In raw SQL, it is possible to look through a database to find rows with a column for which the contents are written in all capital letters; that question was answered here.
Is there a way to accomplish the same thing using the Django ORM and without resorting to .raw() ?
You can use regular expression match in Django ORM. Link to documentation - https://docs.djangoproject.com/en/dev/ref/models/querysets/#iregex
Example:
Entry.objects.get(title__regex=r'^(An?|The) +')
SQL equivalents:
SELECT ... WHERE title REGEXP BINARY '^(An?|The) +'; -- MySQL
SELECT ... WHERE REGEXP_LIKE(title, '^(an?|the) +', 'c'); -- Oracle
SELECT ... WHERE title ~ '^(An?|The) +'; -- PostgreSQL
SELECT ... WHERE title REGEXP '^(An?|The) +'; -- SQLite
Using raw strings (e.g., r'foo' instead of 'foo') for passing in the regular expression syntax is recommended.
EDIT:
You can add the regular expression like:
Entry.objects.get(title__regex=r'^[[:upper:]]+$') #not tested
Figured it out. Looks like the best way is to use extra. For example,
MyModel.objects.extra(where=['title = UPPER(title)'])
Related
In Postgresql database I have a column called names where I have some names which need to be parsed using regex to clean up punctuation parts. I am able to get a clean name using regexp_replace as follows:
select regexp_replace(name,'\.COM|''[A-Z]|[^a-zA-Z0-9 -]+|\s(?=&)|(?<!\w\w)(?:\s+|-)(?!\w\w)','','g')
from tableA
However, I would like to compare with some strings that are also cleaned of punctuation. How can I use similar to with the formed regular expression?
select name
from tableA
where (lower(name) ~ '\.COM|''[A-Za-z]|[^a-zA-Z0-9 -]+|\s(?=&)|(?<!\w\w)(?:\s+|-)(?!\w\w)') as nameParsed similar to '(fg )%' and
(lower(name) ~ '\.COM|''[A-Za-z]|[^a-zA-Z0-9 -]+|\s(?=&)|(?<!\w\w)(?:\s+|-)(?!\w\w)') as nameParsed similar to '%( cargo| carrier| cartage )%'
With the previous query I am getting this error:
LINE 3: ...-zA-Z0-9 -]+|\s(?=&)|(?<!\w\w)(?:\s+|-)(?!\w\w)') as namePar...
I have tried in where clause like this and it seems to be working:
select name
from tableA
where (select lower(regexp_replace(name,'\.COM|''[A-Z]|[^a-zA-Z0-9 -]+|\s(?=&)|(?<!\w\w)(?:\s+|-)(?!\w\w)','','g'))) similar to '(fg )%'
Is this the best approach? The execution time went to 46 seconds :(
Thanks in advance
You're trying to get a column name in a WHERE clause (is a comparison, not a column). So, you can use as follows:
SELECT name
FROM "tableA"
WHERE (regexp_replace(name,'\.COM|''[A-Z]|[^a-zA-Z0-9 -]+|\s(?=&)|(?<!\w\w)(?:\s+|-)(?!\w\w)','','g') similar to '(fg )%'
OR regexp_replace(name,'\.COM|''[A-Z]|[^a-zA-Z0-9 -]+|\s(?=&)|(?<!\w\w)(?:\s+|-)(?!\w\w)','','g') similar to '%( cargo| carrier| cartage )%');
Alternatively, you can use ilike instead of similar to if you want to find a specific word.
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 require a select query that adds a space to the data based on the placement of the capital letters i.e. 'HelpMe' using this query would be displayed as 'Help Me' . Note i cannot use a stored function to do this the it must be done in the query itself. The Data is of variable length and query must be in SQL. Any Help will be appreciated.
Thanks
You need to use user defined function for this until MS give us support for regular expressions. Solution would be something like:
SELECT col1, dbo.RegExReplace(col1, '([A-Z])',' \1') FROM Table
Aldo this would produce leading space that you can remove with TRIM.
Replace regular expresion function:
http://connect.microsoft.com/SQLServer/feedback/details/378520
About dbo.RegexReplace you can read at:
TSQL Replace all non a-z/A-Z characters with an empty string
Assume if you are using Oracle RDBMS, you use the following,
REGEX_REPLACE
SELECT REGEXP_REPLACE('ILikeToWatchCSIMiami',
'([A-Z.])', ' \1')
AS RX_REPLACE
FROM dual
;
Managed to get this output: * SQLFIDDLE
But as you see it doesn't treat well on words such as CSI though.
