I'm trying to use GLOB operator to determine if some characters are presented in a string:
SELECT *
FROM Test
WHERE num GLOB '*[~!?.;:+=()<>_#%&/\\]*'
It works fine with '][' at the beginning of the pattern:
WHERE num GLOB '*[][~!?.;:+=()<>_#%&/\\]*'
But it returns nothing when placing '[]' anywhere in the pattern:
WHERE num GLOB '*[[]~!?.;:+=()<>_#%&/\\]*'
What's the reason of such behaviour?
The ] character ends a character list. So [[] is a character list that contains only [.
An empty character list would not make sense, so as an exception, putting the ] directly behind the opening [ ([]...]) can be used to include the ] character in the list. But that is the only exception.
(For similar reasons, - as a literal character must be the last one in the list.)
Related
I'm trying to replace newline etc kind of values using regexp_replace. But when I open the result in query result window, I can still see the new lines in the text. Even when I copy the result, I can see new line characters. See output for example, I just copied from the result.
Below is my query
select regexp_replace('abc123
/n
CHAR(10)
头疼,'||CHR(10)||'allo','[^[:alpha:][:digit:][ \t]]','') from dual;
/ I just kept for testing characters.
Output:
abc123
/n
CHAR(10)
头疼,
allo
How to remove the new lines from the text?
Expected output:
abc123 /nCHAR(10)头疼,allo
There are two mistakes in your code. One of them causes the issue you noticed.
First, in a bracket expression, in Oracle regular expressions (which follow the POSIX standard), there are no escape sequences. You probably meant \t as escape sequence for tab - within the bracket expression. (Note also that in Oracle regular expressions, there are no escape sequences like \t and \n anyway. If you must preserve tabs, it can be done, but not like that.)
Second, regardless of this, you include two character classes, [:alpha:] and [:digit:], and also [ \t] in the (negated) bracket expression. The last one is not a character class, so the [ as well as the space, the backslash and the letter t are interpreted as literal characters - they stand in for themselves. The closing bracket, on the other hand, has special meaning. The first of your two closing brackets is interpreted as the end of the bracket expression; and the second closing bracket is interpreted as being an additional, literal character that must be matched! Since there is no such literal closing bracket anywhere in the string, nothing is replaced.
To fix both mistakes, replace [ \t] with the [:blank:] character class, which consists exactly of space and tab. (And, note that [:alpha:][:digit:] can be written more compactly as [:alnum:].)
Hi I tried using Regex_replace and it is still not working.
select CASE WHEN sbbb <> ' ' THEN regexp_replace(sbbb,'[a-zA-Z _-#]','']
ELSE sbbb
AS ABCDF
from Table where sccc=1;
This is the query which I am using to remove alphabets and specials characters from string and have only numbers. but it doesnot work. Query returns me the complete string with numbers,characters and special characters .What is wrong in the above query
I am working on a sql query. There is a column in database which contains characters,special characters and numbers. I want to only keep the numbers and remove all the special characters and alphabets. How can I do it in query of DB2. If a use PATINDEX it is not working. please help here.
The allowed regular expression patterns are listed on this page
Regular expression control characters
Outside of a set, the following must be preceded with a backslash to be treated as a literal
* ? + [ ( ) { } ^ $ | \ . /
Inside a set, the follow must be preceded with a backslash to be treated as a literal
Characters that must be quoted to be treated as literals are [ ] \
Characters that might need to be quoted, depending on the context are - &
So for you, this should work
regexp_replace(sbbb,'[a-zA-Z _\-#]','')
I was testing a regular expression in Oracle SQL and found something I could not understand:
-- NO MATCH
SELECT 1 FROM DUAL WHERE REGEXP_LIKE ('Professor Frank', '(^|\s)Prof[^\s]*(\s|$)');
Above doesn't match, while the following matches:
-- MATCH
SELECT 1 FROM DUAL WHERE REGEXP_LIKE ('Professor Frank', '(^|\s)Prof\S*(\s|$)');
In other regex flavors, It will be like \bProf[^\s]*\b versus \bProf\S*\b and have similar results. Note: Oracle SQL regex does not have \b or word boundary.
Question: Why don't [^\s]* and \S* work the same way in Oracle SQL?
I notice if I remove the (\s|$) at the end, the first regex will match.
In Oracle regular expressions, \s is indeed the escape sequence for a space, but NOT in a matching character set (that is, [.....], or [^....] for excluding one character). In a matching character set, only two characters have a special meaning, - for ranges and ] for closing the set enumeration. They can't be escaped; if needed in the matching set, ] must always be the first character right after the opening [ (it is the ONLY position in which a closing ] stands for itself as a character, and does not denote the end of the matching set), and - must be first or last (best to leave it always to the end of the matching set) - anywhere else it is seen as a range marker. To include (or exclude, if using the [^.....] syntax) a space, just type an actual physical space in the matching set.
Edit: What I said above is not entirely right. There is another special character in a matching set, namely ^. If it is used in the first position, it means "match any character OTHER THAN." In any other position it stands for itself. For example, '[^^]' will match any single character OTHER THAN ^ (the first ^ has special meaning, the second stands in for itself). And, a closing bracket ] stands for itself if it is the second character in brackets, if the first character is ^ (with its SPECIAL meaning). That is, to match any single character OTHER THAN ], we can use the matching pattern '[^]]'.
As a lexer rule I'd like to match a string according to these rules:
must not contain tabs (\t) or line breaks (\r, \n)
must not contain two succeeding spaces
can contain all other characters, including single spaces
I came up with:
STRING: ~[\t\r\n]*
But I don't know how to prevent succeeding spaces.
