I've been learning ksh for quite some time but I still cannot understand %(pattern-pair(s)) in the manual. Anyone can give a simple meaningful example?
A pattern of the form %(pattern-pair(s)) is a sub-pattern that can be
used to match nested character expressions. Each pattern-pair is a two
character sequence which cannot contain & or |. The first pattern-pair
specifies the starting and ending characters for the match. Each sub-
sequent pattern-pair represents the beginning and ending characters of
a nested group that will be skipped over when counting starting and
ending character matches. The behavior is unspecified when the first
character of a pattern-pair is alpha-numeric except for the following:
D Causes the ending character to terminate the search for
this pattern without finding a match.
E Causes the ending character to be interpreted as an
escape character.
L Causes the ending character to be interpreted as a quote
character causing all characters to be ignored when look-
ing for a match.
Q Causes the ending character to be interpreted as a quote
character causing all characters other than any escape
character to be ignored when looking for a match.
I suppose it's useful for JSON parsing:
json='{"foo":{"bar":"baz"} }'
#remove all quoted values nested in punctuation except the first
jsonroot=${json%%[[:punct:]][[:punct:]]%(\"\")*}
#remove initial curly
jsonroot=${jsonroot#?}
Related
I am new here. I wanted to ask a question on using REGEX for an entity in DialogFlow
I wanted the entity to accept all text and spaces except for the symbol *
I have tried to use [A-Za-z0-9 ][^*], but it is not working. Any advice. thanks!
In your Regex expression, [^*] means "capture any character at the start of the line." To refer to a literal asterisk rather than matching any character, you need to use \*
If you want to match a line of letters or numbers as in the [A-Za-z0-9] example you give, but only if that string does not include an asterisk, then this expression should work for you:
^[a-zA-Z0-9]+$
This means "match a whole line of text if it only contains one or more of the characters a-z, A-Z, or 0-9".
If you want to match any character or group of characters in a line except for the asterisk, then you could use something like this:
(?!\*)([a-zA-Z0-9]+)(?<!\*)
The first part is called a "negative lookahead," and it looks forward to ensure we're not matching the asterisk. The last part is called a "negative lookbehind," and it looks backwards to make sure we're not matching the asterisk. The middle part is your "capture group," and confirms that you're matching any letters or numbers in a given string, but excluding the * character.
If this Regex gets input like *abc, it will capture abc. If it encounters abc*, it will still capture abc. If it encounters abc*def, it will capture abc and def separately in two capture groups, because it will break around the asterisk.
This link explains the concept of lookarounds in Regex. You can also use this Regex tester to get started practicing your Regular Expressions with explanations of what each block of characters does.
EDITED TO ADD If you're just interested in matching single characters rather than groups of characters, you can use [A-Za-z0-9] and match any upper or lowercase letter and any single digit. You don't need to exclude the * character, because the character group is already exclusive.
This is a slight duplicate of the question below, so responses here may also help you. Hope this helps!
How can I exclude asterisk in a regex expression
[A-Za-z0-9 ][^*]
What you regex will do is match 2 consecutive characters. First, it will look for anything A-Za-z0-9 . Then, it will look at the negated set that includes *, and will match ANY character except *.
You can type your regex into https://regexr.com/ to see a breakdown of how it matches and test some strings.
For example, your regex would match these:
Aa
AA
a&
A1
0_
But would not match these:
A*
a*
1*
And WOULD NOT match anything longer than 2 characters. If you really want to match any string with any characters except *, this should work:
[^\*]+
What that will do is match any number of consecutive characters that are not *. (The + means match 1 or more characters in the set). It is also a good idea to escape * because it is also a reserved character in regex. Even though most regex parsers are smart enough to know that inside a group you probably mean the literal char *, it is still a best practice to escape it. (And by that same token, you would want to use \s instead of the blank space in your original regex.)
I want to replace a line, that represents a part of mathematical equation:
f(x,z,time,temp)=-(2.0)/(exp(128*((x-2.5*time)*(x-2.5*time)+(z-0.2)*(z-0.2))))+(
with a new one similar to the above. Both new and old lines are saved in bash variables.
Main problem is that mathematical equation is full with special characters that do not allow proper search and replace in bash mode, even when I used as delimiter special character that is not used in equation.
I used
sed -n "s|$OLD|$NEW|g" restart.k
and
sed -i "s|$OLD|$NEW|g" restart.k
but all times I get wrong results.
Any idea to solve this?
