I have a huge JSON output and I just need to delete everything except a small string in each line.
The string has the format
"title": "someServerName"
the "someServerName" (the section within quotes) can vary wildly.
The closest I've come is this:
:%s/\("title":\s"*"\)
But this just manages to delete
"title": "
The only thing I want left in each line is
"title": "someServerName"
EDIT to answer the posted question:
The Text I'm going to be working with will have a format similar to
{"_links": {"self": {"href": "/api/v2/servers/32", "title": "someServerName"},tons_of_other_json_crap_goes_here
All I want left at the end is:
"title": "someServerName"
It should be .* rather than * to match a group of any characters. This does the job:
%s/^.*\("title":\s".*"\).*$/\1/
Explanation of each part:
%s/ Substitute on each matching line.
^.* Ignore any characters starting from beginning of line.
\("title":\s".*"\) Capture the title and server name. ".*" will match any characters between quotes.
.*$ Ignore the rest of the line.
/\1/ The result of the substitution will be the first captured group. The group was captured by parentheses \(...\).
This sounds like a job for grep.
:%!grep -o '"title":\s*"[^"]*"'
For more help with Vim's filtering see :h :range!.
See man grep for more information on the -o/--only-matching flag.
It's quit convenient if you break the replace command into two steps as below. (p.s. I learned this skill from good guide book 《Practical Vim》 recently).
Step 1： Search the contents that you want to keep
\v"title":\s.*"
This will match "title": "someServerName". You can try again and again with command q/ to open the search command window and modify the regular expression (This is the most excellent part I think).
\v^.*("title":\s.*").*$
Then add bracket for latter use and add .* to match other parts that you wish to delete.
Step 2: Replace the matched contents
:%s//\1/g Note the original string in this substitute command is the matched part in last search (Very good feature in vim). And \1 means using the matched group which is actually the part you wish to keep.
Hope you can find it more convenient than the long obscure substitute command.
%s/\v.*("title":\s".*").*/\1
Explanation and source here.
I think the most elegant way to solve this is using ':g'
More comprehensive details in this link below!
Power fo G
It may not fully archive what you're looking, but damn close ;)
Related
This is what I got:
.*([1-30000]#.*#).*
However, I would like to follow these rules without limiting to only 1 line.
For example, using: .*([1-30000]#.*#).* I could find:
5173#bunch of text here#
And what I would like to find:
5173#bunch of text here
of, bunch here, text
text here, bunch of
#
Hope I managed to be clear about my problem, thanks for the help.
Edit:
\b(?:[12]?\d{1,4}|30{4})#[^#]+#
Seems to be working, now the "challenge" is another, i want to save the number before the # (5173#) and replace what i got into another file where the same number is found.
You may use this regex:
(?<=\d#)[^#]+
Enable . matches newline and regex in your NP++ search box.
This matches text preceded by only a single digit followed by a pound and succeeded by a pound since NP++ doesn't support variable-length lookbehinds.
With the contribution of all, that is, joining what was answered by you. I got:
\b(?:[12]?\d{1,4}|30{4})#[^#]+#
I'm not sure if there's any mistake, I'm not familiar with all of this. :)
https://regex101.com/r/t7xBXk/1
On the basis of some html editing I've came up with need for help from some VIM master out there.
I wan't to achieve simple task - I have html file with mangled urls.
Just description Just description
...
Just description
Unfortunately it's not "one url per line".
I am aware of three approaches:
I would like to be able to replace only within '"http://[^"]*"' regex (similar like replace only in matching lines - but this time not whole lines but only matching pattern should be involved)
Or use sed-like labels - I can do this task with sed -e :a -e 's#\("http://[^" ]*\) \([^"]*"\)#\1_\2#g;ta'
Also I know that there is something like "\#<=" but I am non native speaker and vim manual on this is beyond my comprehension.
All help is greatly appreciated.
If possible I would like to know answer on all three problems (as those are pretty interesting and would be helpful in other tasks) but either of those will do.
Re: 1. You can replace recursively by combining vim's "evaluate replacement as an expression" feature (:h :s\=) with the substitute function (:h substitute()):
:%s!"http://[^"]*"!\=substitute(submatch(0), ' ', '_', 'g')!g
Re: 2. I don't know sed so I can't help you with that.
Re: 3. I don't see how \#<= would help here. As for what it does: It's equivalent to Perl's (?<=...) feature, also known as "positive look-behind". You can read it as "if preceded by":
:%s/\%(foo\)\#<=bar/X/g
"Replace bar by X if it's preceded by foo", i.e. turn every foobar into fooX (the \%( ... \) are just for grouping here). In Perl you'd write this as:
s/(?<=foo)bar/X/g;
More examples and explanation can be found in perldoc perlretut.
