I want extract C-like comments from source, f.e. from
(updated example)
/**
* base comment
* (c) SOMEBODY SOMETIME
* something
*/
///<!-- ------metadata-XML------- -->
/// <module type="javascript"> A
///<desc> some desc
/// </desc>
(function( a /* param A */) { // programmers comment ... enclosure
/*! user doc
this module ....
* reguired
.....
*/
var b={}; // programmers in line comment
// single line comments
// The cookie spec says up to 4k per cookie, so at ~50 bytes per entry
// that gives a maximum of around 80 items as a max value for this field
b.a=a;
var str = " tttt \/\/this is not comment ! tttt "
var str2 = " tttt \/\* this is not comment too ! \
.............. \*\/ ttt ";
global.b = b;
}(global);
///</module>
regexp which I use is
^\s*\/\*(.*[\r\n]*)*\*\/
Problem is that this regexp stops (kills) regexp engine. RegexCouch becomes unresponsible,
using in browser causes unresponsible page.
What is wrong with this regexp ? How is possible, that regexp engine cannot solve it ?
Are there some regexp-es (syntactically correct, I think) which cannot be used ?
This is called Catastrophic Backtracking. Your regex has to check to many possibilities, because you are nesting quantifiers:
^\s*\/\*(.*[\r\n]*)*\*\/
^^ ^ ^
A better approach would be this:
/^\s*\/\*.*?\*\//gms
See it here in action.
You need the s option to make the . match the newline, the m option to make the ^ matches the start of he row.
.*? is matching as less characters as possible.
(/\*([^*]|[\r\n]|(\*+([^*/]|[\r\n])))*\*+/)|(//.*)
this will work for c-like comments match
if you use pcre-like regex you can use this:
\s*+\/\*(?>[^*]++|\*++(?!\/))*\*\/
if your regex flavor doesn't support atomic groups and possessive quantifiers, use this:
\s*\/\*(?:[^*]+|\*+(?!\/))*\*\/
Related
I am trying to delete a bunch of comments that are all in the following format:
/**
* #ngdoc
... comment body (delete me, too!)
*/
I have tried using this command: %s/\/**\n * #ngdoc.\{-}*\///g
Here is the regex without the patterns: %s/pattern1.\{-}pattern2//g
Here are the individual patterns: \/**\n * #ngdoc and *\/
When I try my pattern in vim I get the following error:
E871: (NFA regexp) Can't have a multi follow a multi !
E61: Nested *
E476: Invalid command
Thanks for any help with this regexp nightmare!
Instead of trying to cram this into one complex regex, it's much easier to search for the start of a comment and delete from there on to the end of a comment
:g/^\/\*\*$/,/\*\/$/d_
This breaks down into
:g start a global command
/^\/\*\*$/ search for start of a comment: <sol>/**<eol>
,/^\*\/$/ extend the range to the end of a comment: <sol>*/<eol>
d delete the range
_ use the black hole register (performance optimization)
Your problem is you have \{-} followed by * which are the multis referenced in the error message. Quote the *:
%s/\/\*\*\n \* #ngdoc\_.\{-}\*\/\n//g
Using embedded newlines in the pattern is the wrong approach. You should instead use an address range. Something like:
sed '\#^/\*\*$#,\#^\*/$#d' file
This will delete all lines starting from one that matches /** anchored at column 1 to the line matching */ anchored at column 1. If your comments are well behaved (eg, no trailing space after /**), this should do what you want.
Try this using gc to be careful when deleting
%s/\v\/\*\*\n\s\*\s\#ngdoc\n((\s*\n)?(\s\*.*\n)?){-}\s?\*\///gc
Match comments like
/**
* #ngdoc
* ... comment body (delete me, too!)
*
*/
My approached consists of using a macro:
qa/\/\*\*<enter><shift-v>/\*\/<enter>d
qa ........ starts recording macro "a"
/\/\*\* ... searches for the comment beginning
<Enter> ... use Ctrl-v Enter
V ......... starts visual block (until...)
/\*\/ ..... end of your comment
<Enter> ... Ctrl-v Enter agai
d ......... it will delete selected area
In order to isert etc presse followed by the keyword you want.
I need to check an input field for a German IBAN. The user should be allowed to leave in white spaces and input should be validated to have a starting DE and then exact 20 characters numbers and letters.
