I'm trying to to a trim to some values using replaceregexp. Everything looks great when I try it in software like EditPad Pro.
Here's a sample of what I want to accomplish:
mf.version.impl = 2.01.00
mf.version.spec= 2.01.00
Notice the extra spaces after the last digit.
Then I'm using this pattern:
[0-9]+.[0-9]+.[0-9]+[ ]*
But it doesn't work in Netbeans.
Here's my ant command for it:
<!--If postfix is empty, remove the empty space-->
<replaceregexp file="../Xinco/nbproject/project.properties"
match="mf.version.spec?=?[0-9]+.[0-9]+.[0-9]+[ ]*"
replace="mf.version.spec = ${version_high}.${version_mid}.${version_low}"
byline="false"/>
<replaceregexp file="../Xinco/nbproject/project.properties"
match="mf.version.impl?=?[0-9]+.[0-9]+.[0-9]+[ ]*"
replace="mf.version.impl = ${version_high}.${version_mid}.${version_low}"
byline="true"/>
${version_high}.${version_mid}.${version_low} are variables already defined that correspond to 2.01.00 respectively.
It results in
mf.version.impl = 2.01.00
mf.version.spec = 2.01.00
Notice one extra space after the last digit.
I did debug the ant calls and it seems like the above command is not executing like a match didn't occur.
Any idea?
Since you don't care about the value, you don't have to match it explicitly. Try:
^mf\.version\.impl\s*=.*$
Meaning:
^ - start of the line (on multiline mode)
mf\.version\.impl - the string "mf.version.impl" literally, with the dots escaped.
\s* - zero or more spaces
.* - anything else (we can ignore the version, since you change it with a constant), all the way through to the...
$ - end of the line
Bonus track:
Looking at the specs, it looks like you can catch both lines with a single regex (not sure it works though):
^(mf\.version\.(impl|spec))\s*=.*$
and the replace rule:
replace="\1 = ${version_high}.${version_mid}.${version_low}"
This will replace \1 with the value it captured before, so again, you only need a single rule. (for trivia, usually $1 is used in replaces, but not here)
You should probably escape your .'s and use a capture group
e.g. (Perl regex for example)
s/([0-9]+\.[0-9]+\.[0-9]+)[ ]*/$1/
Related
I'm using Java. So I have a comma separated list of strings in this form:
aa,aab,aac
aab,aa,aac
aab,aac,aa
I want to use regex to remove aa and the trailing ',' if it is not the last string in the list. I need to end up with the following result in all 3 cases:
aab,aac
Currently I am using the following pattern:
"aa[,]?"
However it is returning:
b,c
If lookarounds are available, you can write:
,aa(?![^,])|(?<![^,])aa,
with an empty string as replacement.
demo
Otherwise, with a POSIX ERE syntax you can do it with a capture:
^(aa(,|$))+|(,aa)+(,|$)
with the 4th group as replacement (so $4 or \4)
demo
Without knowing your flavor, I propose this solution for the case that it does know the \b.
I use perl as demo environment and do a replace with "_" for demonstration.
perl -pe "s/\baa,|,aa\b/_/"
\b is the "word border" anchor. I.e. any start or end of something looking like a word. It allows to handle line end, line start, blank, comma.
Using it, two alternatives suffice to cover all the cases in your sample input.
Output (with interleaved input, with both, line ending in newline and line ending in blank):
aa,aab,aac
_aab,aac
aab,aa,aac
aab_,aac
aab,aac,aa
aab,aac_
aa,aab,aac
_aab,aac
aab,aa,aac
aab_,aac
aab,aac,aa
aab,aac_
If the \b is unknown in your regex engine, then please state which one you are using, i.e. which tool (e.g. perl, awk, notepad++, sed, ...). Also in that case it might be necessary to do replacing instead of deleting, i.e. to fine tune a "," or "" as replacement. For supporting that, please show the context of your regex, i.e. the replacing mechanism you are using. If you are deleting, then please switch to replacing beforehand.
(I picked up an input from comment by gisek, that the cpaturing groups are not needed. I usually use () generously, including in other syntaxes. In my opinion not having to think or look up evaluation orders is a benefit in total time and risks taken. But after testing, I use this terser/eleganter way.)
If your regex engine supports positive lookaheads and positive lookbehinds, this should work:
,aa(?=,)|(?<=,)aa,|(,|^)aa(,|$)
You could probably use the following and replace it by nothing :
(aa,|,aa$)
Either aa, when it's in the begin or the middle of a string
,aa$ when it's at the end of the string
Demo
As you want to delete aa followed by a coma or the end of the line, this should do the trick: ,aa(?=,|$)|^aa,
see online demo
I am trying to get version number of my assembly. I use regex for it and here is my pattern.
