I am trying to tokenize a jsonpath string. For example, given a string like the following:
$.sensor.subsensor[0].foo[1][2].temp
I want ["$", "sensor", "subsensor", "0", "foo", "1", "2", "temp"]
I am terrible with regexs but I managed to come up with the following which according to regexr matches "." and "[" and "]". Assume the jsonpath string does not contain slices, wildcards, unions nor recursive descent.
[\.\[\]]
I am planning to match all "." and "[" and "]" chars and replace them with ";". Then i will split on ";".
The problem with the above regex is that I will get in certain instances ";;".
$;sensor;subsensor;0;;foo;1;;2;;temp
Is there a way I can in a single regex replace ".", "[", "]" as well as ".[" and "]." or "][" with ";"? Do I need to check for these groups explicitly or do I need to run the sequence through 2 regexs?
No needs to transform .[] into ;, just split directly:
console.log('$.sensor.subsensor[0].foo[1][2].temp'.split(/[.[\]]+/));
You can use this code to omit double semicolon:
console.log(
'$.sensor.subsensor[0].foo[1][2].temp'.replace(/[\.\[\]]+/g, ';')
)
Got a decent solution.
/(\.|\].|\[)/g
Apparently when you use [] as part of your regex that matches only a single character, which is why groups like "]." become ";;". Using () allows you to specify character groups, and the above group just enumerates the possibilities.
Related
my use case is the follow:
String text_string: "text1:message1,text3:message3,text2:message,..."
select regexp_replace(text_string, '[^:]*:([^,]*(,|$))', '$1')
Correct output: message1,message3,message2,...
The pattern work, but the problem is that if there is a character ":" o "," in the message the replace doesn't work.
So I tried to use "::" and ",," characters as a separators in the string
String text_string: "text1::message1,,text3::message3,,text2::message2,..."
select regexp_replace(text_string, '[^::]*::([^,,]*(,,|$))', '$1')
Correct output: message1,,message3,,message2,,...
but also in this case, if there is one ":" or "," character in the string (in the text or in the message) the replace command doesn't work.
How should the regular expression be modified to work?
Delimiters cannot be characters that are likely to be in the data. Since you have control over it, use pipes '|' or tildes '~' maybe. Only you can come up with the right characters by analyzing the data.
If you can't do that, then you'll need to put quotes around the data that contains the delimiter character and come up with a way to deal with that.
I need to extract all matches from a huge text that start with [" and end with "]. These special characters separate each record from database. I need to extract all records.
Inside this record there are letters, numbers and special characters like -, ., &, (), /, {space} or so.
I'm writing this in Office VBA.
The pattern I have come so far looks like this: .Pattern = "[[][""][a-z|A-Z|w|W]*".
With this pattern, I am able to extract the first word from each record, with the starting characters [". The count of found matches is correct.
Example of one record:
["blabla","blabla","blabla","\u00e1no","nie","\u00e1no","\u00e1no","\u00e1no","\u003Ca class=\u0022btn btn-default\u0022 href=\u0022\u0026#x2F;siea\u0026#x2F;suppliers\u0026#x2F;42\u0022\u003E\u003Ci class=\u0022fa fa-pencil\u0022\u003E\u003C\/i\u003E Upravi\u0165\u003C\/a\u003E \u003Ca class=\u0022btn btn-default\u0022 href=\u0022\u0026#x2F;siea\u0026#x2F;suppliers\u0026#x2F;form\u0026#x2F;42\u0022\u003E\u003Ci class=\u0022fa fa-file-pdf-o\u0022\u003E\u003C\/i\u003E Zmluva\u003C\/a\u003E \u003Ca class=\u0022btn btn-default\u0022 href=\u0022\u0026#x2F;siea\u0026#x2F;suppliers\u0026#x2F;crz-form\u0026#x2F;42\u0022\u003E\u003Ci class=\u0022fa fa-file-pdf-o\u0022\u003E\u003C\/i\u003E Zmluva CRZ\u003C\/a\u003E"]
The question is : How can I extract the all records starting with [" and ending with "]?
I don't necessary need the starting and ending characters, but I can clean that up later.
Thanks for help.
The easiest way is to get rid of the initial and trailing [" and "] with either Replace or Left/Right/Mid functions, and then Split with "," (in VBA, """,""").
