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
Related
I made a regex which inserts a space where ever there is any of the characters
-:\*_/;, present for example JET*AIRWAYS\INDIA/858701/IDBI 05/05/05;05:05:05 a/c should beJET* AIRWAYS\ INDIA/ 858701/ IDBI 05/05/05; 05:05:05 a/c
The regex I used is (?!a\/c|w\/d|m\/s|s\/w|m\/o)(\D-|\D:|\D\*|\D_|\D\\|\D\/|\D\;)
I have added some words exceptions like a/c w/d etc. \D conditions given to avoid date/time values getting separated, but this created an issue, the numbers followed by the above mentioned characters never get split.
My requirement is
1. Insert a space after characters -:\*_/;,
2. but date and time should not get split which may have / :
3. need exception on words like a/c w/d
The following is the full code
Private Function formatColon(oldString As String) As String
Dim reg As New RegExp: reg.Global = True: reg.Pattern = "(?!a\/c|w\/d|m\/s|s\/w|m\/o)(\D-|\D:|\D\*|\D_|\D\\|\D\/|\D\;)" '"(\D:|\D/|\D-|^w/d)"
Dim newString As String: newString = reg.Replace(oldString, "$1 ")
formatColon = XtraspaceKill(newString)
End Function
I would use 3 replacements.
Replace all date and time special characters with a special macro that should never be found in your text, e.g. for 05/15/2018 4:06 PM, something based on your name:
05MANUMOHANSLASH15MANUMOHANSLASH2018 4MANUMOHANCOLON06 PM
You can encode exceptions too, like this:
aMANUMOHANSLASHc
Now run your original regex to replace all special characters.
Finally, unreplace the macros MANUMOHANSLASH and MANUMOHANCOLON.
Meanwhile, let me tell you why this is complicated in a single regex.
If trying to do this in a single regex, you have to ask, for each / or :, "Am I a part of a date or time?"
To answer that, you need to use lookahead and lookbehind assertions, the latter of which Microsoft has finally added support for.
But given a /, you don't know if you're between the first and second, or second and third parts of the date. Similar for time.
The number of cases you need to consider will render your regex unmaintainably complex.
So please just use a few separate replacements :-)
I have string like
a;b;"aaa;;;bccc";deef
I want to split string based on delimiter ; only if ; is not inside double quotes. So after the split, it will be
a
b
"aaa;;;bccc"
deef
I tried using look-behind, but I'm not able to find a correct regular expression for splitting.
Regular expressions are probably not the right tool for this. If possible you should use a CSV library, specify ; as the delimiter and " as the quote character, this should give you the exact fields you are looking for.
That being said here is one approach that works by ensuring that there are an even number of quotation marks between the ; we are considering the split at and the end of the string.
;(?=(([^"]*"){2})*[^"]*$)
Example: http://www.rubular.com/r/RyLQyR8F19
This will break down if you can have escaped quotation marks within a string, for example a;"foo\"bar";c.
Here is a much cleaner example using Python's csv module:
import csv, StringIO
reader = csv.reader(StringIO.StringIO('a;b;"aaa;;;bccc";deef'),
delimiter=';', quotechar='"')
for row in reader:
print '\n'.join(row)
Regular expression will only get messier and break on even minor changes. You are better off using a csv parser with any scripting language. Perl built in module (so you don't need to download from CPAN if there are any restrictions) called Text::ParseWords allows you to specify the delimiter so that you are not limited to ,. Here is a sample snippet:
#!/usr/local/bin/perl
use strict;
use warnings;
use Text::ParseWords;
my $string = 'a;b;"aaa;;;bccc";deef';
my #ary = parse_line(q{;}, 0, $string);
print "$_\n" for #ary;
Output
a
b
aaa;;;bccc
deef
This is kind of ugly, but if you don't have \" inside your quoted strings (meaning you don't have strings that look like this ("foo bar \"badoo\" goo") you can split on the " first and then assume that all your even numbered array elements are, in fact, strings (and split the odd numbered elements into their component parts on the ; token).
If you *do have \" in your strings, then you'll want to first convert those into some other temporary token that you'll convert back later after you've performed your operation.
