Can someone please help me to understand the following code:
REPLACE(LTRIM(RTRIM(dbo.UFN_SEPARATES_COLUMNS(CompletionDetails, 3, ','))), '.', '') AS BuildRequestID,
Does it say remove all trailing and leading spaces, then replace 3 with comma. Next, if there is ., replace it with ' '?
It does not at any point replace 3 with ,.
We can make all this easier to follow by formatting the full expression to cover multiple lines:
REPLACE(
LTRIM(RTRIM(
dbo.UFN_SEPARATES_COLUMNS(CompletionDetails, 3, ',')
))
,'.', ''
) AS BuildRequestID,
Expressions like this have to read from the inside out. So we start with this inner-most part:
dbo.UFN_SEPARATES_COLUMNS(CompletionDetails, 3, ',')
This UFN_SEPARATES_COLUMNS() function is not part of Sql Server, but was added by someone at your organization or as part of the vendor software package for the database you're looking at. But I'm confident based on inferences and the link (found via Google) it will treat CompletionDetails as delimited text, where the delimiter is a comma (based on the 3rd ',' argument) and returns the 3rd field (based on the 2nd 3 argument, where counting starts at 1 rather than 0). As CSV parsers go, this one is particularly naive, so be very careful what you expect from it.
Then we use LTRIM() and RTRIM() to remove both leading and trailing blanks from the field. Not all whitepsace is removed; only space characters. Tabs, line feeds, etc are not trimmed. Sql Server 2017 has a new TRIM() function that can handle wider character sets and do both sides of the string with one call.
The code then uses the REPLACE() function to remove all . characters from the result (replaces them with an empty string).
The code is trimming the leading and trailing spaces via the LTRIM() and RTRIM() functions of whatever is returned from the function dbo.x_COLUMNS... i.e. dbo.x_COLUMNS(CompletionDetails, 3, ','). LTRIM is left, RTRIM is right.
It then is replacing all periods (.) with nothing via the REPLACE() function.
So in summary, it's removing all periods from the string and the leading and trailing spaces.
The LTRIM removes leading spaces. RTRIM removes trailing spaces. REPLACE removes the period.
Declare #Val Char(20) = ' Frid.ay '
Select REPLACE(
LTRIM(
RTRIM(
#Val --dbo.x_COLUMNS(CompletionDetails, 3, ',')
)
), '.', ''
)
Result
BuildRequestID
--------------
Friday
remove all trailing and leading spaces, then replace 3 with comma.
Next, if there is ., replace it with ' '
No it does not say that.
But this does:
REPLACE(REPLACE(LTRIM(RTRIM(CompletionDetails)), '3', ','), '.', ' ')
it's not clear if you want . replaced by ' ' or ''.
I used the 1st case, you can change it as you like.
It's easier to understand like this:
remove all trailing and leading spaces: LTRIM(RTRIM(CompletionDetails))
replace 3 with comma: REPLACE( ?, '3', ',')
replace it with ' ': REPLACE(? , '.', ' ') or REPLACE(? , '.', '')
Related
I have a column in a table ident_nums that contains different types of ids. I need to remove special characters(e.g. [.,/#&$-]) from that column and replace them with space; however, if the special characters are found at the beginning of the string, I need to remove it without placing a space. I tried to do it in steps; first, I removed the special characters and replaced them with space (I used
REGEXP_REPLACE) then found the records that contain spaces at the beginning of the string and tried to use the TRIM function to remove the white space, but for some reason is not working that.
Here is what I have done
Select regexp_replace(id_num, '[:(),./#*&-]', ' ') from ident_nums
This part works for me, I remove all the unwanted characters from the column, however, if the string in the column starts with a character I don't want to have space in there, I would like to remove just the character, so I tried to use the built-in function TRIM.
update ident_nums
set id_num = TRIM(id_num)
I'm getting an error ORA-01407: can't update ident_nums.id_num to NULL
Any ideas what I am doing wrong here?
It does work if I add a where clause,
update ident_nums
set id_num = TRIM(id_num) where id = 123;
but I need to update all the rows with the white space at the beginning of the string.
Any suggestions are welcome.
Or if it can be done better.
The table has millions of records.
Thank you
Regexp can be slow sometimes so if you can do it by using built-in functions - consider it.
