My data looks like
ID LPNumber
1 30;#TEST123
2 302;#TEST1232
How can I update MyText to drop everything before the # and including the #, so I'm left with the following:
ID LPNumber
1 TEST123
2 TEST1232
I've looked at SQL Server Replace, but can't think of a viable way of checking for the ";"
On the MSDN REPLACE page, the menu on the left gives the complete list of string functions available.
UPDATE
MyTable
SET
LPNumber = SUBSTRING(LPNumber, CHARINDEX('#', LPNumber)+1, 8000);
I'll let you work out (from MSDN) the filter needed in case there is no # in the column...
Edit:
Why 8000?
The longest non-LOB string length is 8000 so it is shorthand for "until end of string". You can use 2147483647 too for max columns or to make it consistent.
Also, LEN can bollix you.
SET ANSI_PADDING is ON by default
LEN ignores trailing spaces
You'd need to use DATALENGTH but then you need to know the data type because this counts bytes, not characters. See https://stackoverflow.com/a/2557843/27535 for example
So using a magic number is perhaps a lesser evil...
Use CHARINDEX(), LEN() and RIGHT() instead.
RIGHT(LPNumber, LEN(LPNumber) - CHARINDEX('#', LPNumber, 0))
Related
I need some help with the next. I have a field text in SQL, this record a list of times sepparates with '|'. For example
'14613|15474|3832|148|5236|5348|1055|524' Each value is a time in milliseconds. This field could any length, for example is perfect correct '3215|2654' or '4565' (only 1 value). I need get this field and replace all number with -1000 value.
So '14613|15474|3832|148|5236|5348|1055|524' will be '-1000|-1000|-1000|-1000|-1000|-1000|-1000|-1000'
Or '3215|2654' => '-1000|-1000' Or '4565' => '-1000'.
I try use regexp_replace(times_field,'[[:digit:]]','-1000','g') but it replace each digit, not the complete number, so in this example:
'3215|2654' than must be '-1000|-1000', i get:
'-1000-1000-1000-1000|-1000-1000-1000-1000', I try with other combinations and more options of regexp but i'm done.
Please need your help, thanks!!!.
We can try using REGEXP_REPLACE here:
UPDATE yourTable
SET times_field = REGEXP_REPLACE(times_field, '\y[0-9]+\y', '-1000', 'g');
If instead you don't really want to alter your data but rather just view your data this way, then use a select:
SELECT
times_field,
REGEXP_REPLACE(times_field, '\y[0-9]+\y', '-1000', 'g') AS times_field_replace
FROM yourTable;
Note that in either case we pass g as the fourtb parameter to REGEXP_REPLACE to do a global replacement of all pipe separated numbers.
[[:digit:]] - matches a digit [0-9]
+ Quantifier - matches between one and unlimited times, as many times as possible
your regexp must look like
regexp_replace(times_field,'[[:digit:]]+','-1000','g')
I am using Snowflake SQL. I would like to remove characters from a string after a special character ~. How can I do that?
here is the whole scenario. Let me explain. I do have a string like 'CK#123456~fndkjfgdjkg'. Now, i want only the number after #.And not anything after ~. This is number length varies for that field value. It might be 1 or 5 or 3. And i want to add the condition in where class where this number is equal to check_num from other table after joining. I am trying REGEXP_SUBSTR(A.SRC_TXT, '(?<=CK#)(.+?\b)') = C.CHK_NUM in the where condition. I am getting the error as 'No repititive argument after ?'
You can use a regex for this
-- To remove just the character after a ~
select regexp_replace('fo~o bar','~.', '');
-- returns 'fo bar'
--If you want to keep the ~
select regexp_replace('fo~o bar','~.', '~');
-- returns 'fo~ bar'
--If you want to remove everything after the ~
select regexp_replace('fo~o bar','~.*', '');
--returns 'fo'
If you need to remove other specific character sets after a ~, you can probably do this with a slightly more complicated regex, but I'd need examples of your desired input/output to help with that.