Without the white space allowance, I tried
^[DE]{2}([0-9a-zA-Z]{20})$
but I cannot find where and how I can add "white spaces anywhere allowed.
This should be simple, but I simply cannot find a solution.
Thanks for help!
Because you should use the right tool for the right task: you should not rely on regexps to validate IBAN numbers, but instead use the IBAN checksum algorithm to check the whole code is actually correct, making any regexp superfluous and redundant. i.e.: remove all spaces, rearrange the code, convert to integers, and compute remainder, here it's best explained.
Though, there am I trying to answer your question, for the fun of it:
what about:
^DE([0-9a-zA-Z]\s?){20}$
which only difference is allowing a whitespace (or not) after each occurence of a alphanumeric character.
here is the visualization:
edit: for the OP's information, the only difference is that this regexp, from #ulugbex-umirov: (?:\s*[0-9a-zA-Z]\s*) does a lookahead check to see if there's a space between the iso country code and the checksum (which only made of numerical digits), which I do not support on purpose.
And actually to support a correct IBAN syntax, which is formed of groups of 4 characters, as the wikipedia page says:
^DE\d{2}\s?([0-9a-zA-Z]{4}\s?){4}[0-9a-zA-Z]{2}$
example
If your UI is in Javascript, you can use that library for doing IBAN validation:
<script src="iban.js"></script>
<script>
// the API is now accessible from the window.IBAN global object
IBAN.isValid('hello world'); // false
IBAN.isValid('BE68539007547034'); // true
</script>
so you know this is a valid IBAN, and can validate it before the data is ever even sent to the backend. Simpler, lighter and more elegant… Why do something else?
Here is a list of IBANs from 70 Countries. I generated it with a python script i wrote based on this https://en.wikipedia.org/wiki/International_Bank_Account_Number
AL[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){2}([a-zA-Z0-9]{4}\s?){4}\s?
AD[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){2}([a-zA-Z0-9]{4}\s?){3}\s?
AT[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}\s?
AZ[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{4}\s?){1}([0-9]{4}\s?){5}\s?
BH[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([a-zA-Z0-9]{4}\s?){3}([a-zA-Z0-9]{2})\s?
BY[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{4}\s?){1}([0-9]{4}\s?){5}\s?
BE[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){3}\s?
BA[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}\s?
BR[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}([0-9]{3})([a-zA-Z]{1}\s?)([a-zA-Z0-9]{1})\s?
BG[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([0-9]{4}\s?){1}([0-9]{2})([a-zA-Z0-9]{2}\s?)([a-zA-Z0-9]{4}\s?){1}([a-zA-Z0-9]{2})\s?
CR[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}([0-9]{2})\s?
HR[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}([0-9]{1})\s?
CY[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){2}([a-zA-Z0-9]{4}\s?){4}\s?
CZ[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}\s?
DK[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){3}([0-9]{2})\s?
DO[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([0-9]{4}\s?){5}\s?
TL[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}([0-9]{3})\s?
EE[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}\s?
FO[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){3}([0-9]{2})\s?
FI[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){3}([0-9]{2})\s?
FR[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){2}([0-9]{2})([a-zA-Z0-9]{2}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{1})([0-9]{2})\s?
GE[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{2})([0-9]{2}\s?)([0-9]{4}\s?){3}([0-9]{2})\s?
DE[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}([0-9]{2})\s?
GI[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([a-zA-Z0-9]{4}\s?){3}([a-zA-Z0-9]{3})\s?
GR[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){1}([0-9]{3})([a-zA-Z0-9]{1}\s?)([a-zA-Z0-9]{4}\s?){3}([a-zA-Z0-9]{3})\s?
GL[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){3}([0-9]{2})\s?
GT[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{4}\s?){1}([a-zA-Z0-9]{4}\s?){5}\s?
HU[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){6}\s?
IS[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}([0-9]{2})\s?
IE[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{4}\s?){1}([0-9]{4}\s?){3}([0-9]{2})\s?
IL[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}([0-9]{3})\s?
IT[a-zA-Z0-9]{2}\s?([a-zA-Z]{1})([0-9]{3}\s?)([0-9]{4}\s?){1}([0-9]{3})([a-zA-Z0-9]{1}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{3})\s?
