All I have what seems to be a pretty straight forward question, that I haven't been able to figure out. For example, if I have a text string like T6L 7H5. Using SQL I need to remove the inner white space from this string so that it displays like T6L7H5.
Things to consider:
Teradata (v.13.10) is my RDBMS, so REPLACE('T6L 7H5', ' ', '') is not an
option here.
On this particular server I am a business user w/ very limited
permissions so creating a UDF is not an option either.
Can't test this, so spit-balling, but you should be able to leverage Substring and Position:
SELECT SUBSTRING('T6L 7H5', 1, POSITION (' ' IN 'T6L 7H5')-1) || SUBSTRING('T6L 7H5', POSITION (' ' IN 'T6L 7H5')+1, CHARACTER_LENGTH('T6L 7H5') - POSITION (' ' IN 'T6L 7H5') )
If the field is consistently formatted like your example then:
substring('T6L 7H5',1,3)||substring('T6L 7H5',4,3)
For single white space you can use POSITION() and SUBSTRING(). You have to offset for the whitespace location that is returned by the POSITION() function.
WITH CTE(FieldName) AS
(SELECT 'TB7 TCH' AS FieldName)
SELECT SUBSTRING(FieldName FROM 1 FOR (POSITION(' ' IN FieldName) - 1))
|| SUBSTRING(FieldName FROM (POSITION(' ' IN FieldName) + 1))
FROM CTE;
try using substr( );
Ex : 4= start at position 4 && 1 = delete one char.
substr(Table1.fieldname, 4,1)
Related
What will be the Postgres equivalent for the below code from Oracle.
Select instr('abc de fgh',' ', -1) from dual;
returns: 7
Original code: substr(country, instr(country,' ', -1)+1);
I want to create the same logic in Postgres but position & reverse function didn't work
You apparently want the last element from the string where elements are delimited by a space character.
If you are using Postgres 14, you can use:
split_part('abc de fgh', ' ', -1);
which returns fgh
Or with a column reference: split_part(country, ' ', -1)
In earlier versions, split_part() doesn't allow negative offsets, so you would need something different:
(string_to_array('abc de fgh', ' '))[cardinality(string_to_array('abc de fgh', ' '))]
It's a bit shorter with a column reference:
(string_to_array(country, ' '))[cardinality(string_to_array(country, ' '))]
This first converts the string into an array, then picks the last element of the array (which in turn is determined using the cardinality() function that returns the length of the array)
Since you mention REVERSE, you can use:
SELECT RIGHT(value, POSITION(' ' in REVERSE(value)) - 1) AS country,
LENGTH(value) - POSITION(' ' in REVERSE(value)) + 1 AS pos
FROM (SELECT 'abc de fgh' AS value) v;
Which outputs:
country
pos
fgh
7
db<>fiddle here
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.
How can I remove any white space from substring of string?
For example I have this number '+370 650 12345'. I need all numbers to have this format country_code rest_of_the_number or in that example: +370 65012345. How could you achieve that with PostgreSQL?
I could use trim() function, but then it would remove all whitespace.
Assuming the column is named phone_number:
left(phone_number, strpos(phone_number, ' '))
||regexp_replace(substr(phone_number, strpos(phone_number, ' ') + 1), ' ', '', 'g')
It first takes everything up to the first space and then concatenates it with the result of replacing all spaces from the rest of the string.
If you also need to deal with other whitespace than just a space, you could use '\s' for the search value in regexp_replace()
If you are able to assume that a country code will always be present, you could try using a regular expression to capture the parts of interest. Assuming that your phone numbers are stored in a column named content in a table named numbers, you could try something like the following:
SELECT parts[1] || ' ' || parts[2] || parts[3]
FROM (
SELECT
regexp_matches(content, E'^\\s*(\\+\\d+)\\s+(\\d+)\\s+(\\d+)\\s*$') AS parts
FROM numbers
) t;
The following will work even if the country code is absent (see SQL Fiddle Demo here):
SELECT TRIM(REPLACE(REPLACE(REGEXP_REPLACE('+370 650 12345', '^((\+\d+)\s+)?(.*)$', '\1|\3'), ' ', ''), '|', ' '));
Returns: +370 65012345
SELECT TRIM(REPLACE(REPLACE(REGEXP_REPLACE('370 650 12345', '^((\+\d+)\s+)?(.*)$', '\1|\3'), ' ', ''), '|', ' '));
Returns: 37065012345
It looks for a country code (a set of numbers starting with a + sign) at the beginning, and replaces any whitespace following that code with a pipe |. It then replaces all the spaces in the resulting string with the empty string, then replaces occurrences of | with spaces! The choice of | is arbitrary, I suppose it could be any non-digit character.
The string contains many words separated by spaces e.g.
employee_first_nm = "John Walker"
I want to retrieve the first part alone ("John"). this part I have done using the following code:
SUBSTR(employee_first_nm, 1, INSTR(employee_first_nm, ' '));
In some cases the string has only one word e.g. "sonia", this is where I got a problem. Here if there is only one word the function doesn't retrieve any value at all. But I want it to get the full string in this case i.e. "sonia".
Please help
Simply ensure there always will be a space;
... INSTR(employee_first_nm + ' ', ' ')
If there is already a space in the string then stuffing another one on the end makes no difference as the 1st one will be found, if there is no existing space adding one makes your logic work.
(Usually you also need to INSTR(..)-1 to strip the trailing space)
SUBSTR(employee_first_nm, 1, INSTR(CONCAT(employee_first_nm,' '), ' '));
Alternatively, you could check that INSTR returns a correct value:
(case when INSTR(employee_first_nm, ' ') > 0
then SUBSTR(employee_first_nm, 1, INSTR(employee_first_nm, ' '))
else employee_first_nm
end)
Why not harness the power of regular expressions here.
REGEXP_SUBSTR(employee_first_nm, '^[a-zA-Z]*\s?')
This should return the substring you desire.
You can try:
SUBSTRING(select employee_first_nm,0, charindex(' ',employee_first_nm)
Tested with:
select substring('John walker',0, charindex(' ','John walker more'))
returns -> 'John'
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