E.g.
input = "YOUR_STRING"
input = Replace(Replace(input, """]", ""), "[""", "")
result = Split(input, """,""")
If you plan to use Regex, you can use \["[\s\S]*?"] pattern, but it is not that efficient with long inputs and may even freeze the macro if timeout issue occurs. You can unroll it as
\["[^"]*(?:"(?!])[^"]*)*"]
See the regex demo. In VBA, Pattern = "\[""[^""]*(?:""(?!])[^""]*)*""]"
Note that with this unrolled pattern, you do not even need to use the workarounds for dot matching newline issue (negated character class [^"] matches any char but ", including a newline).
Pattern details:
\[" - [" literally
[^"]* - zero or more characters other than "
(?:"(?!])[^"]*)* - zero or more sequences of
"(?!]) - " not followed with ]
[^"]* - zero or more characters other than "
"] - literal character sequence "]
I need to hide part of the string. Hide all before some ending part.
It easy to implement by regexp like this:
replace("123-134-04", ".(?=.*-)", " ")
replace any symbol if future part of string contains "-".
So result is: " -04"
It is important to keep spaces.
But, I can't use lookahead or lookbehind.
I can catch the group before ending part, but how to replace this for right number of spaces?
Or maybe some other ways to resolve this with regex?
Tnanks in advance!
If the number of to be replaced characters does not differ too much, and you have a means to match the part to be preserved, you could run through a series of search and replace:
replace("12-14-04", "^.{5}(-[^-]+)$", " \1")
replace("123-134-04", "^.{7}(-[^-]+)$", " \1")
replace("adfasd-adf-da7474-04", "^.{17}(-[^-]+)$", " \1")
Or you do:
split the string at the position, where the to be preserved part begins,
run the replace("ALL OF THIS SHOULD BECOME BLANKS", ".", " ") on the first part, and
join them up again.
Beginner RegExp question. I have lines of JSON in a textfile, each with slightly different Fields, but there are 3 fields I want to extract for each line if it has it, ignoring everything else. How would I use a regex (in editpad or anywhere else) to do this?
Example:
"url":"http://www.netcharles.com/orwell/essays.htm",
"domain":"netcharles.com",
"title":"Orwell Essays & Journalism Section - Charles' George Orwell Links",
"tags":["orwell","writing","literature","journalism","essays","politics","essay","reference","language","toread"],
"index":2931,
"time_created":1345419323,
"num_saves":24
I want to extract URL,TITLE,TAGS,
/"(url|title|tags)":"((\\"|[^"])*)"/i
I think this is what you're asking for. I'll provide an explanation momentarily. This regular expression (delimited by / - you probably won't have to put those in editpad) matches:
"
A literal ".
(url|title|tags)
Any of the three literal strings "url", "title" or "tags" - in Regular Expressions, by default Parentheses are used to create groups, and the pipe character is used to alternate - like a logical 'or'. To match these literal characters, you'd have to escape them.
":"
Another literal string.
(
The beginning of another group. (Group 2)
(
Another group (3)
\\"
The literal string \" - you have to escape the backslash because otherwise it will be interpreted as escaping the next character, and you never know what that'll do.
|
or...
[^"]
Any single character except a double quote The brackets denote a Character Class/Set, or a list of characters to match. Any given class matches exactly one character in the string. Using a carat (^) at the beginning of a class negates it, causing the matcher to match anything that's not contained in the class.
)
End of group 3...
*
The asterisk causes the previous regular expression (in this case, group 3), to be repeated zero or more times, In this case causing the matcher to match anything that could be inside the double quotes of a JSON string.
)"
The end of group 2, and a literal ".
I've done a few non-obvious things here, that may come in handy:
Group 2 - when dereferenced using Backreferences - will be the actual string assigned to the field. This is useful when getting the actual value.
The i at the end of the expression makes it case insensitive.
Group 1 contains the name of the captured field.
EDIT: So I see that the tags are an array. I'll update the regular expression here in a second when I've had a chance to think about it.
Your new Regex is:
/"(url|title|tags)":("(\\"|[^"])*"|\[("(\\"|[^"])*"(,"(\\"|[^"])*")*)?\])/i
All I've done here is alternate the string regular expression I had been using ("((\\"|[^"])*)"), with a regular expression for finding arrays (\[("(\\"|[^"])*"(,"(\\"|[^"])*")*)?\]). No so easy to Read, is it? Well, substituting our String Regex out for the letter S, we can rewrite it as:
\[(S(,S)*)?\]
Which matches a literal opening bracket (hence the backslashes), optionally followed by a comma separated list of strings, and a closing bracket. The only new concept I've introduced here is the question mark (?), which is itself a type of repetition. Commonly referred to as 'making the previous expression optional', it can also be thought of as exactly 0 or 1 matches.
