So we have a table with a field that contains strings.
These strings can contain wildcards.
For example:
id | name
---+----------------
1 | thomas
2 | san*
3 | *max*
Now I want to select from that table with respect to these wildcards.
For example something like this:
SELECT * FROM table WHERE name = 'sandra'.
That SELECT should fetch the record with ID = 2 from my table.
Note that it would be ok to use % instead of * as the wildcard character in the table.
Any way to achieve this in OpenSQL?
You can use wildcards, just the sign (like Matecki said), is %.
Take a look here:
https://scn.sap.com/thread/1418148
Additionally You could create and use a ranges table in the where clause. If You do not know, what it is, and how it can be done, just tell me. Populate the ranges table like this: OPTION = CP, SIGN = I, LOW = san.
Ok for You?
UPDATE:
I was wrong and changed the answer
Related
I have a problem where I have two tables. One table constains urls and their information and another groups of urls that should be grouped by a pattern.
Urls table:
------------------------------------------------
| url | files |
| https://myurl1/test/one/es/main.html | 530 |
| https://myurl1/test/one/en/main.html | 530 |
| https://myurl1/test/one/ar/main.html | 530 |
------------------------------------------------
Urls patterns table:
---------------------------------------------
| group | url_pattern |
| group1 | https://myurl1/test/one/(es|en)/%|
| group2 | https://myurl1/test/one/(ar)/% |
---------------------------------------------
I have tried something like this bearing in mind that url_patterns will only have one row per group.
SELECT * FROM urls_table
WHERE url SIMILAR TO (SELECT MAX (url_pattern) FROM url_patterns WHERE group='group1')
LIMIT 10
The main problem here is that it seems that applying SIMILAR TO with a column argument is not working.
Could anyone give me some advices?
Thanks in advance.
You are running into the requirement that regexp patterns are compiled and that SIMILAR TO is a layer on regexp. So what you are trying to do won't work. I believe there are a number of other ways to do this.
I) Change to LIKE pattern matching: LIKE patterns aren't precompiled so can use dynamic patterns. The downside is that they are more limited but I think you can still do what you want. Just change your patterns to be set of pattern columns (if the number of patterns is limited) and test for all the patterns. Unneeded patterns can just be a value that can never match. Definitely a brute force hack.
II) Change to LIKE pattern matching w/ SQL to provide OR behavior: have multiple LIKE patterns in the url_pattern column separated by '|' (for example). Then use split_part to match each sub-pattern - a bit complex and possible slow but works. Like this:
SELECT url
FROM urls_table
LEFT JOIN (SELECT split_part(pattern, '|', part_no::int) as pattern
FROM url_patterns
CROSS JOIN (SELECT row_number() over () as part_no FROM urls_table)
WHERE "group" = 'group1'
)
ON url LIKE pattern
WHERE p.pattern IS NOT NULL;
You will also need to change your pattern strings to use the simpler LIKE format and use '|' for multiple possibilities - Ex: Group1 pattern becomes 'https://myurl1/test/one/es/%|https://myurl1/test/one/en/%'
III) Use some front-end query modification to find the pattern for the group and apply it to query BEFORE it is sent to the compiler. This could be an external tool or a stored procedure on Redshift. Get the pattern in one query and use it to issue the second query.
Do you want exists?
SELECT u.*
FROM urls_table u
WHERE EXISTS (SELECT 1
FROM url_patterns p
WHERE u.url SIMILAR TO p.url_pattern AND
p.group = 'group1'
)
LIMIT 10;
I have a table with a column Document that is FullText Index.
Let say I have this in this table:
| ID | Document |
| 1 | WINTER SUMMER SPRING OTHER |
My requirement is to find rows that contains 'ER'.
For this I am querying like this:
SELECT TOP 100
[FullTextSearch].[Document], [FullTextSearch].[ID]
FROM
[FullTextSearch]
WHERE
CONTAINS(Document, '"*ER*"')
But this is not working.
Please suggest what should be best way to do this using FullTextSearch.
I am expecting id 1 should be returned.
You can user LIKE operator to find the value.
The LIKE operator is used in a WHERE clause to search for a specified pattern in a column.
There are two wildcards used in conjunction with the LIKE operator:
% - The percent sign represents zero, one, or multiple characters
_ - The underscore represents a single character
Syntax,
SELECT column1, column2, ...
FROM table_name
WHERE columnN LIKE pattern;
This query can help to find the result.
