Detect \n character saved in SQLite table? - sql

I designed a table with a column whose data contains \n character (as the separator, I used this instead of comma or anything else). It must save the \n characters OK because after loading the table into a DataTable object, I can split the values into arrays of string with the separator '\n' like this:
DataTable dt = LoadTable("myTableName");
DataRow dr = dt.Rows[0]; //suppose this row has the data with \n character.
string[] s = dr["myColumn"].ToString().Split(new char[]{'\n'}, StringSplitOptions.RemoveEmptyEntries);//This gives result as I expect, e.g an array of 2 or 3 strings depending on what I saved before.
That means '\n' does exist in my table column. But when I tried selecting only rows which contain \n character at myColumn, it gave no rows returned, like this:
--use charindex
SELECT * FROM MyTable WHERE CHARINDEX('\n',MyColumn,0) > 0
--use like
SELECT * FROM MyTable WHERE MyColumn LIKE '%\n%'
I wonder if my queries are wrong?
I've also tested with both '\r\n' and '\r' but the result was the same.
How can I detect if the rows contain '\n' character in my table? This is required to select the rows I want (this is by my design when choosing '\n' as the separator).
Thank you very much in advance!

Since \n is the ASCII linefeed character try this:
SELECT *
FROM MyTable
WHERE MyColumn LIKE '%' || X'0A' || '%'
Sorry this is just a guess; I don't use SQLite myself.

Maybe you should just be looking for carriage returns if you arent storing the "\n" literal in the field. Something like
SELECT *
FROM table
WHERE column LIKE '%
%'
or select * from table where column like '%'+char(13)+'%' or column like '%'+char(10)+'%'
(Not sure if char(13) and 10 work for SQLite
UPDATED: Just found someone's solution here They recommend to replace the carriage returns
So if you want to replace them and strip the returns, you could
update yourtable set yourCol = replace(yourcol, '
', ' ');

The following should do it for you
SELECT *
FROM your_table
WHERE your_column LIKE '%' + CHAR(10) + '%'
If you want to test for carriage return use CHAR(13) instead or combine them.

I've found a solution myself. There is few way (with some dedicated function) to convert ascii code to symbol in SQLite at the moment (CHAR function is not support and using '\n' or '\r' directly doesn't work). But we can convert using CAST function and passing in a Hex string (specified by append X or x before the string) in SQLite like this:
-- use CHARINDEX
SELECT * FROM MyTable WHERE CHARINDEX(CAST(x'0A' AS text),MyColumn,0) > 0
-- use LIKE
SELECT * FROM MyTable WHERE MyColumn LIKE '%' || CAST(x'0A' AS text) || '%'
The Hex string '0A' is equal to 10 in ascii code (\r). I've tried with '0D' (13 or '\n') but it won't work. Maybe the \n character is turned to \r after being saved in to SQLite table.
Hope this helps others! Thanks!

Related

In SQL Server, how can I search for column with 1 or 2 whitespace characters?

So I need to filter column which contains either one, two or three whitespace character.
CREATE TABLE a
(
[col] [char](3) NULL,
)
and some inserts like
INSERT INTO a VALUES (' ',' ', ' ')
How do I get only the row with one white space?
Simply writing
SELECT *
FROM a
WHERE column = ' '
returns all rows irrespective of one or more whitespace character.
Is there a way to escape the space? Or search for specific number of whitespaces in column? Regex?
Use like clause - eg where column like '%[ ]%'
the brackets are important, like clauses provide a very limited version of regex. If its not enough, you can add a regex function written in C# to the DB and use that to check each row, but it won't be indexed and thus will be very slow.
The other alternative, if you need speed, is to look into full text search indexes.
Here is one approach you can take:
DECLARE #data table ( txt varchar(50), val varchar(50) );
INSERT INTO #data VALUES ( 'One Space', ' ' ), ( 'Two Spaces', ' ' ), ( 'Three Spaces', ' ' );
;WITH cte AS (
SELECT
txt,
DATALENGTH ( val ) - ( DATALENGTH ( REPLACE ( val, ' ', '' ) ) ) AS CharCount
FROM #data
)
SELECT * FROM cte WHERE CharCount = 1;
RETURNS
+-----------+-----------+
| txt | CharCount |
+-----------+-----------+
| One Space | 1 |
+-----------+-----------+
You need to use DATALENGTH as LEN ignores trailing blank spaces, but this is a method I have used before.
NOTE:
This example assumes the use of a varchar column.
Trailing spaces are often ignored in string comparisons in SQL Server. They are treated as significant on the LHS of the LIKE though.
To search for values that are exactly one space you can use
select *
from a
where ' ' LIKE col AND col = ' '
/*The second predicate is required in case col contains % or _ and for index seek*/
Note with your example table all the values will be padded out to three characters with trailing spaces anyway though. You would need a variable length datatype (varchar/nvarchar) to avoid this.
The advantage this has over checking value + DATALENGTH is that it is agnostic to how many bytes per character the string is using (dependant on datatype and collation)
DB Fiddle
How to get only rows with one space?
SELECT *
FROM a
WHERE col LIKE SPACE(1) AND col NOT LIKE SPACE(2)
;
Though this will only work for variable length datatypes.
Thanks guys for answering.
So I converted the char(3) column to varchar(3).
This seemed to work for me. It seems sql server has ansi padding that puts three while space in char(3) column for any empty or single space input. So any search or len or replace will take the padded value.