With our same S Notation, here's the whole dirty Regular Expression:
/"(url|title|tags)":(S|\[(S(,S)*)?\])/i
If it helps to see it in action, here's a view of it in action.
This question is a bit older, but I have had browsed a bit on my PC and found that expression. I passed him as GIST, could be useful to others.
EDIT:
# Expression was tested with PHP and Ruby
# This regular expression finds a key-value pair in JSON formatted strings
# Match 1: Key
# Match 2: Value
# https://regex101.com/r/zR2vU9/4
# http://rubular.com/r/KpF3suIL10
(?:\"|\')(?<key>[^"]*)(?:\"|\')(?=:)(?:\:\s*)(?:\"|\')?(?<value>true|false|[0-9a-zA-Z\+\-\,\.\$]*)
# test document
[
{
"_id": "56af331efbeca6240c61b2ca",
"index": 120000,
"guid": "bedb2018-c017-429E-b520-696ea3666692",
"isActive": false,
"balance": "$2,202,350",
"object": {
"name": "am",
"lastname": "lang"
}
}
]
the json string you'd like to extract field value from
{"fid":"321","otherAttribute":"value"}
the following regex expression extract exactly the "fid" field value "321"
(?<=\"fid\":\")[^\"]*
Please try below expression:
/"(url|title|tags)":("([^""]+)"|\[[^[]+])/gm
Explanation:
1st Capturing Group (url|title|tags): This is alternatively capturing the characters 'url','title' and 'tags' literally (case sensitive).
2nd Capturing Group ("([^""]+)"|[[^[]+]):
1st Alternative "([^""]+)" is matches all words within " and " including " and "
2nd Alternative [[^[]+] is matches all words within [ and ] including [ and ]
I have tested here
I adapted regex to work with JSON in my own library. I've detailed algorithm behavior below.
First, stringify the JSON object. Then, you need to store the starts and lengths of the matched substrings. For example:
"matched".search("ch") // yields 3
For a JSON string, this works exactly the same (unless you are searching explicitly for commas and curly brackets in which case I'd recommend some prior transform of your JSON object before performing regex (i.e. think :, {, }).
Next, you need to reconstruct the JSON object. The algorithm I authored does this by detecting JSON syntax by recursively going backwards from the match index. For instance, the pseudo code might look as follows:
find the next key preceding the match index, call this theKey
then find the number of all occurrences of this key preceding theKey, call this theNumber
using the number of occurrences of all keys with same name as theKey up to position of theKey, traverse the object until keys named theKey has been discovered theNumber times
return this object called parentChain
With this information, it is possible to use regex to filter a JSON object to return the key, the value, and the parent object chain.
You can see the library and code I authored at http://json.spiritway.co/
if your json is
{"key1":"abc","key2":"xyz"}
then below regex will extract key1 or key2 based on a key that you pass in regex
"key2(.*?)(?=,|}|$)
you can verify it here - regex101.com
Why does it have to be a Regular Expression object?
Here we can just use a Hash object first and then go search it.
mh = {"url":"http://www.netcharles.com/orwell/essays.htm","domain":"netcharles.com","title":"Orwell Essays & Journalism Section - Charles' George Orwell Links","tags":["orwell","writing","literature","journalism","essays","politics","essay","reference","language","toread"],"index":2931,"time_created":1345419323,"num_saves":24}
The output of which would be
=> {:url=>"http://www.netcharles.com/orwell/essays.htm", :domain=>"netcharles.com", :title=>"Orwell Essays & Journalism Section - Charles' George Orwell Links", :tags=>["orwell", "writing", "literature", "journalism", "essays", "politics", "essay", "reference", "language", "toread"], :index=>2931, :time_created=>1345419323, :num_saves=>24}
Not that I want to avoid using Regexp but don't you think it would be easier to take it a step at a time until your getting the data you want to further search through? Just MHO.
mh.values_at(:url, :title, :tags)
The output:
["http://www.netcharles.com/orwell/essays.htm", "Orwell Essays & Journalism Section - Charles' George Orwell Links", ["orwell", "writing", "literature", "journalism", "essays", "politics", "essay", "reference", "language", "toread"]]
Taking the pattern that FrankieTheKneeman gave you:
pattern = /"(url|title|tags)":"((\\"|[^"])*)"/i
we can search the mh hash by converting it to a json object.