JO[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([0-9]{4}\s?){5}([0-9]{2})\s?
KZ[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){3}([0-9]{1})([a-zA-Z0-9]{3}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{2})\s?
XK[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){1}([0-9]{4}\s?){2}([0-9]{2})([0-9]{2}\s?)\s?
KW[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([a-zA-Z0-9]{4}\s?){5}([a-zA-Z0-9]{2})\s?
LV[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([a-zA-Z0-9]{4}\s?){3}([a-zA-Z0-9]{1})\s?
LB[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){1}([a-zA-Z0-9]{4}\s?){5}\s?
LI[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){1}([0-9]{1})([a-zA-Z0-9]{3}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{1})\s?
LT[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}\s?
LU[a-zA-Z0-9]{2}\s?([0-9]{3})([a-zA-Z0-9]{1}\s?)([a-zA-Z0-9]{4}\s?){3}\s?
MK[a-zA-Z0-9]{2}\s?([0-9]{3})([a-zA-Z0-9]{1}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{1})([0-9]{2})\s?
MT[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([0-9]{4}\s?){1}([0-9]{1})([a-zA-Z0-9]{3}\s?)([a-zA-Z0-9]{4}\s?){3}([a-zA-Z0-9]{3})\s?
MR[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}([0-9]{3})\s?
MU[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([0-9]{4}\s?){4}([0-9]{3})([a-zA-Z]{1}\s?)([a-zA-Z]{2})\s?
MC[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){2}([0-9]{2})([a-zA-Z0-9]{2}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{1})([0-9]{2})\s?
MD[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{2})([a-zA-Z0-9]{2}\s?)([a-zA-Z0-9]{4}\s?){4}\s?
ME[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}([0-9]{2})\s?
NL[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([0-9]{4}\s?){2}([0-9]{2})\s?
NO[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){2}([0-9]{3})\s?
PK[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{4}\s?){1}([0-9]{4}\s?){4}\s?
PS[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{4}\s?){1}([0-9]{4}\s?){5}([0-9]{1})\s?
PL[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){6}\s?
PT[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}([0-9]{1})\s?
QA[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([a-zA-Z0-9]{4}\s?){5}([a-zA-Z0-9]{1})\s?
RO[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([a-zA-Z0-9]{4}\s?){4}\s?
SM[a-zA-Z0-9]{2}\s?([a-zA-Z]{1})([0-9]{3}\s?)([0-9]{4}\s?){1}([0-9]{3})([a-zA-Z0-9]{1}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{3})\s?
SA[a-zA-Z0-9]{2}\s?([0-9]{2})([a-zA-Z0-9]{2}\s?)([a-zA-Z0-9]{4}\s?){4}\s?
RS[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){4}([0-9]{2})\s?
SK[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}\s?
SI[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){3}([0-9]{3})\s?
ES[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}\s?
SE[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}\s?
CH[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){1}([0-9]{1})([a-zA-Z0-9]{3}\s?)([a-zA-Z0-9]{4}\s?){2}([a-zA-Z0-9]{1})\s?
TN[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){5}\s?
TR[a-zA-Z0-9]{2}\s?([0-9]{4}\s?){1}([0-9]{1})([a-zA-Z0-9]{3}\s?)([a-zA-Z0-9]{4}\s?){3}([a-zA-Z0-9]{2})\s?
AE[a-zA-Z0-9]{2}\s?([0-9]{3})([0-9]{1}\s?)([0-9]{4}\s?){3}([0-9]{3})\s?
GB[a-zA-Z0-9]{2}\s?([a-zA-Z]{4}\s?){1}([0-9]{4}\s?){3}([0-9]{2})\s?
VA[a-zA-Z0-9]{2}\s?([0-9]{3})([0-9]{1}\s?)([0-9]{4}\s?){3}([0-9]{2})\s?
VG[a-zA-Z0-9]{2}\s?([a-zA-Z0-9]{4}\s?){1}([0-9]{4}\s?){4}\s?