SQL- Remove Trailing space and add it to the beginning of the Name

Was working on SQL-EX.ru exercises.
There is one question for DML that I could not do, but I cannot proceed to the next one, until this one is done.
the question itself: All the trailing spaces in the name column of the Battles table remove and add them at the beginning of the name.
My code:
Update Battles
set name=concat(' ',(LTRIM(RTRIM(name))))
The system does not let it go through, I understand that I am using ' ' for the concat, whereas I need to use the stuff that got trimmed. And I have no idea how...
Any help would be very much appreciated
Try Something Like:-
set name = lpad(trim(name), length(trim(name))+4, ' ')
Here use TRIM to remove space from both side. use LPAD to add something on left side with n (4) chars
I'm not familiar with SQL-EX.ru, but if it's Oracle compatible and you can use regular expressions (or you are at that point in the training) here's a way. Maybe it'll give you an idea at least. The first part is just setup and uses a WITH clause to create a table (like a temp table in memory, actually called a Common Table Expression or CTE) called battles containing a name column with 2 rows. Each name column datum has a different number of spaces at the end. Next select from that column using a regular expression that uses 2 "remembered" groups surrounded by parentheses, the first containing the string up to until but not including the first space, the second containing 0 or more space characters anchored to the end of the line. Replace that with the 2nd group (the spaces) first, followed by the first group (the first part of the string). This is surrounded by square brackets just to prove in the output the same spaces were moved to the front of the string.
SQL> with battles(name) as (
select 'test2 ' from dual union
select 'test1 ' from dual
)
select '[' || regexp_replace(name, '(.*?)([ ]*)$', '\2\1') || ']' fixed
from battles;
FIXED
----------------------------------------------------------------------------
[ test1]
[ test2]
SQL>
I hope this solution can be applied to your problem or at least give you some ideas.
Try this:
set name = case when len(name) > len(rtrim(name))
then replicate(' ', len(name) - len(rtrim(name))) + rtrim(name)
else name
end
update battles
set name = case when (len(name+'a')-1) > len(rtrim(name))
then
replicate(' ',
(len(name+'a')-1) - len(rtrim(name))) + rtrim(name)
else name
end
Len() doesn't count trailing spaces. So using (len(name+'a')-1).
Simplest answer:
UPDATE Battles
SET name = SPACE(DATALENGTH(name)-DATALENGTH(RTRIM(name))) + RTRIM(name)
But only works because name is VARCHAR.
More generic is to do:
UPDATE Battles
SET name = SPACE(len(name+'x')-1-len(RTRIM(name))) + RTRIM(name)
simple example below ... enjoy :)
update battles set name =
Space( DATALENGTH(name) - DATALENGTH(rtrim(name))) + rtrim(name)
where date in ( select date from battles)

How to show leading/trailing whitespace in a PostgreSQL column?

