I'm having trouble with what seems like a simple regex capture. I'm using AutoIt's stringRegExp() function.
The source string is:
1 U:\some text here\more text over here\06-17-2011\Folder 2\161804\abc9831\xyz10007569.JPG
I'm trying to capture "abc9831" and "161804". The "abc" part can be "abc", "def", or "ghi", followed by a string of digits. The "161804" can be replaced with any string of text. Everything is case insensitive. I'm currently using the following regex pattern:
(?i)\\\\.+\\\\((abc\d+)|(def\d+)|(ghi\d+))
But it's only capturing the "abc9831" part. How do I pick up the text string preceding it?
When the regex below is used in AutoIt's StringRegExp() function (using the flag "1" to return an array of matches), it returns 161804\abc9831. Is this what you're wanting to return?
.*\\([^\\]+\\[a-z]{3}\d+)\\.*
Here's an example you can run yourself:
#include <Array.au3>
$string = 'U:\some text here\more text over here\06-17-2011\Folder 2\161804\abc9831\xyz10007569.JPG'
$capture = StringRegExp($string,'.*\\([^\\]+\\[a-z]{3}\d+)\\.*',1)
_ArrayDisplay($capture)
(?i)\\\\(.+\\\\(abc\d+)|(def\d+)|(ghi\d+))
should do the trick if you want it all in one string (with a \ in between).
If you want two separate captures:
(?i)\\\\(.+)\\\\((abc\d+)|(def\d+)|(ghi\d+))
Edit: New version...
The raw regex is \b(\d+)\\((?:abc|def|ghi)\d+). The escaped string is \\b(\\d+)\\\\((?:abc|def|ghi)\\d+)
Related
i am trying to build one regex expression for the below sample text in which i need to replace the bold text. So far i could achieve this much
((\|)).*(\|) which is selecting the whole string between the first and last pip char. i am bound to use apache or java regex.
Sample String: where text length between pipes may vary
1.1|ProvCM|111111111111|**10.15.194.25**|10.100.10.3|10.100.10.1|docsis3.0
To match part after nth occurrence of pipe you can use this regex:
/^(?:[^|]*\|){3}([^|]*)/
Here n=3
It will match 10.15.194.25 in matched group #1
RegEx Demo
^((?:[^|]*\\|){3})[^|]+
You can use this.Replace by $1<anything>.See demo.
https://regex101.com/r/tP7qE7/4
This here captures from start of string to | and then captures 3 such groups and stores it in $1.The next part of string till | is what you want.Now you can replace it with anything by $1<textyouwant>.
Here's how you can do the replacement:
String input = "1.1|ProvCM|111111111111|10.15.194.25|10.100.10.3|10.100.10.1|docsis3.0";
int n = 3;
String newValue = "new value";
String output = input.replaceFirst("^((?:[^|]+\\|){"+n+"})[^|]+", "$1"+newValue);
This builds:
"1.1|ProvCM|111111111111|new value|10.100.10.3|10.100.10.1|docsis3.0"
I need to come up with a regular expression to parse my input string. My input string is of the format:
[alphanumeric].[alpha][numeric].[alpha][alpha][alpha].[julian date: yyyyddd]
eg:
A.A2.ABC.2014071
3.M1.MMB.2014071
I need to substring it from the 3rd position and was wondering what would be the easiest way to do it.
Desired result:
A2.ABC.2014071
M1.MMB.2014071
(?i) will be considered as case insensitive.
(?i)^[a-z\d]\.[a-z]\d\.[a-z]{3}\.\d{7}$
Here a-z means any alphabet from a to z, and \d means any digit from 0 to 9.
Now, if you want to remove the first section before dot, then use this regex and replace it with $1 (or may be \1)
(?i)^[a-z\d]\.([a-z]\d\.[a-z]{3}\.\d{7})$
Another option is replace below with empty:
(?i)^[a-z\d]\.
If the input string is just the long form, then you want everything except the first two characters. You could arrange to substitute them with nothing:
s/^..//
Or you could arrange to capture everything except the first two characters:
/^..(.*)/
If the expression is part of a larger string, then the breakdown of the alphanumeric components becomes more important.
The details vary depending on the language that is hosting the regex. The notations written above could be Perl or PCRE (Perl Compatible Regular Expressions). Many other languages would accept these regexes too, but other languages would require tweaks.
Use this regex:
\w.[A-Z]\d.[A-Z]{3}.\d{7}
Use the above regex like this:
String[] in = {
"A.A2.ABC.2014071", "3.M1.MMB.2014071"
};
Pattern p = Pattern.compile("\\w.[A-Z]\\d.[A-Z]{3}.\\d{7}");
for (String s: in ) {
Matcher m = p.matcher(s);
while (m.find()) {
System.out.println("Result: " + m.group().substring(2));
}
}
Live demo: http://ideone.com/tns9iY
Example data:
029Extract this specific string. Do not capture anything else.
