Regex - Extract a substring from a given string

Question

I have a string here, This is a string: AAA123456789.

So the idea here is to extract the string AAA123456789 using regex.

I am incorporating this with X-Path.

Note: If there is a post to this, kindly lead me to it.

I think, by right, I should substring(myNode, [^AAA\d+{9}]),

I am not really sure bout the regex part.

The idea is to extract the string when met with "AAA" and only numbers but 9 consequent numbers only.

What have you tried? You must have at least one regex you have tried that didn't work (unless you just came in an expected us to do your work for you...) — John3136
– John3136, Commented Sep 20, 2012 at 7:00
Will the string always have the same format? And will you always have exactly nine digits? Because if so, you don't need a regex, just simple substring processing. — Jere Käpyaho
– Jere Käpyaho, Commented Sep 20, 2012 at 7:01
Almost right, just use \d{9} (both + and {9} are repeat operators). — nneonneo
– nneonneo, Commented Sep 20, 2012 at 7:15
The string isnt always the same format. See, it can be "This is a string: AAA123456789 but not an double". So i cant really use common xpath. So the fullproof solution here is to substring the string, AAA123456789. Well I have tried liek the one shown above. Maybe its wrong. The other way is like what Jere mentioned, common but not fullproof method. substring-after(upper-case(myNode), "STRING"). Then again, this is not suitable. Thanks — Vincent
– Vincent, Commented Sep 20, 2012 at 7:24

Dimitre Novatchev · Accepted Answer · 2012-09-20 13:01:15Z

11

Pure XPath solution:

substring-after('This is a string: AAA123456789', ': ')

produces:

AAA123456789

XPath 2.0 solutions:

tokenize('This is a string: AAA123456789 but not an double',
              ' '
              )[starts-with(., 'AAA')]

or:

tokenize('This is a string: AAA123456789 but not an double',
              ' '
              )[matches(., 'AAA\d+')]

or:

replace('This is a string: AAA123456789 but not an double',
              '^.*(A+\d+).*$',
              '$1'
              )

edited Sep 20, 2012 at 13:01

answered Sep 20, 2012 at 12:49

Dimitre Novatchev

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6 Comments

Vincent Over a year ago

Hi Dimitre, can you please explain the replace part. This is what I understand, I think, replace everything else but *(A+\d+).*$ with $1...is this correct...but what does $1 do...

Dimitre Novatchev Over a year ago

@Vincent, This means: Replace the whole string (if it contains a substring of the form A+\d+ ) with just the subexpression that is inside the (1st pair of) brackets. The third argument of replace must contain a string specifying with what to replace every target. It allows "capture references" by number (position). Read more about replace() here: w3.org/TR/xpath-functions/#func-replace

Alptigin Jalayr Over a year ago

Sir, is there an XPath question that you have not answered, or do not know the answer to? :-)

Dimitre Novatchev Over a year ago

@AlptiginJalayr, It can be immediately seen that I haven't attempted to answer all XPath questions at SO. :)

Stefan Jung Over a year ago

The possibility to combine tokenize and matches was new to me. Has helped in my case, thank you very much.

|

Vincent · Accepted Answer · 2012-09-21 09:08:27Z

Alright, after referencing answers and comments by wonderful people here, I summarized my findings with this solution which I opted for. Here goes,

concat("AAA", substring(substring-after(., "AAA"), 1, 9)).

So I firstly, substring-after the string with "AAA" as the 1st argument, with the length of 1 to 9...anything more, is ignored. Then since I used the AAA as a reference, this will not appear, thus, concatenating AAA to the front of the value. So this means that I will get the 1st 9 digits after AAA and then concat AAA in front since its a static data.

This will allow the data to be correct no matter what other contributions there is.

But I like the regex by @Dimitre. The replace part. The tokenize not so as what if there isn't space as the argument. The replace with regex, this is also wonderful. Thanks.

And also thanks to you guys out there to...

Andrew Cheong · Accepted Answer · 2012-09-20 07:44:04Z

1

First, I'm pretty sure you don't mean to have the [^ ... ]. That defines a "negative character class", i.e. your current regex says, "Give me a single character that is not one of the following: A0123456789{}". You probably meant, plainly, "AAA(\d{9})". Now, according to this handy website, XPath does support capture groups, as well as backreferences, so take your pick:

"AAA(\d{9})"

And extracting $1, the first capture group, or:

"(?<=AAA)\d{9}"

And taking the whole match ($0).

answered Sep 20, 2012 at 7:44

Andrew Cheong

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Comments

Stephan · Accepted Answer · 2012-09-20 08:02:52Z

1

Can you try this :

A{3}(\d{9})

answered Sep 20, 2012 at 8:02

Stephan

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