I want to select where 2 strings but without taking
underscore
apostrophe
dash..
Hello !
I want to select an option in my SQL database who look like this :
Chef d'équipe aménagement-finitions
With an original tag who look like this
chef-déquipe-aménagement-finitions
Some results in database had a - too
SELECT *
FROM table
WHERE REPLACE(name, '-', ' ') = REPLACE('chef-déquipe-aménagement-finitions', '-', ' ')
didnt work because of missing '
And a double replace didn't work too.
I want the string be able to compare without taking
underscore
apostrophe
dash
and all things like that
is this possible ?
Thanks for your help
Have good day !
Depends on your rdbms, but here's how I would perform in MySQL 8. If using a different version or rdbms, then first determine how to escape the single quote and modify as needed.
with my_data as (
select 'Chef d''équipe aménagement-finitions' as name
)
select name,
lower(replace(replace(name, '\'', ''), ' ', '-')) as name2
from my_data;
name
name2
Chef d'équipe aménagement-finitions
chef-déquipe-aménagement-finitions
Sql-server and Postgres version:
lower(replace(replace(name, '''', ''), ' ', '-')) as name
After posting, this, I re-read and noticed you are also looking to replace other characters. You could either keep layering the replace function, or, look into other functions.
I would like to keep rows only with Numeric Character i.e. 0-9. My source data can have any type of character e.g. 2,%,( .
Input (postcode)
3453gds sdg3
454232
sdg(*d^
452
Expected Output (postcode)
454232
452
I have tried using WHERE REGEXP_LIKE(postcode, '^[[:digit:]]+$');
however in my version of Oracle I get an error saying
function regexp_like(character varying, "unknown") does not exist
You want regexp_like() and your version should work:
select t.*
from t
where regexp_like(t.postcode, '^[0-9]+$');
However, your error looks more like a Postgres error, so perhaps this will work:
t.postcode ~ '^[0-9]+$'
For Oracle 10 or higher you can use regexp functions. In earlier versions translate function will help you :
SELECT postcode
FROM table_name
WHERE length(translate(postcode,'0123456789','1')) is null
AND postcode IS NOT NULL;
OR
SELECT translate(postcode, '0123456789' || translate(postcode,'x123456789','x'),'0123456789') nums
FROM table_name ;
the above answer also works for me
SELECT translate('1234bsdfs3#23##PU', '0123456789' || translate('1234bsdfs3#23##PU','x123456789','x'),'0123456789') nums
FROM dual ;
Nums:
1234323
For an alternative to the Gordon Linoff answer, we can try using REGEXP_REPLACE:
SELECT *
FROM yourTable
WHERE REGEXP_REPLACE(postcode, '[0-9]+', '') IS NULL;
The idea here is to strip away all digit characters, and then assert that nothing were left behind. For a mixed digit-letter value, the regex replacement would result in a non-empty string.
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!
Hi
I have a column of nvarchar(1000) type. I need to get rid of encode characters from that column and replace them with their special characters.
For Example:
column value is : 'This text values contains this '&' this'.
I have to replace '&' with '&'.
First have to find the record which has '&' in the column (may be using like condition)
And then replace only this word with its special character
How do i do that? Pl. help
This will replace in the entire column
REPLACE(MyColumn, '&', '&')
You'll have to nest other replacements...
REPLACE(REPLACE(MyColumn, '&', '&'), '>', '>')
All together
UPDATE myTable
SET MyColumn = REPLACE(MyColumn, '&', '&')
WHERE MyColumn LIKE '%&%'
UPDATE mytable
SET mycol = REPLACE(mycol, N'&', N'&')
WHERE mycol LIKE '%&%'
EDIT
If you decide to replace multiple html entities in one go, the order of the replacements may change results.
For example:
<
becomes &< if you replace first & with & and then < with <, but the result will be < if you first try to replace < with < and then & with &.
If I have to do that kind of replacement, I usually replace & last for this reason. Sure, an edge case, and not something which happens often, but you never know...
Generic syntax:
UPDATE Table_Name
SET Column_Name = REPLACE(Column_Name, 'Text_To_Be_Replaced', 'New_Text')
WHERE Column_Name LIKE '%Text_To_Be_Replaced%'
Update TABLE
Set field = replace(field, '&', '&');
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