I have AUD0.7195.
I want to mach AUD in $1 and 0.7195in $2.
The amount may vary. It can be positive integer.
I have tried with:
/([A-Z]{3})(\d{+}\.?\d*)/ms;
But it matches nothing. What is wrong? Thanks!
([a-zA-Z]+)(\d+(?:\.\d+)?)
Use this.Yours does not work as \d{+} does not act as quantifier.
\d{+} doesn't make sense and should be \d+
Could you please try this:
my $string = "AUD0.7195";
print "$1\t$2.......\n", if($string=~m/^([\w]{3})([0-9\-\.]*)$/g);
Related
I have a string
K9908098F, G2342D34324/ 234234323, 234-234-234; R324234
How to catch only 234234323 and 234-234-234 in VBA?
This [\d-]+ pattern grabs extra pieces
You are pretty close, just need to add borders: \b[\d-]+\b
Regex demo and explanation
Give this a try:
(\w+),\s+([\w-]+);
This will capture 234234323 in group 1 and 234-234-234 in group 2.
Not too elegant though but would work. Just a small addition to your regex.
[\d-]+[,;]
You can try this too,
[-\d]+(?=[;,. ])
Demo
I am trying to write some form validation, I need one of the inputs to be 1-9999. I know nothing about regular expressions ( never used them before) and here is my first attempt
/^([1-9][1-9]|[1-9]|[1-9]\d|9999)$/
Does not seem to want to work, can anyone help me? Thanks!
Try the below regex,
^(?:[1-9][0-9]{3}|[1-9][0-9]{2}|[1-9][0-9]|[1-9])$
DEMO
This doesn't exclude zero, but /^\d{1,4}$/ should do the trick.
Try using this
^([1-9]|[1-9][0-9]|[1-9][0-9][0-9]|[1-9][0-9][0-9][0-9])$
To exclude zero values but include non-zero ones with leading zeros:
([1-9]\d{0,3})|(\d[1-9])\d{0,2}|(\d{2}[1-9])\d?|(\d{1,3}[1-9])
This Regex should not match numbers that start with 0 a part from 0
Regex: /^(?!(0\d))\d{1,4}$/
Regex (Exclude Zero): /^(?!(0))\d{1,4}$/
Test: https://regex101.com/r/zfCKel/2
100.
101.
102.
guys this is my text and i want to change like this:
<tag>100.</tag>
<tag>101.</tag>
<tag>102.</tag>
My RegEx is:
[0-9][0-9][0-9]\.
Replace with:
<tag>\1</tag>
But it does not work :(
I could not see the numbers and dot sign.
Thanks in advance
You need some (capturing) parentheses;
Re: (\d{3}\.)
Replace <tag>\1</tag>
You're missing the \. in your brackets. Try this $regex = ":[\d\.]+:".
Without the []'s. I'd appreciate any help.
For integers:
/^say \d+.*$/
But you might also want floating point:
/^say (\d+\.)?\d+.*$/
Based on the limited information presented a Perl Regular expression will look like /^say \d.*$/ where \d matches the "number" and .* matches "anything".
POSIX regex:
say [[:digit:]]+.*
Perl regex:
/say \d+.*/
Regexes like those are fun. If your 'any number' is guaranteed to be an integer, you could do this:
/say \d+/
If you wanted to capture the number and the anything, you could do this:
/say (\d+) (.*)/
..which saves them using "groups".
Here is the demo String:
beforeValueAfter
Assume that I know the value I want is between "before" and "After"
I want to extact the "Value" using the regex....
pervious909078375639355544after
Assume that I know the value I want is between "pervious90907" and "55544after"
I want to extact the "83756393" using the regex....
thx in advance.
The answer depends on two things:
If you know exactly what the value consists of (if you know it will be digits, etc., it makes it easier). If it could be anything, the answer is a little harder.
If your system is greedy/ungreedy by default, it affects the way you'd set up the expression. I will assume it is greedy by default.
If it can be anything (the ? will be needed to toggle the .* to ungreedy because ".*" also matches "After":
/before(.*?)After/
If you know it is digits:
/before(\d*)After
If it could be any word characters (0-9, a-z, A-Z, _):
/before(\w*?)After
Try this regular expression:
pervious90907(.*?)55544after
That will get you the shortest string (note the non-greedy *? quantifier) between pervious90907 and 55544after.
The regex should be like this:
previous([0-9]*)after
bash-3.2$ echo pervious909078375639355544after | perl -ne 'print "$1\n" if /pervious90907(.*)55544after/'
83756393