As #Abra suggested TRIM and TRANSLATE is a good choice, but maybe you would prefer LTRIM - removes only leading spaces from string (TRIM removes both - leading and trailing character ). If you want to remove "space" you can ommit defining the trim character parameter, space is default.
select
ltrim(translate('#kdjdj:', '[:(),./#*&-]', ' '))
from dual;
select
ltrim(translate(orginal_string, 'special_characters_to_remove', ' '))
from dual;
Combination of Oracle built-in functions TRANSLATE and TRIM worked for me.
select trim(' ' from translate('#$one,$2-zero...', '#$,-.',' ')) as RESULT
from DUAL
Refer to this dbfiddle
I think trim() is the key, but if you want to keep only alpha numerics, digits, and spaces, then:
select trim(' ' from regexp_replace(col, '[^a-zA-Z0-9 ]', ' ', 1, 0))
regexp_replace() makes it possible to specify only the characters you want to keep, which could be convenient.
Thanks, everyone, It this query worked for me
update update ident_nums
set id_num = LTRIM(REGEXP_REPLACE(id_num, '[:space:]+', ' ')
where REGEXP_LIKE(id_num, '^[ ?]')
this should work for you.
SELECT id_num, length(id_num) length_old, NEW_ID_NUM, length(NEW_ID_NUM) len_NEW_ID_NUM, ltrim(NEW_ID_NUM), length(ltrim(NEW_ID_NUM)) length_after_ltrim
FROM (
SELECT id_num, regexp_replace(id_num, '[:(),./#*&-#]', ' ') NEW_ID_NUM FROM
(
SELECT '1234$%45' as id_num from dual UNION
SELECT '#SHARMA' as id_num from dual UNION
SELECT 'JACK TEST' as id_num from dual UNION
SELECT 'XYZ#$' as id_num from dual UNION
SELECT '#ABCDE()' as id_num from dual -- THe 1st character is space
)
)
Basically translate will change character to character and Replace string to string , and here i have tried to remove spaces using translate to count the number words .
select translate(' #',' ','') from dual;
select replace(' #',' ','') from dual;
select ename , nvl(length(replace(TRANSLATE(upper(trim(ename)),'ABCDEFGHIJKLMNOPQRSTUVWXYZ'' ',' # '),' ',''))+1,1) NOOFWORDs
from emp;
Unfortunately Oracle has made many bizarre choices around null vs. empty string.
One of those has to do with TRANSLATE. TRANSLATE will return NULL if any of its arguments (including the last one) is NULL, no matter what the logical behavior should be.
So, to remove spaces (say) with TRANSLATE, you must add a character you do NOT want to be removed to both the second and the third argument. I added the lower-case letter z, but you could add anything (a dot, the digit 0, whatever - just make sure you add the same character at the beginning of both arguments)
... translate (input_string, 'z ', 'z') ....
For example:
select translate(' #','z ','z') from dual;
TRANSLATE('#','Z','Z')
------------------------
#
select translate(' #',' ','') from dual;
Returns NULL because in Oracle empty strings unfortunately yield NULLs. Therefore it's equivalent to
SELECT translate(' #', ' ', NULL)
FROM dual;
and translate() returns NULL when an argument is null. Actually this is well documented in "TRANSLATE":
(...)
You cannot use an empty string for to_string to remove all characters in from_string from the return value. Oracle Database interprets the empty string as null, and if this function has a null argument, then it returns null.
If you want to replace one character, use replace() as you already did. For a few but more than one characters you can nest the replace()s.
This however gets unhandy, when you want to replace quite a lot of characters. In such a situation, if the replacement character is only one character or the empty string regexp_replace() using a character class or alternates may come in handy.
For example
SELECT regexp_replace('a12b478c01', '[0-9]', '')
FROM dual;
replaces all the digits so just 'abc' remains and
SELECT regexp_replace('ABcc1233', 'c|3', '')
FROM dual;
removes any '3' or 'c' and results in 'AB12'. In your very example
SELECT regexp_replace(' #', ' ', '')
FROM dual;
would also work and give you '#'. Though in the simple case of your example a simple replace() is enough.
I have a column in my SQL Server database and it has white spaces from left and right site of the record. Basically it's a nvarchar(250) column.
