action: 'blah' 'blah'
Need to remove anything that is after action: in a file
sed -i 's/action:\*//g' tes1
This does not do anything.
It appears you are confusing globbing with regular expressions. In your question you have:
Need to remove anything that is after action: in a file
sed -i 's/action:\*//g' tes1
This does not do anything.
Of course it doesn't if the input is "action: 'blah' 'blah'". The only thing that would be matched by your attempt would be a literal:
action:*
(the glob, e.g. wildcard, '*' matches any number of characters as a shell expression, not a regular expression. As a regular expression, it controls repetition)
What you want to do is match the "action:" portion and then match everything that follows it, replacing what follows with nothing. The general match any number of characters until end of line is .*$ Where '.' matches any character, the repetition '*' matches zero-or-more times and finally the '$' anchors the expression to the end of line.
Put another way, you want to match "action:" at the beginning of the line, followed by any number of additional characters, and you want to replace all additional characters with nothing. To anchor action: to the beginning of the line you use the circumflex '^', e.g. "^action:" to ensure you only match "action:" and not "any action:".
Putting it altogether, you could use:
sed -i 's/^action:.*$/action:/' tes1
Which in sum matches "action: and to end of line" and replaces the entire expression with "action:" leaving you with your desired line containing only "action:".
Related
I am a complete Reg-exp noob, so please bear with me. Tried to google this, but haven't found it yet.
What would be an appropriate way of writing a Regular expression matching files starting with a dot, such as .buildpath or .htaccess?
Thanks a lot!
In most regex languages, ^\. or ^[.] will match a leading dot.
The ^ matches the beginning of a string in most languages. This will match a leading .. You need to add your filename expression to it.
^\.
Likewise, $ will match the end of a string.
You may need to substitute the \ for the respective language escape character. However, under Powershell the Regex I use is: ^(\.)+\/
Test case:
"../NameOfFile.txt" -match '^(\\.)+\\\/'
works, while
"_./NameOfFile.txt" -match '^(\\.)+\\\/'
does not.
Naturally, you may ask, well what is happening here?
The (\\.) searches for the literal . followed by a +, which matches the previous character at least once or more times.
Finally, the \\\/ ensures that it conforms to a Window file path.
It depends a bit on the regular expression library you use, but you can do something like this:
^\.\w+
The ^ anchors the match to the beginning of the string, the \. matches a literal period (since an unescaped . in a regular expression typically matches any character), and \w+ matches 1 or more "word" characters (alphanumeric plus _).
See the perlre documentation for more info on Perl-style regular expressions and their syntax.
It depends on what characters are legal in a filename, which depends on the OS and filesystem.
For example, in Windows that would be:
^\.[^<>:"/\\\|\?\*\x00-\x1f]+$
The above expression means:
Match a string starting with the literal character .
Followed by at least one character which is not one of (whole class of invalid chars follows)
I used this as reference regarding which chars are disallowed in filenames.
To match the string starting with dot in java you will have to write a simple expression
^\\..*
^ means regular expression is to be matched from start of string
\. means it will start with string literal "."
.* means dot will be followed by 0 or more characters
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:].)
I try to use awk to remove all byte order marks from a file (I have many of them):
awk '{sub(/\xEF\xBB\xBF/,"")}{print}' f1.txt > f2.txt
It seems to remove all the BOMs that are in the beginning of the line but those in the middle are not removed. I can verify that by:
grep -U $'\xEF\xBB\xBF' f2.txt
Grep returns me one line where BOM is in the middle.
As mentioned sub() will only swap out the leftmost substring, so if global is what you're after then using gsub(), or even better gensub() is the way to go.
sub(regexp, replacement [, target])
Search target, which is treated as a string, for the leftmost, longest
substring matched by the regular expression regexp. Modify the entire
string by replacing the matched text with replacement. The modified
string becomes the new value of target. Return the number of
substitutions made (zero or one).
gsub(regexp, replacement [, target])
Search target for all of the longest, leftmost, nonoverlapping
matching substrings it can find and replace them with replacement. The
‘g’ in gsub() stands for “global,” which means replace everywhere.
gensub(regexp, replacement, how [, target]) #
Search the target string target for matches of the regular expression
regexp. If how is a string beginning with ‘g’ or ‘G’ (short for
“global”), then replace all matches of regexp with replacement.