There is only * in your pattern here that is special for sed, so escape it and do replacement as usual:
sed "s:$(sed 's:[*]:\\&:g' <<<"$old"):$new:" infile
if there are more special characters in your real sample, then you will need to add them inside bracket []; there are some exceptions like:
if ^ character: it can be place anywhere in [] but not first character, because ^ character at first negates the characters within its bracket expression.
if ] character: it should be the first character, because this character is also used to end the bracket expression.
if - character: it should be the first or last character, because this character is also can be used for defining the range of characters too.
I have a line in a source file: [12 13 15]. In vim, I type:
:%s/\([0-90-9]\) /\0, /g
wanting to add a coma after 12 and 13. It works, but not quite, as it inserts an extraspace [12 , 13 , 15].
How can I achieve the desired effect?
Use \1 in the replacement expression, not \0.
\1 is the text captured by the first \(...\). If there were any more pairs of escaped parens in your pattern, \2 would match the text capture between the pair starting at the second \(, \3 at the third \(, and so on.
\0 is the entire text matched by the whole pattern, whether in parentheses or not. In your case this includes the space at the end of your pattern.
Also note that [0-90-9] is the same as [0-9]: each [...] collection matches just one character. It happens to work anyway, because in your data ‘a digit followed by a space’ matches in the same places as ‘2 digits followed by a space’. (If you actually needed to only insert commas after 2 digits, you could write [0-9][0-9].)
"I have a line in a source file:..."
then you type :%s/... this will do the substitution on all lines, if it matched. or that is the single line in your file?
If it is the single line, you don't have to group, or [0-9], just :%s/ \+/,/g will do the job.
The fine answers already point interesting solutions, but here's another one,
making use of the \zs, which marks the start of the match. In this pattern:
/[0-9]\zs /
The searched text is /[0-9] /, but only the space counts as a match. Note
that you can use the class \d to simplify the digit character class, so the
following command shall work for your needs:
:s/\d\d\zs /, /g ; matches only the space, replace by `, '
You said you have multiple lines and these changes are only to certain lines.
You can either visually select the lines to be changed or use the :global
command, which searches for lines matching a pattern and applies a command to
them. Now you'd need to build an expression to match the lines to be changed
in a less precise as possible way. If the lines that begins with optional
spaces, a [ and two digits are the only lines to be matched and no other
ones, then this would work for you:
:g/\s*[\d\d/s/\d\d\zs /, /g
Check the help for pattern.txt for \ze and similar and
:global.
Homework: use the help to understand \zs and see how this works:
:s/\d\d\zs\ze /,/g
I messed something up. In my xml, each non preferred term has a preferred term to use:
Something I have done has created some non preffered terms where the preferred term to use is the exact same name as this non preferred term.
<term>
<termId>127699289611384833453kNgWuDxZEK37Lo4QVWZ</termId>
<termUpdate>Add</termUpdate>
<termName>Adenosquamous Carcinoma</termName>
<termType>Nd</termType>
<termStatus>Active</termStatus>
<termApproval>Approved</termApproval>
<termCreatedDate>20110704T09:41:31</termCreatedDatae>
<termCreatedBy>admin</termCreatedBy>
<termModifiedDate>20110704T09:45:17</termModifiedDate>
<termModifiedBy>admin</termModifiedBy>
<relation>
<relationType>USE</relationType>
<termId>1276992897N1537166632rbr7BISWAI93SarY118G</termId>
<termName>Adenosquamous Carcinoma</termName>
</relation>
Is there a text editor with a find and replace function I can use to tell it that if the in =the of the actual term, to just delete the whole ? I looked at the related queries and they mentioned regular expressions, but I've spent ages trying to build them and they are beyond me,
thanks!
It is nearly 3 years too late answering this question, but there are Perl regular expressions which can be indeed used for this task.
Finding and deleting a term block containing same termName in relation as defined above for the term itself is possible with UltraEdit for Windows v21.10.0.1032 and most likely also with other text editors supporting Perl regular expression using a case-sensitive Perl regular expression Replace with search string:
^[ \t]*<term>(?:(?!</term>)[\S\s])+<termName>([^\r\n]+?)</termName>(?:(?!</term>)[\S\s])+<relation>(?:(?!</term>)[\S\s])+<termName>\1</termName>(?:(?!</term>)[\S\s])+</term>[ \t\r]*\n
The replace string is an empty string.
Explanation:
^ ... start every search at beginning of a line.
[ \t]* ... there can be 0 or more spaces or tabs at beginning of the line.
<term> ... this string must be found next on the line.