Original:
^[DE]{2}([0-9a-zA-Z]{20})$
Debuggex Demo
Modified:
^DE(?:\s*[0-9a-zA-Z]\s*){20}$
Debuggex Demo
This is the correct regex to match DE IBAN account numbers:
DE\d{2}[ ]\d{4}[ ]\d{4}[ ]\d{4}[ ]\d{4}[ ]\d{2}|DE\d{20}
Pass: DE89 3704 0044 0532 0130 00|||DE89370400440532013000
Fail: DE89-3704-0044-0532-0130-00
Most simple solution I can think of:
^DE(\s*[[:alnum:]]){20}\s*$
In particular, your initial [DE]{2} is wrong, as it allows 'DD', 'EE', 'ED' as well as the intended 'DE'.
To allow any amount of spaces anywhere:
^ *D *E( *[A-Za-z0-9]){20} *$
As you want to allow lower letters, also DE might be lower?
^ *[Dd] *[Ee]( *[A-Za-z0-9]){20} *$
^ matches the start of the string
$ end anchor
in between each characters there are optional spaces *
[character class] defines a set/range of characters
To allow at most one space in between each characters, replace the quantifier * (any amount of) with ? (0 or 1). If supported, \s shorthand can be used to match [ \t\r\n\f] instead of space only.
Test on regex101.com, also see the SO regex FAQ
Using Google Apps Script, I pasted Laurent's code from github into a script and added the following code to test.
// Use the Apps Script IDE's "Run" menu to execute this code.
// Then look at the View > Logs menu to see execution results.
function myFunction() {
//https://github.com/arhs/iban.js/blob/master/README.md
// var IBAN = require('iban');
var t1 = IBAN.isValid('hello world'); // false
var t2 = IBAN.isValid('BE68539007547034'); // true
var t3 = IBAN.isValid('BE68 5390 0754 7034'); // true
Logger.log("Test 1 = %s", t1);
Logger.log("Test 2 = %s", t2);
Logger.log("Test 3 = %s", t3);
}
The only thing needed to run the example code was commenting out the require('iban') line:
// var IBAN = require('iban');
Finally, instead of using client handlers to attempt a RegEx validation of IBAN input, I use a a server handler to do the validation.
I have a list of label names in a text file I'd like to manipulate using Find and Replace in Notepad++, they are listed as follows:
MyLabel_01
MyLabel_02
MyLabel_03
MyLabel_04
MyLabel_05
MyLabel_06
I want to rename them in Notepad++ to the following:
Label_A_One
Label_A_Two
Label_A_Three
Label_B_One
Label_B_Two
Label_B_Three
The Regex I'm using in the Notepad++'s replace dialog to capture the label name is the following:
((MyLabel_0)((1)|(2)|(3)|(4)|(5)|(6)))
I want to replace each capture group as follows:
\1 = Label_
\2 = A_One
\3 = A_Two
\4 = A_Three
\5 = B_One
\6 = B_Two
\7 = B_Three
My problem is that Notepad++ doesn't register the syntax of the regex above. When I hit Count in the Replace Dialog, it returns with 0 occurrences. Not sure what's misesing in the syntax. And yes I made sure the Regular Expression radio button is selected. Help is appreciated.
UPDATE:
Tried escaping the parenthesis, still didn't work:
\(\(MyLabel_0\)\((1\)|\(2\)|\(3\)|\(4\)|\(5\)|\(6\)\)\)
Ed's response has shown a working pattern since alternation isn't supported in Notepad++, however the rest of your problem can't be handled by regex alone. What you're trying to do isn't possible with a regex find/replace approach. Your desired result involves logical conditions which can't be expressed in regex. All you can do with the replace method is re-arrange items and refer to the captured items, but you can't tell it to use "A" for values 1-3, and "B" for 4-6. Furthermore, you can't assign placeholders like that. They are really capture groups that you are backreferencing.
To reach the results you've shown you would need to write a small program that would allow you to check the captured values and perform the appropriate replacements.
EDIT: here's an example of how to achieve this in C#
var numToWordMap = new Dictionary<int, string>();
numToWordMap[1] = "A_One";
numToWordMap[2] = "A_Two";
numToWordMap[3] = "A_Three";
numToWordMap[4] = "B_One";
numToWordMap[5] = "B_Two";
numToWordMap[6] = "B_Three";
string pattern = #"\bMyLabel_(\d+)\b";
string filePath = #"C:\temp.txt";
string[] contents = File.ReadAllLines(filePath);
for (int i = 0; i < contents.Length; i++)
{
contents[i] = Regex.Replace(contents[i], pattern,
m =>
{
int num = int.Parse(m.Groups[1].Value);
if (numToWordMap.ContainsKey(num))
{
return "Label_" + numToWordMap[num];
}
// key not found, use original value
return m.Value;
});
}
File.WriteAllLines(filePath, contents);
You should be able to use this easily. Perhaps you can download LINQPad or Visual C# Express to do so.