This will do it:
STRING:
(
~[\t\r\n ] // non-whitespace
| ' ' ~[\t\r\n ] // or single space followed by non-whitespace
)+
' '? // may optionally end in a space (if desired, remote the line otherwise)
;
I have a column eventDate which contains trailing spaces. I am trying to remove them with the PostgreSQL function TRIM(). More specifically, I am running:
SELECT TRIM(both ' ' from eventDate)
FROM EventDates;
However, the trailing spaces don't go away. Furthermore, when I try and trim another character from the date (such as a number), it doesn't trim either. If I'm reading the manual correctly this should work. Any thoughts?
There are many different invisible characters. Many of them have the property WSpace=Y ("whitespace") in Unicode. But some special characters are not considered "whitespace" and still have no visible representation. The excellent Wikipedia articles about space (punctuation) and whitespace characters should give you an idea.
<rant>Unicode sucks in this regard: introducing lots of exotic characters that mainly serve to confuse people.</rant>
The standard SQL trim() function by default only trims the basic Latin space character (Unicode: U+0020 / ASCII 32). Same with the rtrim() and ltrim() variants. Your call also only targets that particular character.
Use regular expressions with regexp_replace() instead.
Trailing
To remove all trailing white space (but not white space inside the string):
SELECT regexp_replace(eventdate, '\s+$', '') FROM eventdates;
The regular expression explained:
\s ... regular expression class shorthand for [[:space:]]
- which is the set of white-space characters - see limitations below
+ ... 1 or more consecutive matches
$ ... end of string
Demo:
SELECT regexp_replace('inner white ', '\s+$', '') || '|'
Returns:
inner white|
Yes, that's a single backslash (\). Details in this related answer:
SQL select where column begins with \
Leading
To remove all leading white space (but not white space inside the string):
regexp_replace(eventdate, '^\s+', '')
^ .. start of string
Both
To remove both, you can chain above function calls:
regexp_replace(regexp_replace(eventdate, '^\s+', ''), '\s+$', '')
Or you can combine both in a single call with two branches.
Add 'g' as 4th parameter to replace all matches, not just the first:
regexp_replace(eventdate, '^\s+|\s+$', '', 'g')
But that should typically be faster with substring():
substring(eventdate, '\S(?:.*\S)*')
\S ... everything but white space
(?:re) ... non-capturing set of parentheses
.* ... any string of 0-n characters
Or one of these:
substring(eventdate, '^\s*(.*\S)')
substring(eventdate, '(\S.*\S)') -- only works for 2+ printing characters
(re) ... Capturing set of parentheses
Effectively takes the first non-whitespace character and everything up to the last non-whitespace character if available.
Whitespace?
There are a few more related characters which are not classified as "whitespace" in Unicode - so not contained in the character class [[:space:]].
These print as invisible glyphs in pgAdmin for me: "mongolian vowel", "zero width space", "zero width non-joiner", "zero width joiner":
SELECT E'\u180e', E'\u200B', E'\u200C', E'\u200D';
'' | '' | '' | ''
Two more, printing as visible glyphs in pgAdmin, but invisible in my browser: "word joiner", "zero width non-breaking space":
SELECT E'\u2060', E'\uFEFF';
'' | ''
Ultimately, whether characters are rendered invisible or not also depends on the font used for display.
To remove all of these as well, replace '\s' with '[\s\u180e\u200B\u200C\u200D\u2060\uFEFF]' or '[\s]' (note trailing invisible characters!).
Example, instead of:
regexp_replace(eventdate, '\s+$', '')
use:
regexp_replace(eventdate, '[\s\u180e\u200B\u200C\u200D\u2060\uFEFF]+$', '')
or:
regexp_replace(eventdate, '[\s]+$', '') -- note invisible characters
Limitations
There is also the Posix character class [[:graph:]] supposed to represent "visible characters". Example:
substring(eventdate, '([[:graph:]].*[[:graph:]])')
It works reliably for ASCII characters in every setup (where it boils down to [\x21-\x7E]), but beyond that you currently (incl. pg 10) depend on information provided by the underlying OS (to define ctype) and possibly locale settings.
Strictly speaking, that's the case for every reference to a character class, but there seems to be more disagreement with the less commonly used ones like graph. But you may have to add more characters to the character class [[:space:]] (shorthand \s) to catch all whitespace characters. Like: \u2007, \u202f and \u00a0 seem to also be missing for #XiCoN JFS.
The manual:
Within a bracket expression, the name of a character class enclosed in
[: and :] stands for the list of all characters belonging to that
class. Standard character class names are: alnum, alpha, blank, cntrl,
digit, graph, lower, print, punct, space, upper, xdigit.
These stand for the character classes defined in ctype.
A locale can provide others.
Bold emphasis mine.
Also note this limitation that was fixed with Postgres 10:
Fix regular expressions' character class handling for large character
codes, particularly Unicode characters above U+7FF (Tom Lane)
Previously, such characters were never recognized as belonging to
locale-dependent character classes such as [[:alpha:]].
It should work the way you're handling it, but it's hard to say without knowing the specific string.
If you're only trimming leading spaces, you might want to use the more concise form:
SELECT RTRIM(eventDate)
FROM EventDates;
This is a little test to show you that it works.
Tell us if it works out!
If your whitespace is more than just the space meta value than you will need to use regexp_replace:
SELECT '(' || REGEXP_REPLACE(eventDate, E'[[:space:]]', '', 'g') || ')'
FROM EventDates;
In the above example I am bounding the return value in ( and ) just so you can easily see that the regex replace is working in a psql prompt. So you'll want to remove those in your code.
SELECT replace((' devo system ') ,' ','');
It gives: devosystem
A tested one that works like a charm:
UPDATE company SET name = TRIM (BOTH FROM name) where id > 0