Here's a fiddle...
http://jsfiddle.net/VW9an/
var str = 'abc;def;ghi"some other dogs say \\"bow; wow; wow\\". yes they do!"and another; and a fifth'
var strCp = str.replace(/\\"/g,"--##--");
var parts = strCp.split(/"/);
var allPieces = new Array();
for(var i in parts){
if(i % 2 == 0){
var innerParts = parts[i].split(/\;/)
for(var j in innerParts)
allPieces.push(innerParts[j])
}
else{
allPieces.push('"' + parts[i] +'"')
}
}
for(var a in allPieces){
allPieces[a] = allPieces[a].replace(/--##--/g,'\\"');
}
console.log(allPieces)
Match All instead of Splitting
Answering long after the battle because no one used the way that seems the simplest to me.
Once you understand that Match All and Split are Two Sides of the Same Coin, you can use this simple regex:
"[^"]*"|[^";]+
See the matches in the Regex Demo.
The left side of the alternation | matches full quoted strings
The right side matches any chars that are neither ; nor "
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 to parse various strings and determine a prefix, number, and suffix. The problem is the strings can come in a wide variety of formats. The best way for me to think about how to parse it is to find the longest number in the string, then take everything before that as a prefix and everything after that as a suffix.
Some examples:
0001 - No prefix, Number = 0001, No suffix
1-0001 - Prefix = 1-, Number = 0001, No suffix
AAA001 - Prefix = AAA, Number = 001, No suffix
AAA 001.01 - Prefix = AAA , Number = 001, Suffix = .01
1_00001-01 - Prefix = 1_, Number = 00001, Suffix = -01
123AAA 001_01 - Prefix = 123AAA , Number = 001, Suffix = _01
The strings can come with any mixture of prefixes and suffixes, but the key point is the Number portion is always the longest sequential list of digits.
I've tried a variety of RegEx's that work with most but not all of these examples. I might be missing something, or perhaps a RegEx isn't the right way to go in this case?
(The RegEx should be .NET compatible)
UPDATE: For those that are interested, here's the C# code I came up with:
var regex = new System.Text.RegularExpressions.Regex(#"(\d+)");
if (regex.IsMatch(m_Key)) {
string value = "";
int length;
var matches = regex.Matches(m_Key);
foreach (var match in matches) {
if (match.Length >= length) {
value = match.Value;
length = match.Length;
}
}
var split = m_Key.Split(new String[] {value}, System.StringSplitOptions.RemoveEmptyEntries);
m_KeyCounter = value;
if (split.Length >= 1) m_KeyPrefix = split(0);
if (split.Length >= 2) m_KeySuffix = split(1);
}
You're right, this problem can't be solved purely by regular expressions. You can use regexes to "tokenize" (lexically analyze) the input but after that you'll need further processing (parsing).
So in this case I would tokenize the input with (for example) a simple regular expression search (\d+) and then process the tokens (parse). That would involve seeing if the current token is longer than the tokens seen before it.
To gain more understanding of the class of problems regular expressions "solve" and when parsing is needed, you might want to check out general compiler theory, specifically when regexes are used in the construction of a compiler (e.g. http://en.wikipedia.org/wiki/Book:Compiler_construction).
You're input isn't regular so, a regex won't do. I would iterate over the all groups of digits via (\d+) and find the longest and then build a new regex in the form of (.*)<number>(.*) to find your prefix/suffix.
Or if you're comfortable with string operations you can probably just find the start and end of the target group and use substr to find the pre/suf fix.
I don't think you can do this with one regex. I would find all digit sequences within the string (probably with a regex) and then I would select the longest with .NET code, and call Split().
This depends entirely on your Regexp engine. Check your Regexp environment for capturing, there might be something in it like the automatic variables in Perl.
OK, let's talk about your question:
Keep in mind, that both, NFA and DFA, of almost every Regexp engine are greedy, this means, that a (\d+) will always find the longest match, when it "stumbles" over it.
Now, what I can get from your example, is you always need middle portion of a number, try this:
/^(.*\D)?(\d+)(\D.*)?$/ig
The now look at variables $1, $2, $3. Not all of them will exist: if there are all three of them, $2 will hold your number in question, the other vars, parts of the prefix. when one of the prefixes is missing, only variable $1 and $2 will be set, you have to see for yourself, which one is the integer. If both prefix and suffix are missing, $1 will hold the number.
