i have columns in two separate tables that i'm using in a join and subsequent update. i want to be able to see if one column has all of its content captured in a second column.
for instance, here is a sample of contents from each column:
city_table1 | city_table2
Portsmouth Portsmouth, New Hampshire, USA
i want to be able to have a where clause in a select statement that will match the two columns based on the contents in city_table1 (but i can't just do a left or right trim based on content variance), so i'm envisioning something like
where city_table1 like ('%' + city_table2 '%')
is my logic off here? should i referse the two fields in this where clause? i've tried this in postgres and got no results when i know it should work if i have the syntax right.
thank you!
Postgres uses || to concatenate strings. So:
where city_table2 like ('%' || city_table1 || '%')
Also, you have the comparison backwards. The shorter string is surrounded by the '%'.
And, this would be simpler with regular expressions:
where city_table2 ~ city_table1
Related
I have a table with a column feature of type text and a text array (text[]) named args. I need to select from the table those rows in which the feature column contains at least one of the elements of the args array.
I've tried different options, including this:
SELECT * FROM myTable WHERE feature LIKE '%' + ANY (args) + '%';
But that does not work.
The simple solution is to use the regular expression match operator ~ instead, which works with strings in arg as is (without concatenating wildcards):
SELECT *
FROM tbl
WHERE feature ~ ANY(args);
string ~ 'pattern' is mostly equivalent to string LIKE '%pattern%', but not exactly, as LIKE uses different (and fewer) special characters than ~. See:
Escape function for regular expression or LIKE patterns
If that subtle difference is not acceptable, here is an exact implementation of what you are asking for:
SELECT *
FROM tbl t
WHERE t.feature LIKE ANY (SELECT '%' || a || '%' FROM unnest(t.args) a);
Unnest the array, pad each element with wildcards, and use LIKE ANY with the resulting set.
See:
IN vs ANY operator in PostgreSQL
I am attempting to make an Ecto query where I concatenate the first and last name of a given query result, then perform an ilike search using a query string. For example, I may want to search the database for all names that start with "Bob J". Currently, my code looks like this:
pending_result_custom_search = from result in pending_result_query,
where: ilike(fragment("CONCAT(?, '',?)", result.first_name, result.last_name), ^query)
(pending_result_query is a previous query that I am composing on top of)
This approach does not work and I continue to get an empty query set. If I perform the query doing something like this
query = "Bob"
pending_result_custom_search = from result in pending_result_query,
where: ilike(fragment("CONCAT(?, '',?)", "%Bob%", ""), ^query)
I get the correct functionality.
What is the proper syntax to get the first approach working properly?
I think in your case I would use only fragment, e.g.
query = "%" <> "Bob" <> "%"
pending_result_custom_search = from result in pending_result_query,
where: fragment("first_name || last_name ILIKE = ?", ^query)
That way you can shift the focus to PostGres and use its functions instead of worrying too much about the Ecto abstractions of them. In the above example, I used || to concatenate column values, but you could use PostGres' CONCAT() if you desired:
pending_result_custom_search = from result in pending_result_query,
where: fragment("CONCAT(first_name, last_name) ILIKE = ?", ^query)
Note that both examples here did not include a space between first_name and last_name. Also, I added the % characters to the search query before binding it.
I need a like expression that will match a character whether or not it exists. It needs to match the following values:
..."value": "123456"...
..."value": "123456"...
"...value":"123456"...
This like statement will almost work: LIKE '%value":%"123456"%'
But there are values like this one that would also match, but I don't want returned:
..."value":"99999", "other":"123456"...
A regex expression to do what I'm looking to do is 'value": *?"123456"'. I need to do this in SQL Server 2008 and I don't believe there is good regex support in that version. How can I match using a like statement?
Remove the whitespace in your compare with REPLACE():
WHERE REPLACE(column,' ','') LIKE '%"value":"123456"%'
May need a double replace for tabs:
REPLACE(REPLACE(column,' ',''),' ','')
I don't think you can with the like operator. You could exclude ones you could match, like if you want to make sure it just doesn't contain other:
[field] LIKE '%value":%"123456"%` AND [field] NOT LIKE '%"other"%'
Otherwise I think you'd have to do some processing on the string. You could write a UDF to take the string and parse it to find the value for 'value' and compare based on that:
dbo.fn_GetValue([field], 'value') = '123456'
The function could find the index of '"' + #name + '"', find the next index of a quote, and the one after that, then get the string between those two quotes and return it.
I'm trying to compose a WHERE statement that will match rows where a column value is a substring of another string.
For example, I might have an event record with a name field of Edward Sharpe. I'd like to do something like:
SELECT * FROM events WHERE(name LIKE 'Edward Sharpe and the Magnetic Zeroes');
This doesn't work. I've also various permutations of:
SELECT * FROM events WHERE('%' || name || '%' LIKE 'Edward Sharpe and the Magnetic Zeroes');
Which also doesn't work.
Your second attempt is painfully close to correct. The LIKE keyword takes a string on its left, and a pattern on its right. Both can be expressions, but % only has a special meaning in the pattern to the right.
Try this:
SELECT * FROM events
WHERE name LIKE '%Edward Sharpe and the Magnetic Zeroes%';
Or rather this:
SELECT * FROM events
WHERE 'Edward Sharpe and the Magnetic Zeroes' LIKE '%' || name || '%';
Also note that all string operations in Postgres are case sensitive by default. To match a pattern ignoring case, use ILIKE in place of LIKE.
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.