I have a Text,
'Me and you against the world' // false
'Can i have an email address' // true
'This is an' // true
'an' //true
I want to check whether the word an is inside my String.
How do I check if a text contains a specific word in SQL? I can't add a full-text catalog. Otherwies i could
SELECT * FROM TABLE WHERE CONTAINS(Text, 'an')
Here's one approach.
DECLARE #table_name table (
column_name varchar(50)
);
INSERT INTO #table_name (column_name)
VALUES ('Me and you against the world')
, ('Can i have an email address')
, ('This is an')
;
SELECT column_name
FROM #table_name
WHERE ' ' + column_name + ' ' LIKE '% an %'
;
There are some way to do this, seem you want find a word and not a part of a word, so you can do in easy way with like operator
You can have 3 cases to found a word
'space'WORD
WORD'space'
'space'WORD'space'
SELECT * FROM TABLE WHERE Field like ' an' OR Field like 'an ' OR
Field like ' an '
Hope it helps
It is perfectly done in MS SQL Server by the CHARINDEX function (it is internal to MS SQL):
if CHARINDEX('an ',#mainString) > 0
begin
--do something
end
The solution was showed before in another post.
The three cases you'll encounter as Luka mentions:
Space before word
Space after word
Space before and after word
To accomplish this, you'll write a query like the following which searches for the whole word, and pads the expression to search with a leading and trailing space to capture words at the start/end of the expression:
Note: I've used a contrived example to make this portable and demonstrable.
select
t.txt
from (
select
'this is an awesome test of awesomeness man' as txt
) t
where
charindex(' an ', ' ' + t.txt + ' ') > 0;
Related
My SQL Server database table has a column text which is a long string of text.
The search list is a string of words separated by comma. I want to grab those rows where the text column contains any one of words in the string.
DECLARE #words_to_search nvarchar(50)
SET #words_to_search = 'apple, pear, orange'
SELECT *
FROM myTbl
WHERE text ??? --how to specify text contains #words_to_search
Thanks a lot in advance.
If you're running SQL Server 2016 or later, you can use STRING_SPLIT to convert the words to search into a single column table, and then JOIN that to your table using LIKE:
DECLARE #words_to_search nvarchar(50)
SET #words_to_search = 'apple,pear,orange'
SELECT *
FROM myTbl
JOIN STRING_SPLIT(#words_to_search, ',') ON text LIKE '%' + value + '%';
Demo on SQLFiddle
Note that as the query is written it will (for example) match apple within Snapple. You can work around that by making the JOIN condition a bit more complex:
SELECT *
FROM myTbl t
JOIN STRING_SPLIT(#words_to_search, ',') v
ON t.text LIKE '%[^A-Za-z]' + value + '[^A-Za-z]%'
OR t.text LIKE value + '[^A-Za-z]%'
OR t.text LIKE '%[^A-Za-z]' + value;
Demo on SQLFiddle
First, I would use exists, unless you want to return the matching word.
Second, you can do this with a single like comparison If words are separated by spaces:
select t.*
from t
where exists (select 1
from string_split(#words_to_search, ',') s
where ' ' + t.text + ' ' like '% ' + value + ' %'
);
For more generic separators, you can use:
select t.*
from t
where exists (select 1
from string_split(#words_to_search, ',') s
where ' ' + t.text + ' ' like '%[^A-Za-z]' + value + '[^A-Za-z]%'
);
Or whatever describes your separators.
Note that your list of words is separated by a comma-space, not just a comma. However, based on your description (not the sample data), I have only used a ',' for the separator.
I am a bit new to this site but I have looked an many possible answers to my question but none of them has answered my need. I have a feeling it's a good challenge. Here it goes.
In one of our tables we list what is used to run a report this can mean that we can have a short EXEC [svr1].[dbo].[stored_procedure] or "...From svr1.dbo.stored_procedure...".
My goal is to get the stored procedure name out of this string (column). I have tried to get the string between '[' and ']' but that breaks when there are no brackets. I have been at this for a few days and just can't seem to find a solution.
Any assistance you can provide is greatly appreciated.