If your files are too large this might be an inefficient approach, in which case you could use a StreamReader and StreamWriter to read from the original file and write it to another, respectively.
Also be aware that my sample code writes back to the original file. For testing purposes you can change that path to another file so it isn't overwritten.
Bar bar bar - Notepad++ thinks you're a barbarian.
(obsolete - see update below.) No vertical bars in Notepad++ regex - sorry. I forget every few months, too!
Use [123456] instead.
Update: Sorry, I didn't read carefully enough; on top of the barhopping problem, #Ahmad's spot-on - you can't do a mapping replacement like that.
Update: Version 6 of Notepad++ changed the regular expression engine to a Perl-compatible one, which supports "|". AFAICT, if you have a version 5., auto-update won't update to 6. - you have to explicitly download it.
A regular expression search and replace for
MyLabel_((01)|(02)|(03)|(04)|(05)|(06))
with
Label_(?2A_One)(?3A_Two)(?4A_Three)(?5B_One)(?6B_Two)(?7B_Three)
works on Notepad 6.3.2
The outermost pair of brackets is for grouping, they limit the scope of the first alternation; not sure whether they could be omitted but including them makes the scope clear. The pattern searches for a fixed string followed by one of the two-digit pairs. (The leading zero could be factored out and placed in the fixed string.) Each digit pair is wrapped in round brackets so it is captured.
In the replacement expression, the clause (?4A_Three) says that if capture group 4 matched something then insert the text A_Three, otherwise insert nothing. Similarly for the other clauses. As the 6 alternatives are mutually exclusive only one will match. Thus only one of the (?...) clauses will have matched and so only one will insert text.
The easiest way to do this that I would recommend is to use AWK. If you're on Windows, look for the mingw32 precompiled binaries out there for free download (it'll be called gawk).
BEGIN {
FS = "_0";
a[1]="A_One";
a[2]="A_Two";
a[3]="A_Three";
a[4]="B_One";
a[5]="B_Two";
a[6]="B_Three";
}
{
printf("Label_%s\n", a[$2]);
}
Execute on Windows as follows:
C:\Users\Mydir>gawk -f test.awk awk.in
Label_A_One
Label_A_Two
Label_A_Three
Label_B_One
Label_B_Two
Label_B_Three
i have: \£\d+\.\d\d
should find: £6.95 £16.95 etc
+ is one or more
\. is the dot
\d is for a digit
am i wrong? :(
JavaScript for Greasemonkey
// ==UserScript==
// #name CurConvertor
// #namespace CurConvertor
// #description noam smadja
// #include http://www.zavvi.com/*
// ==/UserScript==
textNodes = document.evaluate(
"//text()",
document,
null,
XPathResult.UNORDERED_NODE_SNAPSHOT_TYPE,
null);
var searchRE = /\£[0-9]\+.[0-9][0-9];
var replace = 'pling';
for (var i=0;i<textNodes.snapshotLength;i++) {
var node = textNodes.snapshotItem(i);
node.data = node.data.replace(searchRE, replace);
}
when i change the regex to /Free for example it finds and changes. but i guess i am missing something!
Had this written up for your last question just before it was deleted.
Here are the problems you're having with your GM script.
You're checking absolutely every
text node on the page for some
reason. This isn't causing it to
break but it's unnecessary and slow.
It would be better to look for text
nodes inside .price nodes and .rrp
.strike nodes instead.
When creating new regexp objects in
this way, backslashes must be
escaped, ex:
var searchRE = new
RegExp('\\d\\d','gi');
not
var
searchRE = new RegExp('\d\d','gi');
So you can add the backslashes, or
create your regex like this:
var
searchRE = /\d\d/gi;
Your actual regular expression is
only checking for numbers like
##ANYCHARACTER##, and will ignore £5.00 and £128.24
Your replacement needs to be either
a string or a callback function, not
a regular expression object.
