I need help with writing a query which will find Replacement Character in SQL table.
I have multiple cells which contain that character and I want to find all those cells. This is how the value of cell looks like this:
Thank you for your help!
The UNICODE suggestion didn't work for me - the � character was being treated as a question mark, so the query was finding all strings with question marks, but not those with �.
The fix posted by Tom Cooper at this link worked for me: https://social.msdn.microsoft.com/forums/sqlserver/en-US/2754165e-7ab7-44b0-abb4-3be487710f31/black-diamond-with-question-mark
-- Find rows with the character
Select * From [MyTable]
Where CharIndex(nchar(65533) COLLATE Latin1_General_BIN2, MyColumn) > 0
-- Update rows replacing character with a !
Update [MyTable]
set MyColumn = Replace(MyColumn, nchar(65533) COLLATE Latin1_General_BIN2, '!')
Use the Unicode function:
DECLARE #TEST TABLE (ID INT, WORDS VARCHAR(10))
INSERT INTO #TEST VALUES (1, 'A�AA')
INSERT INTO #TEST VALUES (2, 'BBB')
INSERT INTO #TEST VALUES (3, 'CC�C')
INSERT INTO #TEST VALUES (4, 'DDD')
SELECT * FROM #TEST WHERE WORDS LIKE '%' + NCHAR(UNICODE('�')) + '%'
UPDATE #TEST
SET WORDS = REPLACE(WORDS, NCHAR(UNICODE('�')), 'X')
SELECT * FROM #TEST WHERE WORDS LIKE '%' + NCHAR(UNICODE('�')) + '%'
SELECT * FROM #TEST
In Sql you can able to replace the black diamond symbol.
Input : You�ll be coached through alternating
Output : You will be coached through alternating
select replace(description, nchar(65533) COLLATE Latin1_General_BIN2,' wi') from [Fitnesstable]
where description LIKE '%' + '. You'+NCHAR(55296) +'ll'+ '%'
Try the following code to search the query for the character you wish to find.
select field_name from tbl_name where instr(field_name, 'charToBeSearched') > 0;
This query will find and selects the records which has replacement character.
Related
I have a very large table of words that are stored in a column called "English".
Is there a way that I could update the rows in this table such that any word that starts with an uppercase character A-Z is changed to the same character but lowercase.
Please try following.
UPDATE StudentMasters
SET StdName = LOWER(LEFT(StdName,1)) + SUBSTRING(StdName,2,LEN(StdName))
WHERE LEFT(StdName,1) = UPPER(LEFT(StdName,1)) COLLATE Latin1_General_CS_AI
In this we have table name 'StudentMasters' and field name 'StdName'. above query will update your field value as per your expectation.
try this with other sample data,
declare #t table(col1 varchar(50))
insert into #t values ('DF454fdgd'),('45345dfgfgdf')
SELECT *
,replace(col1, SUBSTRING(col1, 1, 2)
, LOWER(SUBSTRING(col1, 1, 2))) newCol
FROM #t
WHERE col1 LIKE '[A-Z]%'
For making the first letter of the column to lowercase ,use the below script.
UPDATE y
SET YourColumn=LOWER(LEFT(YourColumn,1))+SUBSTRING(YourColumn,2,LEN(YourColumn))
FROM YourTable y
I have a table like this:
DECLARE #T TABLE
(note VARCHAR (50))
INSERT #T
SELECT 'Amplifier'
UNION ALL SELECT ';'
UNION ALL SELECT 'Regulator'
How can I replace the semicolon (';') with blank ('').
Expected Output:
Amplifier
'' -- here semicolon replace with blank
Regulator
If you want to replace ALL semicolons from any outputted cell you can use REPLACE like this:
SELECT REPLACE(note,';','') AS [note] FROM #T
Fetching from the given table, use a CASE statement:
SELECT CASE WHEN note = ';' THEN '' ELSE note END AS note FROM #T;
replace() would replace all occurrences of the character. Doesn't seem like you'd want that. This expression only replaces exact matches of the whole string.