$pattern = '\[assembly: AssemblyVersion\("(.*)"\)\]'
It works good but in AssemblyInfo.cs and AssemblyInfo.vb there are some special characters as example
in cs file
// by using the '*' as shown below:
// [assembly: AssemblyVersion("1.0.*")]
[assembly: AssemblyVersion("3.7.1.0")]
in .vb file
' <Assembly: AssemblyVersion("1.0.*")>
<Assembly: AssemblyVersion("3.2.0.0")>
<Assembly: AssemblyFileVersion("1.0.0.0")>
So I want to exclude // and ' charachters in my pattern. I tried to exclude it with [^//] but it does not work. I tried something else but it did not work either.
And the second question is
in .vb file, there are different starting.
<Assembly: AssemblyVersion("3.2.0.0")>
and in c# file there are different starting
[assembly: AssemblyVersion("3.7.1.0")]
How i can include also vb version into my pattern?
Where is the problem?
You can use negative lookbehind if your library supports it
(?<!\/\/\s|\'\s)\[assembly: AssemblyVersion\("(.*)"\)\]
Edit:
For matching different brackets you can just use [variant1|variant2] syntax
(?<!\/\/\s|\'\s)[<\[][Aa]ssembly: AssemblyVersion\("(.*)"\)[>\]]
You want to "exclude" rows which start either with / or '.
Start from setting m (multi-line) flag in you regex.
It ensures that ^ matches start of each line (not the whole string).
Then start the regex from:
^ - acting now (in multi-line mode) as a row start marker,
(?!\/|') - a negative lookahead group with two variants
inside (do not allow that the line starts from either / or '),
\s* - an optional sequence of spaces.
and then goes your regex.
So the whole regex should look like below:
^(?!\/|')\s*\[assembly: AssemblyVersion\("(.*)"\)\]
(remember about the m flag).
Negative lookbehind solution mentioned in other answers has such a flaw
that even if a row starts from ' or / but has no space thereafter,
such a regex will fail.
I hava a string, like this:
{"content":(uint32)123", "id":(uint64)111, "test":{"hi":"(uint32)456"}}
I want to get result:
(uint32)123
(uint64)111
so I write regex like this:
[^(?!\")](\(uint32\)|\(uint64\))(\d)+[^(?!\")$]
but the result is:
:(uint32)123
:(uint64)111,
here the result adds : and ,
I hope that the regex does not begin with " and does not end with " , now I should how change my regex?
(\(uint(?:32|64)\)\d+) Works for me. It captures the entire string (uint[32/64])<any number of digits\> without bothering about the characters that come before or after.
Tested the following one in python
(?<!\")(\(uint32\)|\(uint64\))\d+(?!(\"|\d))
It looked like you was trying to use negative lookahead and negative lookbehind checks. But you did couple of mistakes:
You put them inside symbol group like this: [^(?!\")] what this regexp really mean - not any of symbols inside square bracket (^ - stands for not). How it should be instead: (?!\") - which mean symbol after current position shouldn't be quote (note: this will also work if there is no symbol after
To check symbol before you need to use look ahead check which have syntax (?<!some_regexp). So it would be (?<!\")
You don't need checks for start or end of the line. If you do you can put then into separate negative look ahead/behind statement.
Here is corrected example without line start/end checks:
(?<!\")(\(uint32\)|\(uint64\))(\d)+(?!\")(?!\d)
Note: you need to add (?!\d) at the end, cause otherwise it would match everything except last digit if there is quote.
Here is example with start/end of line checks:
(?<!^)(?<!\")(\(uint32\)|\(uint64\))(\d)+(?!\")(?!\d)(?!$)
P.S.: depending on language you using - you might not need to escape quote - you do need to escape quote only in case it is string escape sequence not regexp escape sequence.
I'm trying to use regex to find single quotes (so I can turn them all into double quotes) anywhere in a line that starts with mySqlQueryToArray (a function that makes a query to a SQL DB). I'm doing the regex in Sublime Text 3 which I'm pretty sure uses Perl Regex. I would like to have my regex match with every single quote in a line so for example I might have the line:
mySqlQueryToArray($con, "SELECT * FROM Template WHERE Name='$name'");
I want the regex to match in that line both of the quotes around $name but no other characters in that line. I've been trying to use (?<=mySqlQueryToArray.*)' but it tells me that the look behind assertion is invalid. I also tried (?<=mySqlQueryToArray)(?<=.*)' but that's also invalid. Can someone guide me to a regex that will accomplish what I need?
To find any number of single quotes in a line starting with your keyword you can use the \G anchor ("end of last match") by replacing:
(^\h*mySqlQueryToArray|(?!^)\G)([^\n\r']*)'
With \1\2<replacement>: see demo here.