Putting it all together
textNodes = document.evaluate(
"//p[contains(#class,'price')]/text() | //p[contains(#class,'rrp')]/span[contains(#class,'strike')]/text()",
document,
null,
XPathResult.UNORDERED_NODE_SNAPSHOT_TYPE,
null);
var searchRE = /£(\d+\.\d\d)/gi;
var replace = function(str,p1){return "₪" + ( (p1*5.67).toFixed(2) );}
for (var i=0,l=textNodes.snapshotLength;i<l;i++) {
var node = textNodes.snapshotItem(i);
node.data = node.data.replace(searchRE, replace);
}
Changes:
Xpath now includes only p.price and p.rrp span.strke nodes
Search regular expression created with /regex/ instead of new RegExp
Search variable now includes target currency symbol
Replace variable is now a function that replaces the currency symbol with a new symbol, and multiplies the first matched substring with substring * 5.67
for loop sets a variable to the snapshot length at the beginning of the loop, instead of checking textNodes.snapshotLength at the beginning of every loop.
Hope that helps!
[edit]Some of these points don't apply, as the original question changed a few times, but the final script is relevant, and the points may still be of interest to you for why your script was failing originally.
You are not wrong, but there are a few things to watch out for:
The £ sign is not a standard ASCII character so you may have encoding issue, or you may need to enable a unicode option on your regular expression.
The use of \d is not supported in all regular expression engines. [0-9] or [[:digit:]] are other possibilities.
To get a better answer, say which language you are using, and preferably also post your source code.
£[0-9]+(,[0-9]{3})*\.[0-9]{2}$
this will match anything from £dd.dd to £d[dd]*,ddd.dd. So it can fetch millions and hundreds as well.
The above regexp is not strict in terms of syntaxes. You can have, for example: 1123213123.23
Now, if you want an even strict regexp, and you're 100% sure that the prices will follow the comma and period syntaxes accordingly, then use
£[0-9]{1,3}(,[0-9]{3})*\.[0-9]{2}$
Try your regexps here to see what works for you and what not http://tools.netshiftmedia.com/regexlibrary/
It depends on what flavour of regex you are using - what is the programming language?
some older versions of regex require the + to be escaped - sed and vi for example.
Also some older versions of regex do not recognise \d as matching a digit.
Most modern regex follow the perl syntax and £\d+\.\d\d should do the trick, but it does also depend on how the £ is encoded - if the string you are matching encodes it differently from the regex then it will not match.
Here is an example in Python - the £ character is represented differently in a regular string and a unicode string (prefixed with a u):
>>> "£"
'\xc2\xa3'
>>> u"£"
u'\xa3'
>>> import re
>>> print re.match("£", u"£")
None
>>> print re.match(u"£", "£")
None
>>> print re.match(u"£", u"£")
<_sre.SRE_Match object at 0x7ef34de8>
>>> print re.match("£", "£")
<_sre.SRE_Match object at 0x7ef34e90>
>>>
£ isn't an ascii character, so you need to work out encodings. Depending on the language, you will either need to escape the byte(s) of £ in the regex, or convert all the strings into Unicode before applying the regex.
In Ruby you could just write the following
/£\d+.\d{2}/
Using the braces to specify number of digits after the point makes it slightly clearer
I've build a complex (for me) regex to parse some file names, and it broadly works, except for a case where there are additional inside brackets.
(?'field'F[0-9]{1,4})(?'term'\(.*?\))(?'operator'_(OR|NOT|AND)_)?
In the following examples, I need to get the groups after the comment, but in the 3rd example, I am getting ((brackets) instead of ((brackets)are valid).
For the life of me I can't work out how to extend it to search for the final bracket.