/#{pattern}/.match(mh.to_json)
The output:
=> #<MatchData "\"url\":\"http://www.netcharles.com/orwell/essays.htm\"" 1:"url" 2:"http://www.netcharles.com/orwell/essays.htm" 3:"m">
Of course this is all done in Ruby which is not a tag that you have but relates I hope.
But oops! Looks like we can't do all three at once with that pattern so I will do them one at a time just for sake.
pattern = /"(title)":"((\\"|[^"])*)"/i
/#{pattern}/.match(mh.to_json)
#<MatchData "\"title\":\"Orwell Essays & Journalism Section - Charles' George Orwell Links\"" 1:"title" 2:"Orwell Essays & Journalism Section - Charles' George Orwell Links" 3:"s">
pattern = /"(tags)":"((\\"|[^"])*)"/i
/#{pattern}/.match(mh.to_json)
=> nil
Sorry about that last one. It will have to be handled differently.
How do I get the substring " It's big \"problem " using a regular expression?
s = ' function(){ return " It\'s big \"problem "; }';
/"(?:[^"\\]|\\.)*"/
Works in The Regex Coach and PCRE Workbench.
Example of test in JavaScript:
var s = ' function(){ return " Is big \\"problem\\", \\no? "; }';
var m = s.match(/"(?:[^"\\]|\\.)*"/);
if (m != null)
alert(m);
This one comes from nanorc.sample available in many linux distros. It is used for syntax highlighting of C style strings
\"(\\.|[^\"])*\"
As provided by ePharaoh, the answer is
/"([^"\\]*(\\.[^"\\]*)*)"/
To have the above apply to either single quoted or double quoted strings, use
/"([^"\\]*(\\.[^"\\]*)*)"|\'([^\'\\]*(\\.[^\'\\]*)*)\'/
Most of the solutions provided here use alternative repetition paths i.e. (A|B)*.
You may encounter stack overflows on large inputs since some pattern compiler implements this using recursion.
Java for instance: http://bugs.java.com/bugdatabase/view_bug.do?bug_id=6337993
Something like this:
"(?:[^"\\]*(?:\\.)?)*", or the one provided by Guy Bedford will reduce the amount of parsing steps avoiding most stack overflows.
/(["\']).*?(?<!\\)(\\\\)*\1/is
should work with any quoted string
"(?:\\"|.)*?"
Alternating the \" and the . passes over escaped quotes while the lazy quantifier *? ensures that you don't go past the end of the quoted string. Works with .NET Framework RE classes
/"(?:[^"\\]++|\\.)*+"/
Taken straight from man perlre on a Linux system with Perl 5.22.0 installed.
As an optimization, this regex uses the 'posessive' form of both + and * to prevent backtracking, for it is known beforehand that a string without a closing quote wouldn't match in any case.
This one works perfect on PCRE and does not fall with StackOverflow.
"(.*?[^\\])??((\\\\)+)?+"
Explanation:
Every quoted string starts with Char: " ;
It may contain any number of any characters: .*? {Lazy match}; ending with non escape character [^\\];
Statement (2) is Lazy(!) optional because string can be empty(""). So: (.*?[^\\])??
Finally, every quoted string ends with Char("), but it can be preceded with even number of escape sign pairs (\\\\)+; and it is Greedy(!) optional: ((\\\\)+)?+ {Greedy matching}, bacause string can be empty or without ending pairs!
An option that has not been touched on before is:
Reverse the string.
Perform the matching on the reversed string.
Re-reverse the matched strings.
This has the added bonus of being able to correctly match escaped open tags.
Lets say you had the following string; String \"this "should" NOT match\" and "this \"should\" match"
Here, \"this "should" NOT match\" should not be matched and "should" should be.
On top of that this \"should\" match should be matched and \"should\" should not.
First an example.