Next the tricky expression follows which is required to match any character up to next string of interest, but with avoiding matching something in next term block if the remaining expression does not return a positive result on current term block.
(?:(?!</term>)[\S\s])+ ... this expression finds any character because of [\S\s] matching any non whitespace character or any whitespace character. There must be at least 1 character before next fixed string because of the +, but it can be also more characters. Additionally the Perl regular expression must make look ahead on every character matched to check if NOT </term> follows. If right of the currently matched character there is the string </term>, the Perl regexp engine must stop matching any character at current position in stream and continue with next part of the search string. So this expression can match any character, but not beyond </term> and therefore only characters between <term> and </term>. Because of ?: nothing is captured/marked for back referencing by this expression.
<termName> ... this fixed string within a term block must be found next.
([^\r\n]+?) ... matches the characters of the name of the term and captures/marks this string for back referencing. Instead of the negative character class expression [^\r\n], it would be also possible to use another class definition, or just . if a dot does not match new line characters. Also possible would be ([^<]+) if it is not possible that a not encoded opening angle bracket is part of the term name. Character < must be encoded with < according to XML specification within an element's value except within a CDATA block.
</termName> ... this fixed string within a term block must be found next.
(?:(?!</term>)[\S\s])+ ... again any character within a term block up to next fixed string.
<relation> ... this fixed string within a term block must be found next.
(?:(?!</term>)[\S\s])+ ... again any character within a term block up to next fixed string.
<termName> ... this fixed string within a term block must be found next.
\1 ... this expression back references the captured/marked term name and therefore the next string must be the same as the name of the term defined above.
</termName> ... this fixed string within a term block must be found next.
(?:(?!</term>)[\S\s])+ ... again any character within a term block up to next fixed string.
</term> ... this fixed string marking end of a term block must be found next.
[ \t\r]*\n ... matches 0 or more spaces, tabs and carriage returns and next a line-feed. So this expression works for a DOS/Windows (CR+LF) and a Unix (only LF) text file.
Also possible with UltraEdit is:
(?s)^[ \t]*<term>(?:(?!</term>).)+<termName>([^<]+?)</termName>(?:(?!</term>).)+<relation>(?:(?!</term>).)+<termName>\1</termName>(?:(?!</term>).)+</term>[ \t\r]*\n
(?s) ... this expression at beginning of the search string changes the behavior of . from matching any character except line terminators to really any character and therefore . is now like [\S\s].
I am reading this tutorial, and I encountered that bash script uses [...] as a wild card character. So what exactly [...] stands in a bash script?
It's a regex-style character matching syntax; from the Bash Reference Manual, §3.5.8.1 (Pattern Matching):
[...]
Matches any one of the enclosed characters. A pair of characters separated by a hyphen denotes a range expression; any character that sorts between those two characters, inclusive, using the current locale's collating sequence and character set, is matched. If the first character following the ‘[’ is a ‘!’ or a ‘^’ then any character not enclosed is matched. A ‘−’ may be matched by including it as the first or last character in the set. A ‘]’ may be matched by including it as the first character in the set. The sorting order of characters in range expressions is determined by the current locale and the value of the LC_COLLATE shell variable, if set.
For example, in the default C locale, ‘[a-dx-z]’ is equivalent to ‘[abcdxyz]’. Many locales sort characters in dictionary order, and in these locales ‘[a-dx-z]’ is typically not equivalent to ‘[abcdxyz]’; it might be equivalent to ‘[aBbCcDdxXyYz]’, for example. To obtain the traditional interpretation of ranges in bracket expressions, you can force the use of the C locale by setting the LC_COLLATE or LC_ALL environment variable to the value ‘C’.
Within ‘[’ and ‘]’, character classes can be specified using the syntax [:class:], where class is one of the following classes defined in the posix standard:
alnum alpha ascii blank cntrl digit graph lower
print punct space upper word xdigit
A character class matches any character belonging to that class. The word character class matches letters, digits, and the character ‘_’.
Within ‘[’ and ‘]’, an equivalence class can be specified using the syntax [=c=], which matches all characters with the same collation weight (as defined by the current locale) as the character c.
Within ‘[’ and ‘]’, the syntax [.symbol.] matches the collating symbol symbol.
(emphasis added to the most common usage patterns)
It is used in the tutorial to speak about regular expressions in addition to globbing ('*' and '?'). For example [a-z] regular expression will match one lowercase character.
Actually, what is a wildcard is [abc] for example. It matches one of the three letters.