I can't see leading/trailing whitespace in the following following SQL statement executed with psql:
select name from my_table;
Is there a pragmatic way to see leading/trailing whitespace?
One of option is to use format() function.
With given query case: select format( '"%s"', name ) from my_table;
PoC:
SELECT format( '"%s"', name )
FROM ( VALUES ( ' a ' ), ( ' b ' ) ) v(name);
format
--------
" a "
" b "
(2 rows)
Turn off "aligned mode" in psql: \a
\a
select * from my_table;
id|col1|col2
12|foo|bar
I'd append surrounding quotes:
select '"' || name || '"' from my_table;
If you don't mind substituting all whitespace characters whether or not they are leading/trailing, something like the following will do it:
SELECT REPLACE(REPLACE(REPLACE(REPLACE(txt, ' ', '_'),
E'\t', '\t'),
E'\r', '\r'),
E'\n', '\n') AS txt
FROM test;
This is using an underscore to mark the spaces but of course you are free to choose your own. See SQL fiddle demo.
If you strictly only want to show up the leading/trailing ones it will get more complex - but if this is really desired, something may be possible using regex_replace.
You can try this:
select name from my_table
where name like '% '

sql prepend entries

I have some entries that are inconstant, they should be prepended by the same string.
some have numbers and other have the dollar sign and the number
so I have a syntax that finds all the entries which do not have the dollar sign
WHERE [mydata] not like '$%'
how do I add the string before each entry?
update table set mydata = '$' + mydata where [mydata] not like '$%'
The + only works in SQLServer; you may need to use the concatenate function otherwise.
If you are looking for just a select statement to return the value with a $, this will work:
Select '$' + field
from [table]
where field not like '$%'
You can run it as a case statement had you wanted both records that have a $ and those without to be returned with the $
Select case when left(field,1) = '$' then '' else '$' end + field
from [table]
where field not like '$%'
edit: Might need to convert the 'field' into a varchar for the + to work, you'll get a syntax error if the field is an int (but you have the $ spuradically in the field, so I assumed it's a varchar

How can I remove leading and trailing quotes in SQL Server?

