wildcards (%):
SELECT * FROM comments WHERE text LIKE '%omg%';