// The input string.
const myString = 'String \\"this "should" NOT match\\" and "this \\"should\\" match"';
// The RegExp.
const regExp = new RegExp(
// Match close
'([\'"])(?!(?:[\\\\]{2})*[\\\\](?![\\\\]))' +
'((?:' +
// Match escaped close quote
'(?:\\1(?=(?:[\\\\]{2})*[\\\\](?![\\\\])))|' +
// Match everything thats not the close quote
'(?:(?!\\1).)' +
'){0,})' +
// Match open
'(\\1)(?!(?:[\\\\]{2})*[\\\\](?![\\\\]))',
'g'
);
// Reverse the matched strings.
matches = myString
// Reverse the string.
.split('').reverse().join('')
// '"hctam "\dluohs"\ siht" dna "\hctam TON "dluohs" siht"\ gnirtS'
// Match the quoted
.match(regExp)
// ['"hctam "\dluohs"\ siht"', '"dluohs"']
// Reverse the matches
.map(x => x.split('').reverse().join(''))
// ['"this \"should\" match"', '"should"']
// Re order the matches
.reverse();
// ['"should"', '"this \"should\" match"']
Okay, now to explain the RegExp.
This is the regexp can be easily broken into three pieces. As follows:
# Part 1
(['"]) # Match a closing quotation mark " or '
(?! # As long as it's not followed by
(?:[\\]{2})* # A pair of escape characters
[\\] # and a single escape
(?![\\]) # As long as that's not followed by an escape
)
# Part 2
((?: # Match inside the quotes
(?: # Match option 1:
\1 # Match the closing quote
(?= # As long as it's followed by
(?:\\\\)* # A pair of escape characters
\\ #
(?![\\]) # As long as that's not followed by an escape
) # and a single escape
)| # OR
(?: # Match option 2:
(?!\1). # Any character that isn't the closing quote
)
)*) # Match the group 0 or more times
# Part 3
(\1) # Match an open quotation mark that is the same as the closing one
(?! # As long as it's not followed by
(?:[\\]{2})* # A pair of escape characters
[\\] # and a single escape
(?![\\]) # As long as that's not followed by an escape
)
This is probably a lot clearer in image form: generated using Jex's Regulex
Image on github (JavaScript Regular Expression Visualizer.)
Sorry, I don't have a high enough reputation to include images, so, it's just a link for now.
Here is a gist of an example function using this concept that's a little more advanced: https://gist.github.com/scagood/bd99371c072d49a4fee29d193252f5fc#file-matchquotes-js
here is one that work with both " and ' and you easily add others at the start.
("|')(?:\\\1|[^\1])*?\1
it uses the backreference (\1) match exactley what is in the first group (" or ').
http://www.regular-expressions.info/backref.html
One has to remember that regexps aren't a silver bullet for everything string-y. Some stuff are simpler to do with a cursor and linear, manual, seeking. A CFL would do the trick pretty trivially, but there aren't many CFL implementations (afaik).
A more extensive version of https://stackoverflow.com/a/10786066/1794894
/"([^"\\]{50,}(\\.[^"\\]*)*)"|\'[^\'\\]{50,}(\\.[^\'\\]*)*\'|“[^”\\]{50,}(\\.[^“\\]*)*”/
This version also contains
Minimum quote length of 50
Extra type of quotes (open “ and close ”)
If it is searched from the beginning, maybe this can work?
\"((\\\")|[^\\])*\"
I faced a similar problem trying to remove quoted strings that may interfere with parsing of some files.
I ended up with a two-step solution that beats any convoluted regex you can come up with:
line = line.replace("\\\"","\'"); // Replace escaped quotes with something easier to handle
line = line.replaceAll("\"([^\"]*)\"","\"x\""); // Simple is beautiful
Easier to read and probably more efficient.
If your IDE is IntelliJ Idea, you can forget all these headaches and store your regex into a String variable and as you copy-paste it inside the double-quote it will automatically change to a regex acceptable format.
example in Java:
String s = "\"en_usa\":[^\\,\\}]+";
now you can use this variable in your regexp or anywhere.
(?<="|')(?:[^"\\]|\\.)*(?="|')
" It\'s big \"problem "
match result:
It\'s big \"problem
("|')(?:[^"\\]|\\.)*("|')
" It\'s big \"problem "
match result:
" It\'s big \"problem "
Messed around at regexpal and ended up with this regex: (Don't ask me how it works, I barely understand even tho I wrote it lol)
"(([^"\\]?(\\\\)?)|(\\")+)+"