It looks like you need to REPLACE all your semicolons:
DECLARE #T TABLE
(note VARCHAR (50))
INSERT INTO #T
SELECT REPLACE(SourceColumn, ';', '')
FROM SourceTable
SQL Server 2017 introduced the function TRANSLATE where you can specifiy a list of characters to be replaced
SELECT TRANSLATE('MAX(0,MIN(h-36,8))', '(,-)', ' ') -->'MAX 0 MIN h 36 8 '
I have a table in sql server 2012, where one column is nvarchar. It contains Romanian characters. We've noticed that only some of the letters 'Ș' do not show in reports at all, so I found that it depends of the keyboard settings.
There are two different keyboard settings for Romanian - Standard and Legacy. Letter 'Ș' - inserted from Rom(Standard) keyboard have ASCII code 63, from Legacy it's 170.
Letter 'Ş' with CHAR(170) - shows in reports, but CHAR(63) - doesn't - even though it's the same letter (should be).
It would be simple if I could replace char(63) with char(170), but I cannot detect rows with character 63. The next select doesn't return rows:
select * from table1 where columnname like '%'+CHAR(63)+'%'
even though if I do select ASCII(SUBSTRING(columnname , 1, 1)) it returns me '63'.
even select charindex(char(63), columnname) - returns me 0
I also tried to do collation:
select * from table1 where columnname COLLATE Latin1_general_CI_AI like N'%s%'
it doesn't help - it returns only rows with 's' and char(170).
Please help me find these rows with wrong 'Ş'
So firstly from my comments, CHAR(63) is misleading as it represents a character that sql server is unable to display:
Unable to replace Char(63) by SQL query
The issue is possibly down to your selected collation, as if I run this sample I get the 2 rows containing the special characters:
CREATE TABLE #temp ( val NVARCHAR(50) )
INSERT INTO #temp
( val )
VALUES ( N'Șome val 1' ),
( N'some val 2' ),
( N'șome other val 3' )
SELECT *
FROM #temp
WHERE val COLLATE Latin1_General_BIN LIKE N'%ș%'
OR val COLLATE Latin1_General_BIN LIKE N'%Ș%'
DROP TABLE #temp
Output
val
=================
Șome val 1
șome other val 3
The specified collation is: Latin1_General_BIN, as found in this post:
replace only matches the beginning of the string
WHERE columnname LIKE N'%'+NCHAR(536)+'%'
This should help you find the character even if it was inserted as an unknown character as in the first insert below.
DECLARE #Table TABLE (text nvarchar(50))
INSERT INTO #Table(text)
SELECT 'Ș'
UNION ALL
SELECT N'Ș'
SELECT UNICODE(text) UNICODE
FROM #Table
Results:
UNICODE
63
536
'Ș' is NCHAR(536) and 'ș' is NCHAR(537).
If you then do:
SELECT * FROM #Table WHERE text LIKE N'%'+NCHAR(536)+'%'
Results:
text
?
Ș
I have following query with LIKE predicate in SQL Server 2012. It replaces white spaces with %. I have two records in the table.
DECLARE #MyTable TABLE (ITMEID INT, ITMDESC VARCHAR(100))
INSERT INTO #MyTable VALUES (1,'Healty and Alive r')
INSERT INTO #MyTable VALUES (2, 'A liver patient')
DECLARE #SearchCriteria VARCHAR(100)
SET #SearchCriteria = 'Alive'
SELECT *
FROM #MyTable
WHERE (ITMDESC LIKE '%'+REPLACE(#SearchCriteria,' ','%')+'%' ESCAPE '\')
I got this query from a friend to consider multiple consequent white spaces as a single space. The challenge is I don't see any reference for this.
Is there a pitfall in the approach?
REPLACE(#SearchCriteria,' ','%') always returns Alive. There is no Alive word in the second row, therefore it's not returned.