The idea is to make the engine "stumble" over the first few characters and start matching a long number in the middle.
Since the modifier /gis present, you can loop through all available combinations, that the machine finds, you can then simply take the one you like most or something.
This example is in PCRE, but I'm sure .NET has a compatible mode.
Is it possible to write a regular expression that matches all strings that does not only contain numbers? If we have these strings:
abc
a4c
4bc
ab4
123
It should match the four first, but not the last one. I have tried fiddling around in RegexBuddy with lookaheads and stuff, but I can't seem to figure it out.
(?!^\d+$)^.+$
This says lookahead for lines that do not contain all digits and match the entire line.
Unless I am missing something, I think the most concise regex is...
/\D/
...or in other words, is there a not-digit in the string?
jjnguy had it correct (if slightly redundant) in an earlier revision.
.*?[^0-9].*
#Chad, your regex,
\b.*[a-zA-Z]+.*\b
should probably allow for non letters (eg, punctuation) even though Svish's examples didn't include one. Svish's primary requirement was: not all be digits.
\b.*[^0-9]+.*\b
Then, you don't need the + in there since all you need is to guarantee 1 non-digit is in there (more might be in there as covered by the .* on the ends).
\b.*[^0-9].*\b
Next, you can do away with the \b on either end since these are unnecessary constraints (invoking reference to alphanum and _).
.*[^0-9].*
Finally, note that this last regex shows that the problem can be solved with just the basics, those basics which have existed for decades (eg, no need for the look-ahead feature). In English, the question was logically equivalent to simply asking that 1 counter-example character be found within a string.
We can test this regex in a browser by copying the following into the location bar, replacing the string "6576576i7567" with whatever you want to test.
javascript:alert(new String("6576576i7567").match(".*[^0-9].*"));
/^\d*[a-z][a-z\d]*$/
Or, case insensitive version:
/^\d*[a-z][a-z\d]*$/i
May be a digit at the beginning, then at least one letter, then letters or digits
Try this:
/^.*\D+.*$/
It returns true if there is any simbol, that is not a number. Works fine with all languages.
Since you said "match", not just validate, the following regex will match correctly
\b.*[a-zA-Z]+.*\b
Passing Tests:
abc
a4c
4bc
ab4
1b1
11b
b11
Failing Tests:
123
if you are trying to match worlds that have at least one letter but they are formed by numbers and letters (or just letters), this is what I have used:
(\d*[a-zA-Z]+\d*)+
If we want to restrict valid characters so that string can be made from a limited set of characters, try this:
(?!^\d+$)^[a-zA-Z0-9_-]{3,}$
or
(?!^\d+$)^[\w-]{3,}$
/\w+/:
Matches any letter, number or underscore. any word character
.*[^0-9]{1,}.*
Works fine for us.
We want to use the used answer, but it's not working within YANG model.
And the one I provided here is easy to understand and it's clear:
start and end could be any chars, but, but there must be at least one NON NUMERICAL characters, which is greatest.
I am using /^[0-9]*$/gm in my JavaScript code to see if string is only numbers. If yes then it should fail otherwise it will return the string.
Below is working code snippet with test cases:
function isValidURL(string) {
var res = string.match(/^[0-9]*$/gm);
if (res == null)
return string;
else
return "fail";
};
var testCase1 = "abc";
console.log(isValidURL(testCase1)); // abc
var testCase2 = "a4c";
console.log(isValidURL(testCase2)); // a4c
var testCase3 = "4bc";
console.log(isValidURL(testCase3)); // 4bc
var testCase4 = "ab4";
console.log(isValidURL(testCase4)); // ab4
var testCase5 = "123"; // fail here
console.log(isValidURL(testCase5));
I had to do something similar in MySQL and the following whilst over simplified seems to have worked for me:
where fieldname regexp ^[a-zA-Z0-9]+$
and fieldname NOT REGEXP ^[0-9]+$
This shows all fields that are alphabetical and alphanumeric but any fields that are just numeric are hidden. This seems to work.
example:
name1 - Displayed
name - Displayed
name2 - Displayed
name3 - Displayed
name4 - Displayed
n4ame - Displayed
324234234 - Not Displayed