Does anyone have a LIKE pattern that matches whole words only?
It needs to account for spaces, punctuation, and start/end of string as word boundaries.
I am not using SQL Full Text Search as that is not available. I don't think it would be necessary for a simple keyword search when LIKE should be able to do the trick. However if anyone has tested performance of Full Text Search against LIKE patterns, I would be interested to hear.
Edit:
I got it to this stage, but it does not match start/end of string as a word boundary.
where DealTitle like '%[^a-zA-Z]pit[^a-zA-Z]%'
I want this to match "pit" but not "spit" in a sentence or as a single word.
E.g. DealTitle might contain "a pit of despair" or "pit your wits" or "a pit" or "a pit." or "pit!" or just "pit".
Full text indexes is the answer.
The poor cousin alternative is
'.' + column + '.' LIKE '%[^a-z]pit[^a-z]%'
FYI unless you are using _CS collation, there is no need for a-zA-Z
you can just use below condition for whitespace delimiters:
(' '+YOUR_FIELD_NAME+' ') like '% doc %'
it works faster and better than other solutions. so in your case it works fine with "a pit of despair" or "pit your wits" or "a pit" or "a pit." or just "pit", but not works for "pit!".
I think the recommended patterns exclude words with do not have any character at the beginning or at the end. I would use the following additional criteria.
where DealTitle like '%[^a-z]pit[^a-z]%' OR
DealTitle like 'pit[^a-z]%' OR
DealTitle like '%[^a-z]pit'
I hope it helps you guys!
Surround your string with spaces and create a test column like this:
SELECT t.DealTitle
FROM yourtable t
CROSS APPLY (SELECT testDeal = ' ' + ISNULL(t.DealTitle,'') + ' ') fx1
WHERE fx1.testDeal LIKE '%[^a-z]pit[^a-z]%'
If you can use regexp operator in your SQL query..
For finding any combination of spaces, punctuation and start/end of string as word boundaries:
where DealTitle regexp '(^|[[:punct:]]|[[:space:]])pit([[:space:]]|[[:punct:]]|$)'
Another simple alternative:
WHERE DealTitle like '%[^a-z]pit[^a-z]%' OR
DealTitle like '[^a-z]pit[^a-z]%' OR
DealTitle like '%[^a-z]pit[^a-z]'
This is a good topic and I want to complement this to someone how needs to find some word in some string passing this as element of a query.
SELECT
ST.WORD, ND.TEXT_STRING
FROM
[ST_TABLE] ST
LEFT JOIN
[ND_TABLE] ND ON ND.TEXT_STRING LIKE '%[^a-z]' + ST.WORD + '[^a-z]%'
WHERE
ST.WORD = 'STACK_OVERFLOW' -- OPTIONAL
With this you can list all the incidences of the ST.WORD in the ND.TEXT_STRING and you can use the WHERE clausule to filter this using some word.
You could search for the entire string in SQL:
select * from YourTable where col1 like '%TheWord%'
Then you could filter the returned rows client site, adding the extra condition that it must be a whole word. For example, if it matches the regex:
\bTheWord\b
Another option is to use a CLR function, available in SQL Server 2005 and higher. That would allow you to search for the regex server-side. This MSDN artcile has the details of how to set up a dbo.RegexMatch function.
Try using charindex to find the match:
Select *
from table
where charindex( 'Whole word to be searched', columnname) > 0
I am aware of the existence of the RLIKE and REGEX operators, but it seems like they cannot be used for that.
Is there a function or an operator that would help me achieve splitting a text field and selecting it as two or more separate fields:
SELECT $1 as `field_a`, $2 as `field_b` FROM `table` WHERE `field` RLIKE '^(.+):(.+)$';
I am writing a log analyzer so it would be very handy to do that in SQL without additional text-crunching.
So you just want to split the string on the first occurrence of ":"?
There are several ways to achieve this in MySQL.
Using your example, here are two approaches off the top of my head. Hopefully they are helpful to you:
select substr(`field`,1,instr(`field`,':')-1) as `field_a`,
substr(`field`,instr(`field`,':')+1) as `field_b`
FROM `table`
WHERE `field` RLIKE '^(.+):(.+)$';
select left(`field`,instr(`field`,':')-1) as `field_a`,
right(`field`,length(`field`)-instr(`field`,':')) as `field_b`
FROM `table`
WHERE `field` RLIKE '^(.+):(.+)$';