EDIT for updated question
This regex replace should get what you need.
select regexp_replace('CK#123456~fndkjfgdjkg','CK#(\\d*)~.*', '\\1');
-- returns 123456
(\\d*) gets ANY number of digits in a row, and the \\1 causes it to replace the match with what was in the first set of parenthesis, which is your list of digits. the CK# and ~.* are there to make sure the whole string gets matched and replaced.
If the CK# can vary as well, you can use .*? like this.
select regexp_replace('ABCD123HI#123456~fndkjfgdjkg','.*?#(\\d*)~.*', '\\1')
-- returns 123456
I'd probably do something like the following, easy enough but not as cool as RegEx type of functions.
set my_string='fooo~12345';
set search_for_me = '~';
SELECT SUBSTR($my_string, 1, DECODE(position($search_for_me, $my_string), 0, length($my_string), position($search_for_me, $my_string)));
I hope this helps...Rich
It looks like lookahead and lookbehinds are not supported in REGEXP functions, they seem to work in the PATTERN clause of a LIST command. Snowflake documentation makes no mention either way of lookahead or lookbehinds.
In your example:
It seems that the query engine is looking for that repeating argument, where you are attempting a lookbehind
You have not specified what you wanted extracted. You have two capture groups, but in this scenario everything would be returned
Since you are looking to remove everything after ~ you have a delimiter, why not use it in your REGEXP_SUBSTR function?
Try the following:
SELECT $1,REGEXP_SUBSTR($1,'\\w+#(.+?)~',1,1,'is',1)
FROM VALUES
('CK#123456~fndkjfgdjkg')
,('QH#128fklj924~fndkjfgdjkg')
;
This looks for:
One or more word characters
Followed by #
Capturing one or more characters upto and not including ~
Returns the characters within the capture group
You can change the .+? to \\d+? to make sure the pattern is only digits. Backslashes must be escaped with a backslash.
The descriptions for each argument of the function can be found here:
https://docs.snowflake.net/manuals/sql-reference/functions/regexp_substr.html
You could check this!!
select substr('CK#123456~fndkjfgdjkg',4,6) from dual;
OUTPUT
123456
https://docs.snowflake.net/manuals/sql-reference/functions/substr.html
I have a database in an oracle, I filled in the "Number" fields with numbers starting from "A14602727" to "A14603000" but it turned out that I accidentally typed the symbol A (in Ukrainian) instead of A (in English). And now I when find the command:
select number from test where number 'А14602727'
...find me nothing. Is it possible somehow with the help of the command to replace all numbers from "A14602727" (Ukrainian) to "A14602727" (English) ?
I will be grateful for your help!)
You can use following trick to convert any character to its base character using accent-insesitive binary sorting.
select your_col,
utl_raw.cast_to_varchar2(nlssort(your_col, 'nls_sort=binary_ai')) converted_col
from your_table
Use it in update statement accordingly.
Cheers!!
You could use regexp_replace():
update test set number = regexp_replace(number, '^Ä', 'A');
Regexp '^Ä' represents character 'Ä' at the beginning of the string. I used 'Ä' to represent the A in Ukrainian: replace this with the correct character that you want to replace.
This can also be done, probably more efficiently, with substr and like:
update test set number = 'A' || substr(number, 2) where number like 'Ä%';
Use the simple Oracle REPLACE function:
UPDATE test
SET number = REPLACE(number, 'A', 'A');
Replace is taking 3 parameters:
String in which you will search for a string to replace (The number column)
The string you are searchin gto replace (Ukrainian A)
The string you will replace it with (English A)
Here you have a DEMO example.
Also please note that the query in your question:
select number from test where number 'А14602727';
Is not valid. It should be something like this:
select number from test where number like 'А14602727';
or like this:
select number from test where number = 'А14602727';
So first do check that!
Cheers!