Thank you in advance for entertaining this question.
almostanexpert
Considering the ending character of your sample sentences is space, or your sentences end without trailing ( whether space or any other character other than given samples ), and assuming you have no other dots before samples, the following would be a clean way which uses substring(), len(), charindex() and replace() together :
with t(str) as
(
select '[svr1].[dbo].[stored_procedure]' union all
select 'before svr1.dbo.stored_procedure someting more' union all
select 'abc before svr1.dbo.stored_procedure'
), t2(str) as
(
select replace(replace(str,'[',''),']','') from t
), t3(str) as
(
select substring(str,charindex('.',str)+1,len(str)) from t2
)
select
substring(
str,
charindex('.',str)+1,
case
when charindex(' ',str) > 0 then
charindex(' ',str)
else
len(str)
end - charindex('.',str)
) as "Result String"
from t3;
Result String
----------------
stored_procedure
stored_procedure
stored_procedure
Demo
With the variability of inputs you seem to have we will need to plan for a few scenarios. The below code assumes that there will be exactly two '.' characters before the stored_procedure, and that [stored_procedure] will either end the string or be followed by a space if the string continues.
SELECT TRIM('[' FROM TRIM(']' FROM --Trim brackets from final result if they exist
SUBSTR(column || ' ', --substr(string, start_pos, length), Space added in case proc name is end of str
INSTR(column || ' ', '.', 1, 2)+1, --start_pos: find second '.' and start 1 char after
INSTR(column || ' ', ' ', INSTR(column || ' ', '.', 1, 2), 1)-(INSTR(column || ' ', '.', 1, 2)+1))
-- Len: start after 2nd '.' and go until first space (subtract 2nd '.' index to get "Length")
))FROM TABLE;
Working from the middle out we'll start with using the SUBSTR function and concatenating a space to the end of the original string. This allows us to use a space to find the end of the stored_procedure even if it is the last piece of the string.
Next to find our starting position, we use INSTR to search for the second instance of the '.' and start 1 position after.
For the length argument, we find the index of the first space after that second '.' and then subtract that '.' index.
From here we have either [stored_procedure] or stored_procedure. Running the TRIM functions for each bracket will remove them if they exist, and if not will just return the name of the procedure.
Sample inputs based on above description:
'EXEC [svr1].[dbo].[stored_procedure]'
'EXEC [svr1].[dbo].[stored_procedure] FROM TEST'
'svr1.dbo.stored_procedure'
Note: This code is written for Oracle SQL but can be translated to mySQL using similar functions.
I'm trying to update a column that could possibly have a single space or multiple spaces into just one single space using a plain sql statement not pl sql
I could do it through update table set column_name='' where column_name like '% %'
However, there could be some data such as abc def in that column. I do not want to disturb the pattern of that data meaning if want to do it only when the column is filled with white space and not touch columns that have any data.
I would recommend using a regular expression to do this, both to do the replacement and to do the matching:
UPDATE mytable
SET mycolumn = REGEXP_REPLACE(mycolumn, '\s{2,}', ' ')
WHERE REGEXP_LIKE(mycolumn, '\s{2,}')
This will replace two or more consecutive whitespace characters (spaces, tabs, etc.) with a single space. If you just want to replace spaces and not tabs, carriage returns, or newlines, use the following:
UPDATE mytable
SET mycolumn = REGEXP_REPLACE(mycolumn, ' {2,}', ' ')
WHERE REGEXP_LIKE(mycolumn, ' {2,}')
The reason for using {2,} is so that we don't bother replacing spaces where it need not be done.
With regular expression:
update table set column=regexp_replace(column, ' +', ' ')
Try:
update mytable
set col = ' '
where replace (col, ' ', null) is null;
I have used regexp_like to solve this
update table set column= ' '
where column in (select column from table where regexp_like('column','^\s+$')
Thanks
Try this:
update product set name = replace(name, ' ', ' ') where name like '% %';
where the second parameter of replace expression and like ('% %') contains two blank spaces.
In my case, the result was:
TENIS NIKE AIR => TENIS NIKE AIR
TENIS NIKE MAN => TENIS NIKE MAN
I'm trying to write a query to find the percentage match of a search string in a notes or TEXT column.
This is what I'm starting with:
SELECT *
FROM NOTES
WHERE UPPER(NARRATIVE) LIKE 'PAID CALLED RECEIVED'
Ultimately, what I want to do is:
Split the search string by spaces and search individually for all words in the string
Order the results descending based on percentage match
For example, in the above scenario, each word in the search string would constitute 33.333% of the total. A NARRATIVE with 3 matches (100%) should be at the top of the results, while a match containing 2 of the keywords (66.666%) would be lower, and a match containing 1 of the keywords (33.333%) would be even lower.