I think what you want to do is to replace all spaces in your http:// url into _.
To achieve the goal, #melpomene's solution is straightforward. You could try it in your vim.
On the other hand, if you want to simulate your sed line, you could try followings.
:let #s=':%s#\("http://[^" ]*\)\#<= #_#g^M'
^M means Ctrl-V then Enter
then
200#s
this works in same way as your sed line (label, do replace, back to label...) and #<= was used as well.
one problem is, in this way, vim cannot detect when all match-patterns were replaced. Therefore a relative big number (200 in my example) was given. And in the end an error msg "E486: Pattern not found..." shows.
A script is needed to avoid the message.
I have a feed in Yahoo Pipes and want to match everything after a question mark.
So far I've figured out how to match the question mark using..
\?
Now just to match everything that is after/follows the question mark.
\?(.*)
You want the content of the first capture group.
Try this:
\?(.*)
The parentheses are a capturing group that you can use to extract the part of the string you are interested in.
If the string can contain new lines you may have to use the "dot all" modifier to allow the dot to match the new line character. Whether or not you have to do this, and how to do this, depends on the language you are using. It appears that you forgot to mention the programming language you are using in your question.
Another alternative that you can use if your language supports fixed width lookbehind assertions is:
(?<=\?).*
With the positive lookbehind technique:
(?<=\?).*
(We're searching for a text preceded by a question mark here)
Input: derpderp?mystring blahbeh
Output: mystring blahbeh
Example
Basically the ?<= is a group construct, that requires the escaped question-mark, before any match can be made.
They perform really well, but not all implementations support them.
\?(.*)$
If you want to match all chars after "?" you can use a group to match any char, and you'd better use the "$" sign to indicate the end of line.
?(.*\n)+
With this you can get everything Even a new line
Check out this site: http://rubular.com/ Basically the site allows you to enter some example text (what you would be looking for on your site) and then as you build the regular expression it will highlight what is being matched in real time.
str.replace(/^.+?\"|^.|\".+/, '');
This is sometimes bad to use when you wanna select what else to remove between "" and you cannot use it more than twice in one string. All it does is select whatever is not in between "" and replace it with nothing.
Even for me it is a bit confusing, but ill try to explain it. ^.+? (not anything OPTIONAL) till first " then | Or/stop (still researching what it really means) till/at ^. has selected nothing until before the 2nd " using (| stop/at). And select all that comes after with .+.
This is my first question, so I hope I didn't mess too much with the title and the formatting.
I have a bunch of file a client of mine sent me in this form:
Name.Of.Chapter.021x212.The.Actual.Title.Of.the.Chapter.DOC.NAME-Some.stuff.Here.ext
What I need is a regex to output just:
212 The Actual Title Of the Chapter
I'm not gonna use it with any script language in particular; it's a batch renaming of files through an app supporting regex (which already "preserves" the extension).
So far, all I was able to do was this:
/.*x(\d+)\.(.*?)\.[A-Z]{3}.*/ -->REPLACE: $1 $2
(Capture everything before a number preceded by an "x", group numbers after the "x", group everything following until a 3 digit Uppercase word is met, then capture everything that follows)
which gives me back:
212 The.Actual.Title.Of.the.Chapter
Having seen the result I thought that something like:
/.*x(\d+)\.([^.]*?)\.[A-Z]{3}.*/ -->REPLACE: $1 $2
(Changed second group to "Capture everything which is not a dot...") would have worked as expected.
Instead, the whole regex fails to match completely.
What am I missing?
TIA
cià
ale
.*x(\d+)\. matches Name.Of.Chapter.021x212.
\.[A-Z]{3}.* matches .DOC.NAME-Some.stuff.Here.ext
But ([^.]*?) does not match The.Actual.Title.Of.the.Chapter because this regex does not allow for any periods at all.
since you are on Mac, you could use the shell
$ s="Name.Of.Chapter.021x212.The.Actual.Title.Of.the.Chapter.DOC.NAME-Some.stuff.Here.ext"
$ echo ${s#*x}
212.The.Actual.Title.Of.the.Chapter.DOC.NAME-Some.stuff.Here.ext
$ t=${s#*x}
$ echo ${t%.[A-Z][A-Z][A-Z].*}
212.The.Actual.Title.Of.the.Chapter
Or if you prefer sed, eg
echo $filename | sed 's|.[^x]*x||;s/\.[A-Z][A-Z][A-Z].*//'
For processing multiple files
for file in *.ext
do
newfile=${file#*x}
newfile=${newfile%.[A-Z][A-Z][A-Z].*}
# or
# newfile=$(echo $file | sed 's|.[^x]*x||;s/\.[A-Z][A-Z][A-Z].*//')
mv "$file" "$newfile"
done
To your question "How can I remove the dots in the process of matching?" the answer is "You can't." The only way to do that is by processing the result of the match in a second step, as others have said. But I think there's a more basic question that needs to be addressed, which is "What does it mean for a regex to match a given input?"