In fact, WHERE clause will look like this: WHERE (ITMDESC LIKE '%Alive%' ESCAPE '\')
The second row doesn't meet it.
Probably, you want something like this:
SELECT *
FROM #MyTable
WHERE (REPLACE(ITMDESC,' ','') LIKE '%'+#SearchCriteria+'%' ESCAPE '\')
you can use as below
DECLARE #MyTable TABLE (ITMEID INT, ITMDESC VARCHAR(100))
INSERT INTO #MyTable VALUES (1,'Healty and Alive r')
INSERT INTO #MyTable VALUES (2, 'A liver patient')
ECLARE #SearchCriteria VARCHAR(100)
SET #SearchCriteria = 'Alive'
SELECT *
FROM #MyTable
WHERE (REPLACE(ITMDESC,' ','') LIKE '%'+#SearchCriteria+'%' ESCAPE '\')
it will return both records as you want
The simplest solution is to replace all spaces with some moniker and then replace that moniker with a single space.
Select Replace(Replace(ItmDesc, ' ', '<z>'), '<z>', ' ')
From MyTable
SQL Fiddle version
I'd like to know how I can identify trailing spaces in a table. I'm using SQL Server 2008 and create the following table as a test
CREATE TABLE first_test_name
(
firstName varchar(255)
)
And then inserted a record like this:
insert into first_test_name (firstName)
values('Bob')
I then tried inserting a space and then adding a new record like this:
insert into first_test_name (firstName)
values('Bob ') -- just 1 space
And for a 3rd time,
insert into first_test_name (firstName)
values('Bob ') -- two spaces used this time.
Now if I query for for 'Bob' (no spaces), I still get a count of 3.
My query was:
select count(*) from first_test_name WHERE firstName = 'Bob'
Shouldn't the answer have been 1?
Also, I used sp_help on this table and the value for "Trim Trailing Blanks" is set to no.
So why am I getting a count of 3? I was expecting just 1.
On a related note, if I search using this query
select * from first_test_name
where firstName like '% '
I then get the right answer of two rows found.
So just to reiterate, the question is why I get a count of 3 when searching for 'Bob'.
Also, what does "Trim Trailing Blanks" mean in this case?
Why I get a count of 3 when searching for 'Bob'?
SQL Server ignores trailing spaces in most string comparisons.
Also, what does "Trim Trailing Blanks" mean in this case?
This tells you the ANSI_PADDING option set when the table was created.
How can I identify those two with 1 or 2 trailing spaces?
Here's one way.
SELECT *
FROM first_test_name
WHERE firstName LIKE 'Bob '
And to find ones with no trailing space
SELECT *
FROM first_test_name
WHERE firstName LIKE 'Bob' AND firstName NOT LIKE 'Bob '
SQL Server will expand strings with whitespace during comparisons.
This is what I would do:
SELECT COUNT(*)
FROM first_test_name
WHERE REPLACE(firstName, ' ', '_') = 'Bob'
SELECT *
FROM USERS
WHERE DATALENGTH(Username) <> DATALENGTH(RTrim(Username))
Another way might be to append something on the string.
declare #test table(
id varchar(4) not null,
firstname varchar(255) not null)
insert into #test values('1', 'Bob')
insert into #test values('2', 'Bob ')
insert into #test values('3', 'Bob ')
insert into #test values('4', ' Bob')
select count((firstname + 'end')) from #test
where (firstname + 'end') not like '% %'
The query will return a count of 1.
A good clean way to do this would be to compare your original string against an Rtrim version of itself where they don't match e.g.:
SELECT *
FROM First_Test_Name
WHERE Firstname <> RTrim(Firstname)
This should return all records where Firstname has trailing spaces (I think ...)
I was looking recently and couldn't find the answer to this. Thought I'd share since coming across one.
https://stackoverflow.com/a/14188944/1953837
LEN trims trailing whitespaces by default. Using the below they are moved to the front and then the field length is counted.
Hope this helps anyone searching in the future.
(LEN(REVERSE(FieldName))