I have the following test table in SQL Server 2005:
CREATE TABLE [dbo].[TestTable]
(
[ID] [int] NOT NULL,
[TestField] [varchar](100) NOT NULL
)
Populated with:
INSERT INTO TestTable (ID, TestField) VALUES (1, 'A value'); -- Len = 7
INSERT INTO TestTable (ID, TestField) VALUES (2, 'Another value '); -- Len = 13 + 6 spaces
When I try to find the length of TestField with the SQL Server LEN() function it does not count the trailing spaces - e.g.:
-- Note: Also results the grid view of TestField do not show trailing spaces (SQL Server 2005).
SELECT
ID,
TestField,
LEN(TestField) As LenOfTestField, -- Does not include trailing spaces
FROM
TestTable
How do I include the trailing spaces in the length result?
This is clearly documented by Microsoft in MSDN at http://msdn.microsoft.com/en-us/library/ms190329(SQL.90).aspx, which states LEN "returns the number of characters of the specified string expression, excluding trailing blanks". It is, however, an easy detail on to miss if you're not wary.
You need to instead use the DATALENGTH function - see http://msdn.microsoft.com/en-us/library/ms173486(SQL.90).aspx - which "returns the number of bytes used to represent any expression".
Example:
SELECT
ID,
TestField,
LEN(TestField) As LenOfTestField, -- Does not include trailing spaces
DATALENGTH(TestField) As DataLengthOfTestField -- Shows the true length of data, including trailing spaces.
FROM
TestTable
You can use this trick:
LEN(Str + 'x') - 1
I use this method:
LEN(REPLACE(TestField, ' ', '.'))
I prefer this over DATALENGTH because this works with different data types, and I prefer it over adding a character to the end because you don't have to worry about the edge case where your string is already at the max length.
Note: I would test the performance before using it against a very large data set; though I just tested it against 2M rows and it was no slower than LEN without the REPLACE...
"How do I include the trailing spaces in the length result?"
You get someone to file a SQL Server enhancement request/bug report because nearly all the listed workarounds to this amazingly simple issue here have some deficiency or are inefficient. This still appears to be true in SQL Server 2012. The auto trimming feature may stem from ANSI/ISO SQL-92 but there seems to be some holes (or lack of counting them).
Please vote up "Add setting so LEN counts trailing whitespace" here:
https://feedback.azure.com/forums/908035-sql-server/suggestions/34673914-add-setting-so-len-counts-trailing-whitespace
Retired Connect link:
https://connect.microsoft.com/SQLServer/feedback/details/801381
There are problems with the two top voted answers. The answer recommending DATALENGTH is prone to programmer errors. The result of DATALENGTH must be divided by the 2 for NVARCHAR types, but not for VARCHAR types. This requires knowledge of the type you're getting the length of, and if that type changes, you have to diligently change the places you used DATALENGTH.
There is also a problem with the most upvoted answer (which I admit was my preferred way to do it until this problem bit me). If the thing you are getting the length of is of type NVARCHAR(4000), and it actually contains a string of 4000 characters, SQL will ignore the appended character rather than implicitly cast the result to NVARCHAR(MAX). The end result is an incorrect length. The same thing will happen with VARCHAR(8000).
What I've found works, is nearly as fast as plain old LEN, is faster than LEN(#s + 'x') - 1 for large strings, and does not assume the underlying character width is the following:
DATALENGTH(#s) / DATALENGTH(LEFT(LEFT(#s, 1) + 'x', 1))
This gets the datalength, and then divides by the datalength of a single character from the string. The append of 'x' covers the case where the string is empty (which would give a divide by zero in that case). This works whether #s is VARCHAR or NVARCHAR. Doing the LEFT of 1 character before the append shaves some time when the string is large. The problem with this though, is that it does not work correctly with strings containing surrogate pairs.
There is another way mentioned in a comment to the accepted answer, using REPLACE(#s,' ','x'). That technique gives the correct answer, but is a couple orders of magnitude slower than the other techniques when the string is large.