C:\Temp\[DB_3][DT_2][F30(green)].vsl // F30 (green)
C:\Temp\[DB_3][DT_2][F21(red)_OR_F21(blue)_NOT_F21(pink)].vsl // F21 (red) _OR_ OR
C:\Temp\[DB_3][DT_2][F21((brackets)are valid)].vsl // F21 ((brackets)are valid)
C:\Temp\[DB_3][DT_2][F21(any old brackets)))))are valid)].vsl // F21 (any old brackets)))))are valid)
C:\Temp\[DB_3][DT_2][F21(brackets))))))_OR_F21(blue)].vsl // F21 (brackets)))))) _OR_ OR
Thanks
UPDATE: I'm using RegExr to experiment, then implementing in C# like this:
Regex r = new Regex(pattern, RegexOptions.Multiline | RegexOptions.IgnorePatternWhitespace);
foreach(Match m in r.Matches(foo))
{
//etc
}
UPDATE 2: I don't need to match up the brackets. Inside the one set of brackets can be any data, I just need it to terminate with the outside bracket.
UPDATE 3:
Another attempt, this works with extra brackets (example 3 and 4), but still fails to split out the extra terms (example 5), but unfortunatly includes the terminating ] in the group. How can I get it to search for (but not include) either )_ or )] as the delimiter, but just include the bracket?
(?'field'F[0-9]{1,4})(?'term'\(.*?\)[\]])(?'operator'_(OR|NOT|AND)_)?
Final update: I've decided it's not worth the effort in trying to parse this stupid format, so I'm going to ditch support for it and do something more productive with my time. Thank you all for your help, I have now seen the light!
Matching nested parenthesis with regex is a) not possible*, or b) results in a regex that is unmaintainable.
If you're simply trying to match the first ( until the last ) (not checking if the opening- and closing-parenthesis properly match), then just remove the ? after .*?.
* depending what regex flavour you're using.
Hmm, this usually isn't possible with most regex engines. Although it is possible in perl:
PerlMonks
By using a recursive regexp:
use strict;
use warnings;
my $textInner =
'(outer(inner(most "this (shouldn\'t match)" inner)))';
my $innerRe;
my $idx=0;
my(#match);
$innerRe = qr/
\(
(
(?:
[^()"]+
|
"[^"]*"
|
(??{$innerRe})
)*
)
\)(?{$match[$idx++]=$1;})
/sx;
$textInner =~ /^$innerRe/g;
print "inner: $match[0]\n";
It's also possible to do it in most regex engines provided that you want to do it to a fixed depth of bracket nesting. I wrote something in java a while ago that would construct a regex that would match brackets up to 6 deep.
Here's my java function for producing the regex:
public static String generateParensMatchStr(int depth, char openParen, char closeParen)
{
if (depth == 0)
return ".*?";
else
return "(?:\\" + openParen + generateParensMatchStr(depth - 1, openParen, closeParen) + "\\" +closeParen + "|.*?)+?";
}
here is my another test results in python
x="""C:\Temp\[DB_3][DT_2][F30(green)].vsl // F30 (green)
C:\Temp\[DB_3][DT_2][F21(red)_OR_F21(blue)_NOT_F21(pink)].vsl // F21 (red) _OR_ OR
C:\Temp\[DB_3][DT_2][F21((brackets)are valid)].vsl // F21 ((brackets)are valid)
C:\Temp\[DB_3][DT_2][F21(any old brackets)))))are valid)].vsl // F21 (any old brackets)))))are valid)
C:\Temp\[DB_3][DT_2][F21(brackets))))))_OR_F21(blue)].vsl // F21 (brackets)))))) _OR_ OR"""
x=re.sub("//.*","",x)
x=re.sub("(_(OR|NOT|AND)_).*?]"," \\1 \\2]",x)
x=re.findall("(?:F[0-9]{1,4}\(.*\).*(?=]))",x)
for x in x:print x
this gives
F30(green)
F21(red) _OR_ OR
F21((brackets)are valid)
F21(any old brackets)))))are valid)
F21(brackets)))))) _OR_ OR
Thats will meet your expected result?
re.findall("((?:F[0-9]{1,4}\(.*\))(?:_(?:OR|NOT|AND)_)?)+?",YOURTEXT)
gots
['F30(green)', 'F21(red)_OR_F21(blue)_NOT_F21(pink)', 'F21((brackets)are valid)', 'F21(any old brackets)))))are valid)', 'F21(brackets))))))_OR_F21(blue)']
in python, what do you think?
Try this
/(F[0-9]{1,4})(\([^_\]]+\))(?:_(OR|NOT|AND)_)?/
tested with PHP, seems to give the expected results (as long as the strings inside round brackets don't contain _ or ]).