awk : set start and end of match - regex

Question

I have a LaTeX-like table like this (columns are delimited by &) :
foobar99 & 68
foobar4 & 43
foobar2 & 73
I want to get the index of the numbers at column 2 by using match.
In Vim, we can use \zs and \ze to set start and end of matching.
Thus, to match accurately number at colum 2, we can use ^.*&\s*\zs[[:digit:]]\+\ze\s*$.
How about awk? Is there an equivalent?
EDIT:
Matching for the first line:
foobar99 & 68
^^
123456789012345678
Expected output : 18.
EDIT2:
I am writing an awk script to deal with block delimited by line break (Hence, FS="\n" and RS=""). The MWE above is just one of these blocks.
A possible way to get the index of number at column 2 is to do something like that
split(line, cases, "&");
index = match(cases[2], /[[:digit:]]\+/);
but I am looking for a beautiful way to do this.
Apologies for the XY problem. But I'm still interested in start/end matching.

Answer 1

Too little context, so a simple guess: have you tried splitting the table into columns? With something like awk -F '\\s*&\\s*' you have your second column in $2.
In fact, you can use split() to retrieve the exact column of a string:
split(s, a[, fs ])
Split the string s into array elements a[1], a[2], ..., a[n], and
return n. All elements of the array shall be deleted before the split is
performed. The separation shall be done with the ERE fs or with the field
separator FS if fs is not given. Each array element shall have a
string value when created and, if appropriate, the array element
shall be considered a numeric string (see Expressions in awk). The
effect of a null string as the value of fs is unspecified.
So your second column is something like
split(s, a, /\s*&\s*/)
secondColumn = a[2]

Answer 2

By default, awk sees three columns in your data, and column 2 contains & only (and column 3 contains numbers). If you change the field delimiter to &, then you have two columns with trailing spaces in column 1 and leading spaces in column 2 (and some trailing spaces, as it happens; try copying the data from the question).
In awk, you could convert column 2 with leading spaces into a number by adding 0: $2 + 0 would force it to be treated as a number. If you use $2 in a numeric context, it'll be treated as a number. Conversely, you can force awk to treat a field as a string by concatenating with the empty string: $2 "" will be a string.
So there's no need for the complexity of regexes to get at the number — if the data is as simple as shown.
You say you want to use match; it is not clear what you need that for.
awk -F'&' '{ printf "F1 [%s], F2 [%10s] = [%d] = [%-6d] = [%06d]\n", $1, $2, $2, $2, $2 }' data
For your data, which has a single blank at the end of the first two lines and a double blank at the end of the third, the output is:
F1 [foobar99 ], F2 [ 68 ] = [68] = [68 ] = [000068]
F1 [foobar4 ], F2 [ 43 ] = [43] = [43 ] = [000043]
F1 [foobar2 ], F2 [ 73 ] = [73] = [73 ] = [000073]
Note that I didn't need to explicitly convert $2 to a number. The printf formats treated it as a string or a number depending on whether I used %s or %d.
If you need to, you can strip trailing blanks of $1 (or, indeed, $2), but without knowing what else you need to do, it's hard to demonstrate alternatives usefully.
So, I think awk does what you need without needing you to jump through much in the way of hoops. For a better explanation, you'd need to provide a better question, describing or showing what you want to do.

Answer 3

You can try this way
awk '{print index($0,$3)}' infile