Given the problems introduced by surrogate pairs on any technique that uses DATALENGTH, I think the safest method that gives correct answers that I know of is the following:
LEN(CONVERT(NVARCHAR(MAX), #s) + 'x') - 1
This is faster than the REPLACE technique, and much faster with longer strings. Basically this technique is the LEN(#s + 'x') - 1 technique, but with protection for the edge case where the string has a length of 4000 (for nvarchar) or 8000 (for varchar), so that the correct answer is given even for that. It also should handle strings with surrogate pairs correctly.
LEN cuts trailing spaces by default, so I found this worked as you move them to the front
(LEN(REVERSE(TestField))
So if you wanted to, you could say
SELECT
t.TestField,
LEN(REVERSE(t.TestField)) AS [Reverse],
LEN(t.TestField) AS [Count]
FROM TestTable t
WHERE LEN(REVERSE(t.TestField)) <> LEN(t.TestField)
Don't use this for leading spaces of course.
You need also to ensure that your data is actually saved with the trailing blanks. When ANSI PADDING is OFF (non-default):
Trailing blanks in character values
inserted into a varchar column are
trimmed.
You should define a CLR function that returns the String's Length field, if you dislike string concatination.
I use LEN('x' + #string + 'x') - 2 in my production use-cases.
If you dislike the DATALENGTH because of of n/varchar concerns, how about:
select DATALENGTH(#var)/isnull(nullif(DATALENGTH(left(#var,1)),0),1)
which is just
select DATALENGTH(#var)/DATALENGTH(left(#var,1))
wrapped with divide-by-zero protection.
By dividing by the DATALENGTH of a single char, we get the length normalised.
(Of course, still issues with surrogate-pairs if that's a concern.)
This is the best algorithm I've come up with which copes with the maximum length and variable byte count per character issues:
ISNULL(LEN(STUFF(#Input, 1, 1, '') + '.'), 0)
This is a variant of the LEN(#Input + '.') - 1 algorithm but by using STUFF to remove the first character we ensure that the modified string doesn't exceed maximum length and remove the need to subtract 1.
ISNULL(..., 0) is added to deal with the case where #Input = '' which causes STUFF to return NULL.
This does have the side effect that the result is also 0 when #Input is NULL which is inconsistent with LEN(NULL) which returns NULL, but this could be dealt with by logic outside this function if need be
Here are the results using LEN(#Input), LEN(#Input + '.') - 1, LEN(REPLACE(#Input, ' ', '.')) and the above STUFF variant, using a sample of #Input = CAST(' S' + SPACE(3998) AS NVARCHAR(4000)) over 1000 iterations
Algorithm
DataLength
ExpectedResult
Result
ms
LEN
8000
4000
2
14
+DOT-1
8000
4000
1
13
REPLACE
8000
4000
4000
514
STUFF+DOT
8000
4000
4000
0
In this case the STUFF algorithm is actually faster than LEN()!
I can only assume that internally SQL looks at the last character and if it is not a space then optimizes the calculation
But that's a good result eh?
Don't use the REPLACE option unless you know your strings are small - it's hugely inefficient
use
SELECT DATALENGTH('string ')
I have a nvarchar field in my table, which contains all sorts of strings.
In case there are strings which contain a number following a non-number sign, I want to insert a space before that number.
That is - if a certain entry in that field is abc123, it should be turned into abc 123, or ab12.34 should become ab 12. 34.I want this to be done throughout the entire table.
What's the best way to achieve it?
You can try something like that:
select left(col,PATINDEX('%[0-9]%',col)-1 )+space(1)+
case
when PATINDEX('%[.]%',col)<>0
then substring(col,PATINDEX('%[0-9]%',col),len(col)+1-PATINDEX('%[.]%',col))
+space(1)+
substring(col,PATINDEX('%[.]%',col)+1,len(col)+1-PATINDEX('%[.]%',col))
else substring(col,PATINDEX('%[0-9]%',col),len(col)+1-PATINDEX('%[0-9]%',col))
end
from tab
It's not simply, but I hope it will help you.
SQL Fiddle
I used functions (link to MSDN):
LEFT, PATINDEX, SPACE, SUBSTRING, LEN
and regular expression.