A regex is usually said to match a string when it describes any substring of that string. If you want to be sure the regex describes the whole string, you need to add the start (^) and end ($) anchors:
/^.*x(\d+)\.(.*?)\.[A-Z]{3}.*$/
But in your case, you don't need to describe the whole string; if you get rid of the .* at either end, it will serve your just as well:
/x(\d+)\.(.*?)\.[A-Z]{3}/
I recommend you not get in the habit of "padding" regexes with .* at beginning and end. The leading .* in particular can change the behavior of the regex in unexpected ways. For example, it there were two places in the input string where x(\d+)\. could match, your "real" match would have started at the second one. Also, if it's not anchored with ^ or \A, a leading .* can make the whole regex much less efficient.
I said "usually" above because some tools do automatically "anchor" the match at the beginning (Python's match()) or at both ends (Java's matches()), but that's pretty rare. Most of the shells and command-line tools available on *nix systems define a regex match in the traditional way, but it's a good idea to say what tool(s) you're using, just in case.
Finally, a word or two about vocabulary. The parentheses in (\d+) cause the matched characters to be captured, not grouped. Many regex flavors also support non-capturing parentheses in the form (?:\d+), which are used for grouping only. Any text that is included in the overall match, whether it's captured or not, is said to have been consumed (not captured). The way you used the words "capture" and "group" in your question is guaranteed to cause maximum confusion in anyone who assumes you know what you're talking about. :D
If you haven't read it yet, check out this excellent tutorial.
Regular Expressions are a complete void for me.
I'm dealing with one right now in TextMate that does what I want it to do...but I don't know WHY it does what I want it to do.
/[[:alpha:]]+|( )/(?1::$0)/g
This is used in a TextMate snippet and what it does is takes a Label and outputs it as an id name. So if I type "First Name" in the first spot, this outputs "FirstName".
Previously it looked like this:
/[[:alpha:]]+|( )/(?1:_:/L$0)/g (it might have been \L instead)
This would turn "First Name" into "first_name".
So I get that the underscore adds an underscore for a space, and that the /L lowercases everything...but I can't figure out what the rest of it does or why.
Someone care to explain it piece by piece?
EDIT
Here is the actual snippet in question:
<column header="$1"><xmod:field name="${2:${1/[[:alpha:]]+|( )/(?1::$0)/g}}"/></column>
This regular expression (regex) format is basically:
/matchthis/replacewiththis/settings
The "g" setting at the end means do a global replace, rather than just restricting the regex to a particular line or selection.
Breaking it down further...
[[:alpha:]]+|( )
That matches an alpha numeric character (held in parameter $0), or optionally a space (held in matching parameter $1).
(?1::$0)
As Roger says, the ? indicates this part is a conditional. If a match was found in parameter $1 then it is replaced with the stuff between the colons :: - in this case nothing. If nothing is in $1 then the match is replaced with the contents of $0, i.e. any alphanumeric character that is not a space is output unchanged.
This explains why the spaces are removed in the first example, and the spaces get replaced with underscores in your second example.
In the second expression the \L is used to lowercase the text.
The extra question in the comment was how to run this expression outside of TextMate. Using vi as an example, I would break it into multiple steps:
:0,$s/ //g
:0,$s/\u/\L\0/g
The first part of the above commands tells vi to run a substitution starting on line 0 and ending at the end of the file (that's what $ means).
The rest of the expression uses the same sorts of rules as explained above, although some of the notation in vi is a bit custom - see this reference webpage.
I find RegexBuddy a good tool for me in dealing with regexs. I pasted your 1st regex in to Buddy and I got the explanation shown in the bottom frame:
I use it for helping to understand existing regexs, building my own, testing regexs against strings, etc. I've become better # regexs because of it. FYI I'm running under Wine on Ubuntu.
it's searching for any alpha character that appears at least once in a row [[:alpha:]]+ or space ( ).
/[[:alpha:]]+|( )/(?1::$0)/g
The (?1 is a conditional and used to strip the match if group 1 (a single space) was matched, or replace the match with $0 if group 1 wasn't matched. As $0 is the entire match, it gets replaced with itself in that case. This regex is the same as:
/ //g
I.e. remove all spaces.
/[[:alpha:]]+|( )/(?1:_:/\L$0)/g
This regex is still using the same condition, except now if group 1 was matched, it's replaced with an underscore, and otherwise the full match ($0) is used, modified by \L. \L changes the case of all text that comes after it, so \LABC would result in abc; think of it as a special control code.