SELECT Document,ID FROM FullTextSearch
WHERE Document LIKE '%ER%';
It's a wildcard query...This should work.
SELECT TOP 100
[FullTextSearch].[Document], [FullTextSearch].[ID]
FROM
[FullTextSearch]
WHERE
Document like '%ER%'
========OR=============
SELECT TOP 100
[FullTextSearch].[Document], [FullTextSearch].[ID]
FROM
[FullTextSearch]
WHERE
CONTAINS(Document, '%ER%')
everyone;
I am working on a database where path information are stored, one simplified table is shown below
path_id | path value
1 //a/b/c/d
2 //a/b/e
3 //a/b
4 //a/bcd
So here is the question, how can I get information where has '//a/b' as the prefix? In this case, the result should be:
path_id | path value
1 //a/b/c/d
2 //a/b/e
3 //a/b
I am seeking for a more elegant and optimized query, other than using logic operators like 'OR'.
Thanks.
SELECT * FROM YourTable WHERE path_value like '//a/b/%' OR path_value = '//a/b'
Note the extra slash before the wild card in the first part of the WHERE statement. This will exclude //a/bcd.
the % acts as a wild card so it anything can follow that part.
SELECT * FROM tableName WHERE path_value like '//a/b/%' OR path_value = '//a/b'
Let's say I have a simple table with two columns: id (int) and name (varchar). In this table I store some names which are in Polish, e.g.:
1 | sępoleński
2 | świecki
3 | toruński
4 | Włocławek
Now, let's say I want to sort the results by name:
SELECT * FROM table ORDER BY name;
If I have C locale, I get:
4 | Włocławek
1 | sępoleński
3 | toruński
2 | świecki
which is wrong, because "ś" should be after "s" and before "t". If I use Polish locale (pl_PL.UTF-8), I get:
1 | sępoleński
2 | świecki
3 | toruński
4 | Włocławek
which is also not what I want, because I would like names starting with capital letters to be first just like in C locale, like this:
4 | Włocławek
1 | sępoleński
2 | świecki
3 | toruński
How can I do this?
If you want a custom sort, you must define some function that modifies your values in some way so that the natural ordering of the modified values fits your requirement.
For example, you can append some character or string it the value starts with uppercase:
CREATE OR REPLACE FUNCTION mysort(text) returns text IMMUTABLE as $$
SELECT CASE WHEN substring($1 from 1 for 1) =
upper( substring($1 from 1 for 1)) then 'AAAA' || $1 else $1 END
;
$$ LANGUAGE SQL;
And then
SELECT * FROM table ORDER BY mysort(name);
This is not foolprof (you might want to change 'AAA' for something more apt) and hurts performance, of course.
If you want it efficient, you'll need to create another column that "naturally" sorts correctly (e.g. even in the C locale), and use that as a sorting criterion. For that, you should use the approach of the strxfrm C library function. As a straight-forward strxfrm table for your approach, replace each letter with two ASCII letters: 's' would become 's0' and 'ś' would become 's1'. Then 'świecki' becomes 's1w0i0e0c0k0i0', and the regular ASCII sorting will sort it correctly.
If you don't want to create a separate column, you can try to use a function in the where clause:
SELECT * FROM table ORDER BY strxfrm(name);
Here, strxfrm needs to be replaced with a proper function. Either you write one yourself, or you use the standard translate function (although this doesn't support replacing a character with two of them, so you'll need some more involved transformation).
I have some table 'TableName' like Id,Name:
1 | something
2 | _something
3 | something like that
4 | some_thing
5 | ...
I want to get all rows from this table where name containes 'some'.
I have 2 ways:
SELECT * FROM TableName WHERE Name like '%some%'
Result is table :
1 | something
2 | _something
3 | something like that
4 | some_thing
But if I use CONTAINS function
SELECT * FROM TableName WHERE CONTAINS(Name,'"*some*"')
I get only
1 | something
3 | something like that
What should I do to make CONTAINS function work properly?
The last time I looked (admittedly SQL Server 2000) CONTAINS didn't support wildcard matching at the beginning of words, only at the end. Also, you might need to check your noise files to see if the "_" character is being ignored.
Also see
How do you get leading wildcard full-text searches to work in SQL Server?
http://doc.ddart.net/mssql/sql70/ca-co_15.htm
If you read this article you will see that * means prefix this means that word must start with this, but like means the word contains key phrase.
Best Regards,
Iordan
Try this:
SELECT * FROM TableName WHERE CONTAINS(Name,'some')