sed replacing without untouching a string - variables
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.
Related
Trino regexp_replace this character in the beginning but not in the middle Trino [duplicate]
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
How to the numbers that start with a $ only Kotlin
I scanned a document in to kotlin and it has words, numbers, values, etc... but I only want the values that start with a $ and have 2 decimal places after the .(so the price) do I use a combination of a substring with other string parses?
Edit: I have looked into Regex and the problem I am having now is I am using this line
val reg = Regex("\$([0-9]*\.[0-9]*)")
to grab all the prices however the portion of *. is saying Invalid escape. However in other languages this works just fine.
You have to use double \ instead of single . It's because the \ is an escape character both in Regex and in Kotlin/Java strings. So when \ appears in a String, Kotlin expects it to be followed by a character that needs to be escaped. But you aren't trying to escape a String's character...you're trying to escape a Regex character. So you have to escape your backslash itself using another backslash, so the backslash is part of the computed String literal and can be understood by Regex.
You also need double \ before your dollar sign for it to behave correctly. Technically, I think it should be triple \ because $ is a special character in both Kotlin and in Regex and you want to escape it in both. However, Kotlin seems smart enough to guess what you're trying to do with a double escape if no variable name or expression follows the dollar sign. Rather than rely on that, I would use the triple escape.
val reg = Regex("\\\$([0-9]*\\.[0-9]*)")
Use variable as key word for grep function
I have successfully used this syntax to assign and locate keywords. I'd like to take the same approach but instead of knowing the word in advance like in the example below, I want to pass the word/expression as a variable, for example replacing options, with a var. How would I do that? phrase <- "(options) ([^ ]+)"
You can use paste0:
phrase <- paste0("(", optionsVar, ") ([^ ]+)")
Also, note that [^ ]+ matches one or more chars other than spaces, but you can probably replace it with \\S+, one or more non-whitespace chars. Same with the literal space: \\s or \\s+ might prove more flexible.
How do i add reg exp to sed?
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:".
Sed script needed to insert LF before each time match in a large single string
I have lengthy string that I need to put a line feed before each instance of a time stamp. 03:38:11,03/07/2017,node,cpu,user,sys,idle,intr/s,ctxt/s,0,0,0,9,91,0,1,0,24,75,0,total,0,17,83,2370,3574,1,0,3,4, 93,1,1,10,4,86,1,total,7,4,89,2922,4653,03:39:11,03/07/2017,node,cpu,user,sys,idle,intr/s,ctxt/s,0,0,4,25,71,0,1,5 ,16,79,0,total,4,21,75,2487,3876,1,0,0,3,97,1,1,1,1,98,1,total,1,2,98,2880,4728,03:40:11,03/07/2017,node,cpu,user, sys,idle,intr/s,ctxt/s,0,0,1,30,69,0,1,1,30,69,0,total,1,30,69,3237,4344,1,0,3,49,47,1,1,10,47,43,1,total,6,48,45, 3920,5702, I need to see about formatting it as such: 03:38:11,03/07/2017,node,cpu,user,sys,idle,intr/s,ctxt/s,0,0,0,9,91,0,1,0,24,75,0,total,0,17,83,2370,3574,1,0,3,4,93,1,1,10,4,86,1,total,7,4,89,2922,4653, 03:39:11,03/07/2017,node,cpu,user,sys,idle,intr/s,ctxt/s,0,0,4,25,71,0,1,5,16,79,0,total,4,21,75,2487,3876,1,0,0,3,97,1,1,1,1,98,1,total,1,2,98,2880,4728, 03:40:11,03/07/2017,node,cpu,user,sys,idle,intr/s,ctxt/s,0,0,1,30,69,0,1,1,30,69,0,total,1,30,69,3237,4344,1,0,3,49,47,1,1,10,47,43,1,total,6,48,45,3920,5702, I am currently trying to use the following: sed -e 's/^[[:digit:]][[:digit:]]\:[[:digit:]][[:digit:]]/\n&/g' cpu.log
The ^ line anchor forces sed to only match the first date stamp. Remove it and you should be fine. To avoid roplacing the first, maybe massage the script to require something before the match (hard-coding a comma would seem to work, based on your sample data); or just post-process the output to remove the first newline.
sed 's/[0-9][0-9]:[0-9][0-9]:[0-9][0-9]/\n&/g'