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
Related
For a project of mine, I want to create 'blocks' with Regex.
\xyz\yzx //wrong format
x\12 //wrong format
12\x //wrong format
\x12\x13\x14\x00\xff\xff //correct format
When using Regex101 to test my regular expressions, I came to this result:
([\\x(0-9A-Fa-f)])/gm
This leads to an incorrect output, because
12\x
Still gets detected as a correct string, though the order is wrong, it needs to be in the order specified below, and in no other order.
backslash x 0-9A-Fa-f 0-9A-Fa-f
Can anyone explain how that works and why it works in that way? Thanks in advance!
To match the \, folloed with x, followed with 2 hex chars, anywhere in the string, you need to use
\\x[0-9A-Fa-f]{2}
See the regex demo
To force it match all non-overlapping occurrences, use the specific modifiers (like /g in JavaScript/Perl) or specific functions in your programming language (Regex.Matches in .NET, or preg_match_all in PHP, etc.).
The ^(?:\\x[0-9A-Fa-f]{2})+$ regex validates a whole string that consists of the patterns like above. It happens due to the ^ (start of string) and $ (end of string) anchors. Note the (?:...)+ is a non-capturing group that can repeat in the string 1 or more times (due to + quantifier).
Some Java demo:
String s = "\\x12\\x13\\x14\\x00\\xff\\xff";
// Extract valid blocks
Pattern pattern = Pattern.compile("\\\\x[0-9A-Fa-f]{2}");
Matcher matcher = pattern.matcher(s);
List<String> res = new ArrayList<>();
while (matcher.find()){
res.add(matcher.group(0));
}
System.out.println(res); // => [\x12, \x13, \x14, \x00, \xff, \xff]
// Check if a string consists of valid "blocks" only
boolean isValid = s.matches("(?i)(?:\\\\x[a-f0-9]{2})+");
System.out.println(isValid); // => true
Note that we may shorten [a-zA-Z] to [a-z] if we add a case insensitive modifier (?i) to the start of the pattern, or just use \p{Alnum} that matches any alphanumeric char in a Java regex.
The String#matches method always anchors the regex by default, we do not need the leading ^ and trailing $ anchors when using the pattern inside it.
I try to look for this answer for a while but no luck (sorry if I could describe it well). I am still newbie with regex. I am trying to match a string with only number and a certain delimiter. For example: the patter would be 8/16/32/64/.... the number will be split by '/' with arbitrary amount of number, I could find a way to match them.
My attempt is \d+/\d+? but couldn't get it to work.
You could remove the '/' delimiter and then test for the existence of a number
Here is some C# as an example:
static void Main(string[] args)
{
string text = "8/16/32/64/";
Console.WriteLine(text);
TestForNum(text);
text = "8/16/32/64/b";
Console.WriteLine(text);
TestForNum(text);
Console.ReadKey();
}
private static void TestForNum(string text)
{
string tmp = Regex.Replace(text, #"/", "");
Match m = Regex.Match(tmp, #"^\d+$");
if(m.Success)
{
Console.WriteLine("\t" + m.Groups[0]);
}
else Console.WriteLine("\tno match");
}
A naive approach would be
[\d/]+
However, this does match //// as well as just 12345. To match only "proper" strings:
\d+(/\d+)+
Reads digits followed by delimiter+digits repeated at least once. If trailing/leading delimiters are allowed, then
/?(\d+/)+\d*
If you're using a flavor that uses slashes to quote the regex (like javascript), you'll need to escape them:
/\d+(\/\d+)+/
You can do:
(\d+)(\D|$)
See this work That will split a list of digits delimited by any non digit, so 1?2!3.4 would match
If you want a specific delimiter, such as /:
(\d+)(?:/|$)
As simple as possible:
(\d+\/?)+
Every digit followed by [a] slash, as many as possible. You may use g flag for all matches.
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$2", RegexOptions.Singleline);
A regular expression can contain multiple capturing groups. Each group must consist of consecutive characters so it's not possible to have a single group that captures what you want, but the groups themselves do not have to be contiguous so you can combine multiple groups to get your desired result.
Regular expression
(ABC).*(DEF)
Captures
ABC
DEF
See it online: rubular
Example C# code
string myStr = "ABC 123 DEF";
Match m = Regex.Match(myStr, "(ABC).*(DEF)");
if (m.Success)
{
string result = m.Groups[1].Value + m.Groups[2].Value; // Gives "ABCDEF"
// ...
}
How do I use regex to convert
11111aA$xx1111xxdj$%%`
to
aA$xx1111xxdj$%%
So, in other words, I want to remove (or match) the FIRST grouping of 1's.
Depending on the language, you should have a way to replace a string by regex. In Java, you can do it like this:
String s = "11111aA$xx1111xxdj$%%";
String res = s.replaceAll("^1+", "");
The ^ "anchor" indicates that the beginning of the input must be matched. The 1+ means a sequence of one or more 1 characters.
Here is a link to ideone with this running program.
The same program in C#:
var rx = new Regex("^1+");
var s = "11111aA$xx1111xxdj$%%";
var res = rx.Replace(s, "");
Console.WriteLine(res);
(link to ideone)
In general, if you would like to make a match of anything only at the beginning of a string, add a ^ prefix to your expression; similarly, adding a $ at the end makes the match accept only strings at the end of your input.
If this is the beginning, you can use this:
^[1]*
As far as replacing, it depends on the language. In powershell, I would do this:
[regex]::Replace("11111aA$xx1111xxdj$%%","^[1]*","")
This will return:
aA$xx1111xxdj$%%
If you only want to replace consecutive "1"s at the beginning of the string, replace the following with an empty string:
^1+
If the consecutive "1"s won't necessarily be the first characters in the string (but you still only want to replace one group), replace the following with the contents of the first capture group (usually \1 or $1):
1+(.*)
Note that this is only necessary if you only have a "replace all" capability available to you, but most regex implementations also provide a way to replace only one instance of a match, in which case you could just replace 1+ with an empty string.
I'm not sure but you can try this
[^1](\w*\d*\W)* - match all as a single group except starting "1"(n) symbols
In Javascript
var str = '11111aA$xx1111xxdj$%%';
var patt = /^1+/g;
str = str.replace(patt,"");