I have a table in a SQL Server database with an NTEXT column. This column may contain data that is enclosed with double quotes. When I query for this column, I want to remove these leading and trailing quotes.
For example:
"this is a test message"
should become
this is a test message
I know of the LTRIM and RTRIM functions but these workl only for spaces. Any suggestions on which functions I can use to achieve this.
I have just tested this code in MS SQL 2008 and validated it.
Remove left-most quote:
UPDATE MyTable
SET FieldName = SUBSTRING(FieldName, 2, LEN(FieldName))
WHERE LEFT(FieldName, 1) = '"'
Remove right-most quote: (Revised to avoid error from implicit type conversion to int)
UPDATE MyTable
SET FieldName = SUBSTRING(FieldName, 1, LEN(FieldName)-1)
WHERE RIGHT(FieldName, 1) = '"'
I thought this is a simpler script if you want to remove all quotes
UPDATE Table_Name
SET col_name = REPLACE(col_name, '"', '')
You can simply use the "Replace" function in SQL Server.
like this ::
select REPLACE('this is a test message','"','')
note: second parameter here is "double quotes" inside two single quotes and third parameter is simply a combination of two single quotes. The idea here is to replace the double quotes with a blank.
Very simple and easy to execute !
My solution is to use the difference in the the column values length compared the same column length but with the double quotes replaced with spaces and trimmed in order to calculate the start and length values as parameters in a SUBSTRING function.
The advantage of doing it this way is that you can remove any leading or trailing character even if it occurs multiple times whilst leaving any characters that are contained within the text.
Here is my answer with some test data:
SELECT
x AS before
,SUBSTRING(x
,LEN(x) - (LEN(LTRIM(REPLACE(x, '"', ' ')) + '|') - 1) + 1 --start_pos
,LEN(LTRIM(REPLACE(x, '"', ' '))) --length
) AS after
FROM
(
SELECT 'test' AS x UNION ALL
SELECT '"' AS x UNION ALL
SELECT '"test' AS x UNION ALL
SELECT 'test"' AS x UNION ALL
SELECT '"test"' AS x UNION ALL
SELECT '""test' AS x UNION ALL
SELECT 'test""' AS x UNION ALL
SELECT '""test""' AS x UNION ALL
SELECT '"te"st"' AS x UNION ALL
SELECT 'te"st' AS x
) a
Which produces the following results:
before after
-----------------
test test
"
"test test
test" test
"test" test
""test test
test"" test
""test"" test
"te"st" te"st
te"st te"st
One thing to note that when getting the length I only need to use LTRIM and not LTRIM and RTRIM combined, this is because the LEN function does not count trailing spaces.
I know this is an older question post, but my daughter came to me with the question, and referenced this page as having possible answers. Given that she's hunting an answer for this, it's a safe assumption others might still be as well.
All are great approaches, and as with everything there's about as many way to skin a cat as there are cats to skin.
If you're looking for a left trim and a right trim of a character or string, and your trailing character/string is uniform in length, here's my suggestion:
SELECT SUBSTRING(ColName,VAR, LEN(ColName)-VAR)
Or in this question...
SELECT SUBSTRING('"this is a test message"',2, LEN('"this is a test message"')-2)
With this, you simply adjust the SUBSTRING starting point (2), and LEN position (-2) to whatever value you need to remove from your string.
It's non-iterative and doesn't require explicit case testing and above all it's inline all of which make for a cleaner execution plan.
The following script removes quotation marks only from around the column value if table is called [Messages] and the column is called [Description].
-- If the content is in the form of "anything" (LIKE '"%"')
-- Then take the whole text without the first and last characters
-- (from the 2nd character and the LEN([Description]) - 2th character)
UPDATE [Messages]
SET [Description] = SUBSTRING([Description], 2, LEN([Description]) - 2)
WHERE [Description] LIKE '"%"'
You can use following query which worked for me-
For updating-
UPDATE table SET colName= REPLACE(LTRIM(RTRIM(REPLACE(colName, '"', ''))), '', '"') WHERE...
For selecting-
SELECT REPLACE(LTRIM(RTRIM(REPLACE(colName, '"', ''))), '', '"') FROM TableName
you could replace the quotes with an empty string...
SELECT AllRemoved = REPLACE(CAST(MyColumn AS varchar(max)), '"', ''),
LeadingAndTrailingRemoved = CASE
WHEN MyTest like '"%"' THEN SUBSTRING(Mytest, 2, LEN(CAST(MyTest AS nvarchar(max)))-2)
ELSE MyTest
END
FROM MyTable
Some UDFs for re-usability.
Left Trimming by character (any number)
CREATE FUNCTION [dbo].[LTRIMCHAR] (#Input NVARCHAR(max), #TrimChar CHAR(1) = ',')
RETURNS NVARCHAR(max)
AS
BEGIN
RETURN REPLACE(REPLACE(LTRIM(REPLACE(REPLACE(#Input,' ','¦'), #TrimChar, ' ')), ' ', #TrimChar),'¦',' ')
END
Right Trimming by character (any number)
CREATE FUNCTION [dbo].[RTRIMCHAR] (#Input NVARCHAR(max), #TrimChar CHAR(1) = ',')
RETURNS NVARCHAR(max)
AS
BEGIN
RETURN REPLACE(REPLACE(RTRIM(REPLACE(REPLACE(#Input,' ','¦'), #TrimChar, ' ')), ' ', #TrimChar),'¦',' ')
END
Note the dummy character '¦' (Alt+0166) cannot be present in the data (you may wish to test your input string, first, if unsure or use a different character).
To remove both quotes you could do this
SUBSTRING(fieldName, 2, lEN(fieldName) - 2)
you can either assign or project the resulting value
You can use TRIM('"' FROM '"this "is" a test"') which returns: this "is" a test
CREATE FUNCTION dbo.TRIM(#String VARCHAR(MAX), #Char varchar(5))
RETURNS VARCHAR(MAX)
BEGIN
RETURN SUBSTRING(#String,PATINDEX('%[^' + #Char + ' ]%',#String)
,(DATALENGTH(#String)+2 - (PATINDEX('%[^' + #Char + ' ]%'
,REVERSE(#String)) + PATINDEX('%[^' + #Char + ' ]%',#String)
)))
END
GO
Select dbo.TRIM('"this is a test message"','"')
Reference : http://raresql.com/2013/05/20/sql-server-trim-how-to-remove-leading-and-trailing-charactersspaces-from-string/
I use this:
UPDATE DataImport
SET PRIO =
CASE WHEN LEN(PRIO) < 2
THEN
(CASE PRIO WHEN '""' THEN '' ELSE PRIO END)
ELSE REPLACE(PRIO, '"' + SUBSTRING(PRIO, 2, LEN(PRIO) - 2) + '"',
SUBSTRING(PRIO, 2, LEN(PRIO) - 2))
END
Try this:
SELECT left(right(cast(SampleText as nVarchar),LEN(cast(sampleText as nVarchar))-1),LEN(cast(sampleText as nVarchar))-2)
FROM TableName