Explanation
( ^\h*mySqlQueryToArray # beginning of line: check the keyword is here
| (?!^)\G ) # if not at the BOL, check we did match sth on this line
( [^\n\r']* ) ' # capture everything until the next single quote
The general idea is to match everything until the next single quote with ([^\n\r']*)' in order to replace it with \2<replacement>, but do so only if this everything is:
right after the beginning keyword (^mySqlQueryToArray), or
after the end of the last match ((?!^)\G): in that case we know we have the keyword and are on a relevant line.
\h* accounts for any started indent, as suggested by Xælias (\h being shortcut for any kind of horizontal whitespace).
https://stackoverflow.com/a/25331428/3933728 is a better answer.
I'm not good enough with RegEx nor ST to do this in one step. But I can do it in two:
1/ Search for all mySqlQueryToArray strings
Open the search panel: ⌘F or Find->Find...
Make sure you have the Regex (.* ) button selected (bottom left) and the wrap selector (all other should be off)
Search for: ^\s*mySqlQueryToArray.*$
^ beginning of line
\s* any indentation
mySqlQueryToArray your call
.* whatever is behind
$ end of line
Click on Find All
This will select every occurrence of what you want to modify.
2/ Enter the replace mode
⌥⌘F or Find->Replace...
This time, make sure that wrap, Regex AND In selection are active .
Them search for '([^']*)' and replace with "\1".
' are your single quotes
(...) si the capturing block, referenced by \1 in the replace field
[^']* is for any character that is not a single quote, repeated
Then hit Replace All
I know this is a little more complex that the other answer, but this one tackles cases where your line would contain several single-quoted string. Like this:
mySqlQueryToArray($con, "SELECT * FROM Template WHERE Name='$name' and Value='1234'");
If this is too much, I guess something like find: (?<=mySqlQueryToArray)(.*?)'([^']*)'(.*?) and replace it with \1"\2"\3 will be enough.
You can use a regex like this:
(mySqlQueryToArray.*?)'(.*?)'(.*)
Working demo
Check the substitution section.
You can use \K, see this regex:
mySqlQueryToArray[^']*\K'(.*?)'
Here is a regex demo.
I am processing a CSV file and want to search and replace strings as long as it is an exact match in the column. For example:
xxx,Apple,Green Apple,xxx,xxx
Apple,xxx,xxx,Apple,xxx
xxx,xxx,Fruit/Apple,xxx,Apple
I want to replace 'Apple' if it is the EXACT value in the column (if it is contained in text within another column, I do not want to replace). I cannot see how to do this with a single expression (maybe not possible?).
The desired output is:
xxx,GRAPE,Green Apple,xxx,xxx
GRAPE,xxx,xxx,GRAPE,xxx
xxx,xxx,Fruit/Apple,xxx,GRAPE
So the expression I want is: match the beginning of input OR a comma, followed by desired string, followed by a comma OR the end of input.
You cannot put ^ or $ in character classes, so I tried \A and \Z but that didn't work.
([\A,])Apple([\Z,])
This didn't work, sadly. Can I do this with one regular expression? Seems like this would be a common enough problem.
It will depend on your language, but if the one you use supports lookarounds, then you would use something like this:
(?<=,|^)Apple(?=,|$)
Replace with GRAPE.
Otherwise, you will have to put back the commas:
(^|,)Apple(,|$)
Or
(\A|,)Apple(,|\Z)
And replace with:
\1GRAPE\2
Or
$1GRAPE$2
Depending on what's supported.
The above are raw regex (and replacement) strings. Escape as necessary.
Note: The disadvatage with the latter solution is that it will not work on strings like:
xxx,Apple,Apple,xxx,xxx
Since the comma after the first Apple got consumed. You'd have to call the regex replacement at most twice if you have such cases.
Oh, and I forgot to mention, you can have some 'hybrids' since some language have different levels of support for lookbehinds (in all the below ^ and \A, $ and \Z, \1 and $1 are interchangeable, just so I don't make it longer than it already is):
(?:(?<=,)|(?<=^))Apple(?=,|$)
For those where lookbehinds cannot be of variable width, replace with GRAPE.
(^|,)Apple(?=,|$)
And the above one for where lookaheads are supported but not lookbehinds. Replace with \1Apple.
This does as you wish:
Find what: (^|,)(?:Apple)(,|$)
Replace with: $1GRAPE$2
This works on regex101, in all flavors.
http://regex101.com/r/iP6dZ8
I wanted to share my original work-around (before the other answers), though it feels like more of a hack.
I simply prepend and append a comma on the string before doing the simpler:
/,Apple,/,GRAPE,/g
then cut off the first and last character.
PHP looks like:
$line = substr(preg_replace($search, $replace, ','.$line.','), 1, -1);
This still suffers from the problem of consecutive columns (e.g. ",Apple,Apple,").