Does anybody knows which formatting rules uses jsmin/jsformatter plugin of Notepad++? I need this because we are forced to use this formatter but I'm using intellij idea to write js code. So having this rules I can import it some how or, at least, apply manually.
Thanks everyone in advance!
The minimising rules applied are listed here:
http://www.crockford.com/javascript/jsmin.html
JSMin is a filter that omits or modifies some characters. This does
not change the behavior of the program that it is minifying. The
result may be harder to debug. It will definitely be harder to read.
JSMin first replaces carriage returns ('\r') with linefeeds ('\n'). It
replaces all other control characters (including tab) with spaces. It
replaces comments in the // form with linefeeds. It replaces comments
in the /* */ form with spaces. All runs of spaces are replaced with a
single space. All runs of linefeeds are replaced with a single
linefeed.
It omits spaces except when a space is preceded and followed by a
non-ASCII character or by an ASCII letter or digit, or by one of these
characters:
\ $ _
It is more conservative in omitting linefeeds, because linefeeds are
sometimes treated as semicolons. A linefeed is not omitted if it
precedes a non-ASCII character or an ASCII letter or digit or one of
these characters:
\ $ _ { [ ( + -
and if it follows a non-ASCII character or an ASCII letter or digit or
one of these characters:
\ $ _ } ] ) + - " '
No other characters are omitted or modified.
There are other custom formatting rules applied according to the plugin developer's page:
http://www.sunjw.us/jsminnpp/
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:].)
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
I have a column in my database that contains a string like this:
"Warning set for 7 days.\nCritical Notice - Last Time Machine backup was 118 days ago at 2012-11-16 20:40:52\nLast Time Machine Destination was FreeAgent GoFlex Drive\n\nDefined Destinations:\nDestination Name: FreeAgent GoFlex Drive\nBackup Path: Not Specified\nLatest Backup: 2012-11-17"
I am displaying this data in an e-mail to users. I have be able to easily format the field in my html e-mails perfectly by doing the following:
simple_format(#servicedata.service_exit_details.gsub('\n', '<br>'))
The above code replaces the "\n" with "<br>" tags and simple_format handles the rest.
My issues arises with how to format it properly in the plain text template. Initially I thought I could just call the column, seeing as it has "\n" I assumed the plain text would interpret and all would be well. However this simply spits out the string with "\n" intact just as displayed above rather than created line breaks as desired.
In an attempt to find a way to parse the string so the line breaks are acknowledged. I have tried:
#servicedata.service_exit_details.gsub('\n', '"\r\n"')
#servicedata.service_exit_details.gsub('\n', '\r\n')
raw #servicedata.service_exit_details
markdown(#servicedata.service_exit_details, autolinks: false) # with all the necessary markdown setup
simple_format(#servicedata.service_exit_details.html_safe)
none of which worked.
Can anyone tell me what I'm doing wrong or how I can make this work?
What I want is for the plain text to acknowledge the line breaks and format the string as follows:
Warning set for 7 days.
Critical Notice - Last Time Machine backup was 118 days ago at 2012-11-16 20:40:52
Last Time Machine Destination was FreeAgent GoFlex Drive
Defined Destinations:
Destination Name: FreeAgent GoFlex Drive
Backup Path: Not Specified\nLatest Backup: 2012-11-17"
I see.
You need to differentiate a literal backslash followed by a letter n as a sequence of two characters, and a LF character (a.k.a. newline) that is usually represented as \n.
You also need to distinguish two different kinds of quoting you're using in Ruby: singles and doubles. Single quotes are literal: the only thing that is interpreted in single quotes specially is the sequence \', to escape a single quote, and the sequence \\, which produces a single backslash. Thus, '\n' is a two-character string of a backslash and a letter n.
Double quotes allow for all kinds of weird things in it: you can use interpolation with #{}, and you can insert special characters by escape sequences: so "\n" is a string containing the LF control character.
Now, in your database you seem to have the former (backslash and n), as hinted by two pieces of evidence: the fact that you're seeing literal backslash and n when you print it, and the fact that gsub finds a '\n'. What you need to do is replace the useless backslash-and-n with the actual line separator characters.
#servicedata.service_exit_details.gsub('\n', "\r\n")
How to search word start \word in vim. I can do it using the find menu. Is there any other short cut for this?
Try:
/\\word
in command mode.
You can search for most anything in your document using regular expressions. From normal mode, type '/' and then start typing your regular expression, and then press enter. '\<' would match the beginning of a word, so
/\<foo
would match the string 'foo' but only where it is at the beginning of a word (preceded by whitespace in most cases).
You can search for the backslash character by escaping it with a backslash, so:
/\<\\foo
Would find the pattern '\foo' at the beginning of a word.
Not directly relevant (/\\word is the the correct solution, and nothing here changes that), but for your information:
:h magic
If you are for a pattern with many characters with special meaning to regexes, you may find "nomagic" and "very nomagic" mode useful.
/\V^.$
will search for the literal string ^.$, instead of "lines of exactly one character" (\v "very magic" and the default \m "magic" modes) or "lines of exactly one period" (\M "nomagic" mode).
The reason searching for something including "\" is different is because "\" is a special character and needs to be escaped (prepended with a backslash)
Similarly, to search for "$100", which includes the special character "$":
Press /
Type \$100
Press return
To search for "abc", which doesn't include a special character:
Press /
Type abc
Press return