I then want to display the resulting percentage match for that row in a column, along with all the other columns from that table (*).
Hopefully, this makes sense and can be done. Any thoughts on how to proceed? This MUST all be done in SQL Server, and I would prefer not to write any CTEs.
Thank you in advance for any guidance.
Here is what I came up with:
DECLARE #VISIT VARCHAR(25) = '999232'
DECLARE #KEYWORD VARCHAR(100) = 'PAID,CALLED,RECEIVED'
DECLARE SPLIT_CURSOR CURSOR FOR
SELECT RTRIM(LTRIM(VALUE)) FROM Rpt_Split(#KEYWORD, ',')
IF OBJECT_ID('tempdb..#NOTES_FF_SEARCH') IS NOT NULL DROP TABLE #NOTES_FF_SEARCH
SELECT N.VISIT_NO
,N.CREATE_DATE
,N.CREATE_BY
,N.NARRATIVE
,0E8 AS PERCENTAGE
INTO #NOTES_FF_SEARCH
FROM NOTES_FF AS N
WHERE N.VISIT_NO = #VISIT
DECLARE #KEYWORD_VALUE AS VARCHAR(255)
OPEN SPLIT_CURSOR
FETCH NEXT FROM SPLIT_CURSOR INTO #KEYWORD_VALUE
WHILE ##FETCH_STATUS = 0
BEGIN
UPDATE #NOTES_FF_SEARCH
SET PERCENTAGE = PERCENTAGE + ( 100 / ##CURSOR_ROWS )
WHERE UPPER(NARRATIVE) LIKE '%' + UPPER(#KEYWORD_VALUE) + '%'
FETCH NEXT FROM SPLIT_CURSOR INTO #KEYWORD_VALUE
END
CLOSE SPLIT_CURSOR
DEALLOCATE SPLIT_CURSOR
SELECT * FROM #NOTES_FF_SEARCH
WHERE PERCENTAGE > 0
ORDER BY PERCENTAGE, CREATE_DATE DESC
There may be a more efficient way to do this but every other road I started down ended in a dead-end. Thanks for your help
If you want to do a "percentage" match, you need to do two things: calculate the number of words in the string and calculate the number of words you care about. Before giving some guidance, I will say that full text search probably does everything you want and much more efficiently.
Assuming the search string has space delimited words, you can count the words with the expression:
(len(narrative) - len(replace(narrative, ' ', '') + 1) as NumWords
You can count the matching words with success replaces. So, for keywords, it would be something like removing each key word, fixing the spaces, and counting the words.
The overall code is best represented with subqueries. The resulting query is something like:
select n.*
from (select n.*,
(len(narrative) - len(replace(narrative, ' ', '') + 1.0) as NumWords,
ltrim(rtrim(replace(replace(replace(narrative + ' ', #keyword1 + ' ', ''),
#keyword2 + ' ', ''),
#keyword3 + ' ', ''))) as NoKeywords
from notes n
) n
order by 1 - (len(NoKeywords) - len(replace(NoKeywords, ' ', '') + 1.0) / NumWords desc;
SQL Server -- as with many databases -- is not particularly good at parsing strings. You can do that outside the query and assign the #keyword variables accordingly.
Database: Sql Server
I have table column named page_text which contains following value
" I Love stackoverflow.com because i can post questions and get
answers at any time. "
Using SQLQUERY i want to search the number of I it has . So in this string it would should return 4.
I should be able to search anything.
declare #search varchar(10) = 'I'
select len(replace(PageText, #search, #search + '#'))
- len(PageText) as Count from YourTable
based on this code: http://www.sql-server-helper.com/functions/count-character.aspx
you create a function:
CREATE FUNCTION [dbo].[ufn_CountSpecificWords] ( #pInput VARCHAR(1000), #pWord VARCHAR(1000) )
RETURNS INT
BEGIN
RETURN (LEN(#pInput) - LEN(REPLACE(#pInput, ' ' + #pWord + ' ', '')))
END
GO
this however implies that you save your strings with a leading and trailing space and you replace any other separators like ',' and '.' with spaces.
and you need to refrase your question if you want the count of words or just the appeareance of a word.
SELECT (LEN(#Text) - LEN(REPLACE(#Text,#SearchString,''))/(Len(#SearchString))