In the example above, I would like to capture the first n characters immediately after the 3 digit entry which defines the value of n. I.E. the 29 characters "Extract this specific string."
I can do this within a loop, but it is slow. I would like (if it is possible) to achieve this with a single regex statement instead, using some kind of backreference. Something like:
(\d{3})(.{\1})
With perl, you can do:
my $str = '029Extract this specific string. Do not capture anything else.';
$str =~ s/^(\d+)(.*)$/substr($2,0,$1)/e;
say $str;
output:
Extract this specific string.
You can not do it with single regex, while you can use knowledge where regex stop processing to use substr. For example in JavaScript you can do something like this http://jsfiddle.net/75Tm5/
var input = "blahblah 011I want this, and 029Extract this specific string. Do not capture anything else.";
var regex = /(\d{3})/g;
var matches;
while ((matches = regex.exec(input)) != null) {
alert(input.substr(regex.lastIndex, matches[0]));
}
This will returns both lines:
I want this
Extract this specific string.
Depending on what you really want, you can modify Regex to match only numbers starting from line beginning, match only first match etc
Are you sure you need a regex?
From https://stackoverflow.com/tags/regex/info:
Fools Rush in Where Angels Fear to Tread
The tremendous power and expressivity of modern regular expressions
can seduce the gullible — or the foolhardy — into trying to use
regular expressions on every string-related task they come across.
This is a bad idea in general, ...
Here's a Python three-liner:
foo = "029Extract this specific string. Do not capture anything else."
substr_len = int(foo[:3])
print foo[3:substr_len+3]
And here's a PHP three-liner:
$foo = "029Extract this specific string. Do not capture anything else.";
$substr_len = (int) substr($foo,0,3);
echo substr($foo,3,substr_len+3);
I know I can exclude outside characters in a string using look-ahead and look-behind, but I'm not sure about characters in the center.
What I want is to get a match of ABCDEF from the string ABC 123 DEF.
Is this possible with a Regex string? If not, can it be accomplished another way?
EDIT
For more clarification, in the example above I can use the regex string /ABC.*?DEF/ to sort of get what I want, but this includes everything matched by .*?. What I want is to match with something like ABC(match whatever, but then throw it out)DEF resulting in one single match of ABCDEF.
As another example, I can do the following (in sudo-code and regex):
string myStr = "ABC 123 DEF";
string tempMatch = RegexMatch(myStr, "(?<=ABC).*?(?=DEF)"); //Returns " 123 "
string FinalString = myStr.Replace(tempMatch, ""); //Returns "ABCDEF". This is what I want
Again, is there a way to do this with a single regex string?
Since the regex replace feature in most languages does not change the string it operates on (but produces a new one), you can do it as a one-liner in most languages. Firstly, you match everything, capturing the desired parts:
^.*(ABC).*(DEF).*$
(Make sure to use the single-line/"dotall" option if your input contains line breaks!)
And then you replace this with:
$1$2
That will give you ABCDEF in one assignment.
Still, as outlined in the comments and in Mark's answer, the engine does match the stuff in between ABC and DEF. It's only the replacement convenience function that throws it out. But that is supported in pretty much every language, I would say.
Important: this approach will of course only work if your input string contains the desired pattern only once (assuming ABC and DEF are actually variable).
Example implementation in PHP:
$output = preg_replace('/^.*(ABC).*(DEF).*$/s', '$1$2', $input);
Or JavaScript (which does not have single-line mode):
var output = input.replace(/^[\s\S]*(ABC)[\s\S]*(DEF)[\s\S]*$/, '$1$2');
Or C#:
string output = Regex.Replace(input, #"^.*(ABC).*(DEF).*$", "$1$2", RegexOptions.Singleline);
A regular expression can contain multiple capturing groups. Each group must consist of consecutive characters so it's not possible to have a single group that captures what you want, but the groups themselves do not have to be contiguous so you can combine multiple groups to get your desired result.
Regular expression
(ABC).*(DEF)
Captures
ABC
DEF
See it online: rubular
Example C# code
string myStr = "ABC 123 DEF";
Match m = Regex.Match(myStr, "(ABC).*(DEF)");
if (m.Success)
{
string result = m.Groups[1].Value + m.Groups[2].Value; // Gives "ABCDEF"
// ...
}
what is the best way to extract last 2 characters of a string using regular expression.
For example, I want to extract state code from the following
"A_IL"
I want to extract IL as string..
please provide me C# code on how to get it..
string fullexpression = "A_IL";
string StateCode = some regular expression code....
thanks
Use the regex:
..$
This will return provide the two characters next to the end anchor.
Since you're using C#, this would be simpler and probably faster:
string fullexpression = "A_IL";
string StateCode = fullexpression.Substring(fullexpression.Length - 2);
Use /(..)$/, then pull group 1 (.groups(1), $1, \1, etc.).
as for the best way, I'd say it's .{2}$
it is more elegant and self-descriptive.