I have tried removing white spaces completely like this:
UPDATE MyTable
SET whitespacecolumn = LTRIM(RTRIM(whitespacecolumn))
But this didn't work out at all, the whitespace is still there. What am I doing wrong here?
Check the below;
Find any special characters like char(10), char(13) etc in the field value.
Check the status of ANSI_PADDING ON. Refer this MSDN article.
I think replace is the way as you are looking to update
UPDATE MyTable SET whitespacecolumn = Replace(whitespacecolumn, ' ', '')
you can try doing select first and then prefer to update
SELECT *, Replace(whitespacecolumn, ' ', '') from MyTable
LTRIM, RTRIM will remove spaces in front and rear of column. In 2016 you can use TRIM function as below to trim special characters as well:
SELECT TRIM( '.,! ' FROM '# test .') AS Result;
Output:
# test
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.
I have a relational database with several fixed length character fields. I need to permanently replace all the embedded spaces with another character like - so JOHN DOE would become JOHN-DOE and ITSY BISTSY SPIDER would become ITSY-BISTSY-SPIDER. I can search before hand to make sure there are no strings that would conflict. I just need to be able to print the requested files with no embedded spaces. I would do the replacement in the C code but I want to make sure that there is never a future case where there is a JANE DOE and JANE-DOE in the DB.
By the way I have already made sure that there are no strings with more than one consecutive embedded space or leading spaces only trailing spaces to fill the fixed length fields.
Edit: thanks for all the help!
It looks like when I cut & pasted my question from Word to StackOverflow the trailing spaces got lost so the meaning my question was lost a bit.
I need to replace only the embedded spaces not the trailing spaces!
Note: I am using middle dot to stand in for spaces that don't show well.
Using:
SELECT REPLACE(operator_name, ' ', '-') FROM operator_info ;
the string JOHN·DOE············ became JOHN-DOE------------.
I need JOHN-DOE············.
I am thinking I need to use aliasing and the TRIM command but not sure how.
With whatever REPLACE function is built into your particular database.
MySQL:
http://dev.mysql.com/doc/refman/5.0/en/string-functions.html#function_replace
Oracle:
http://psoug.org/reference/translate_replace.html
SQLServer:
http://msdn.microsoft.com/en-us/library/ms186862.aspx
Edits below based on your comment.
I've done this in SQLServer syntax so please modify the example as needed. The first example really breaks down what's going on and the second one bunches it all into a single ugly query :D
#output in this case contains your final value.
DECLARE #input VARCHAR (100) = ' some test ';
DECLARE #trimmed VARCHAR (100);
DECLARE #replaced VARCHAR (100);
DECLARE #output VARCHAR (100);
-- Get just the inner text without the preceding / trailing spaces.
SET #trimmed = LTRIM (RTRIM (#input));
-- Replace the spaces *inside* the trimmed text with a dash.
SET #replaced = REPLACE (#trimmed, ' ', '-');
-- Take the original text and replace the trimmed version (with the inner spaces) with the dash version.
SET #output = REPLACE (#input, #trimmed, #replaced);
-- Show each step of the process!
SELECT #input AS INPUT,
#trimmed AS TRIMMED,
#replaced AS REPLACED,
#output AS OUTPUT;
And as a SELECT statement.
DECLARE #inputTable TABLE (Value VARCHAR (100) NOT NULL);
INSERT INTO #inputTable (Value)
VALUES (' some test '),
(' another test ');
SELECT REPLACE (Value,
LTRIM (RTRIM (Value)),
REPLACE (LTRIM (RTRIM (Value)), ' ', '-'))
FROM #inputTable;
If you are using MSSQL:
SELECT REPLACE(field_name,' ','-');
Edit: After the requirement about skipping the trailing spaces.
You can try this one-liner:
SELECT REPLACE(RTRIM(#name), ' ', '-') + SUBSTRING(#name, LEN(RTRIM(#name)) + 1, LEN(#NAME))
However I would recommend that you put it into a user defined function instead.
assuming SQL Server:
update TABLE set column = replace (column, ' ','-')
SELECT REPLACE(field_name,' ','-');
Edit: After the requirement about skipping the trailing spaces. You can try this one-liner:
SELECT REPLACE(RTRIM(#name), ' ', '-') + SUBSTRING(#name, LEN(RTRIM(#name)) + 1;