Otherwise, "how" is treated as a number indicating which match of regexp
to replace. gensub() is a general substitution function. Its purpose is to provide more features than the standard sub() and gsub() functions.
There's tons more helpful information and examples linked below:
↳ The GNU Awk User's Guide: String Functions / 9.1.3 String-Manipulation Functions
I am trying to construct a regular expression to find the text of the following variations.
NSLocalizedString(#"TEXT")
NSLocalizedStringFromTable(#"TEXT")
NSLocalizedStringWithDefaultValue(#"TEXT")
...
The goal is to extract TEXT. I have been able to construct a regex for each individual function or macro, e.g., (?<=NSLocalizedString)\(#"(.*?)". However, I am looking for a solution that does the job no matter what the name of the function as long as it starts with NSLocalizedString.
I assumed it was as simple as (?<=NSLocalizedString\w+)\(#"(.*?)", but that does't seem to do the trick.
How about this one?
/NSLocalizedString\w*\(#"(.*)"\)/
Explanation:
NSLocalizedString 'NSLocalizedString'
\w+ word characters (a-z, A-Z, 0-9, _) (0 or
more times (matching the most amount
possible))
\( '('
#" '#"'
( group and capture to \1:
.* any character except \n (0 or more times
(matching the most amount possible))
) end of \1
" '"'
\) ')'
The only reason your regex doesn't work is because the regex engine doesn't support variable length lookbehinds. The (?<=NSLocalizedString\w+) is variable length so can't be used.
Firstly it needs to be \w* not \w+, to allow your first example string to match.
If you move the \w* outside the lookbehind (?<=NSLocalizedString)\w* it will work just fine.
Alternatively, since you have to use a capturing group to grab the text value anyway, theres no need for the lookbehind at all. Change the (?<= to a (?: and it becomes a non-capturing group (which can be variable length), and then just grab your text value from group 1.
Your attempt was:
(?<=NSLocalizedString\w+)\(#"(.*?)"
Both of these minor changes should make it work:
(?<=NSLocalizedString)\w*\(#"(.*?)"
(?:NSLocalizedString\w*)\(#"(.*?)"
The following is actually not supported in Objective-C:
The solution that will extract exactly TEXT without using any groups is:
NSLocalizedString\w*\(#"\K[^"]*
It avoids the need to use a negative lookbehind (which can't be used for reasons I explain below) by using the \K modifier, which chops off anything before it from the match.
Im trying to replace all lines within files that contains:
/var/www/webxyz/html
to
/home/webxyz/public_html
the string: webxyz is variable: like web1, web232
So only the string before and after webxyz should be replaced.
Tried this without solution:
sed -i 's/"var/www/web*/html"/"home/web*/public_html"/g'
Also i want this should check and replace files (inclusive subdirectory and files),
the * operator don't work.
Within a regular expression, you’ll need to escape the delimiting character that surround them, in your case the /. But you can also use a different delimiter like ,:
sed -i 's,"var/www/web*/html","home/web*/public_html",g'
But to get it working as intended, you’ll also need to remove the " and replace the b* (sed doesn’t understand globbing wildcards) to something like this:
sed -i 's,var/www/web\([^/]*\)/html,home/web\1/public_html,g'
Here \([^/]*\) is used to match anything after web except a /. The matching string is then referenced by \1 in the replacement part.
Here is what your replacement operation should look like (see sed for more info):
sed -i 's/var\/www\(\/.*\/\)html/home\1public_html/g'
Note that \(...\) is a grouping, and specifies a "match variable" which shows up in the replacement side as \1. Note also the .* which says "match any single character (dot) zero or more times (star)". Note further that your / characters must be escaped so that they are not treated as part of the sed control structure.