I want to do some regexp matches on strings with expr, a tool I rarely
used, and I'm puzzled about its correct usage for the intended purpose,
if it's appropriate at all.

The test cases I am actually trying to implement are (in terms of grep)

$ grep -o "%[auUlLcCsS%]" <<< "A%aB%bC%cX"
%a
%c

$ grep -o "%[^auUlLcCsS%]" <<< "A%aB%bC%cX"
%b

or in terms of awk

$ echo "A%aB%bC%cX" |
awk '{ print match($0,/%[^auUlLcCsS%]/)
print substr($0,RSTART,RLENGTH) }'
5
%b

If possible I'd want either the substring or the index of the match in
the string and I thought that expr might serve as a light-weight tool
to avoid grep or awk.

The POSIX specs say: "Usually, the matching operator shall return a
string representing the number of characters matched ('0' on failure)."
so it might not be the appropriate tool. - Or am I missing something?

Janis

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On 15.01.2022 17:40, Axel Reichert wrote:

I got concerned by "achored", tested

expr "A%aB%bC%cX" : "%[^auUlLcCsS%]"

and got a "0" returned, which is not what you want.

Yep, that was also my first test and the result puzzled me.

expr "A%aB%bC%cX" : ".*\(%[^auUlLcCsS%]\)"

But this one is fine. I forgot about the subexpression semantics
and obviously was inattentive when re-reading the man page.

returns "%b", though. So your mileage may vary.

Thanks!

Janis

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Janis Papanagnou <[email protected]> writes:

$ echo "A%aB%bC%cX" |
awk '{ print match($0,/%[^auUlLcCsS%]/)
print substr($0,RSTART,RLENGTH) }'
5
%b

My (BSD) man page mentions

expr1 : expr2
The ``:'' operator matches expr1 against expr2, which must be a basic
regular expression. The regular expression is anchored to the begin-
ning of the string with an implicit ``^''.

If the match succeeds and the pattern contains at least one regular
expression subexpression ``\(...\)'', the string corresponding to
``\1'' is returned; otherwise the matching operator returns the num-
ber of characters matched. If the match fails and the pattern con-
tains a regular expression subexpression the null string is returned;
otherwise 0.

I got concerned by "achored", tested

expr "A%aB%bC%cX" : "%[^auUlLcCsS%]"

and got a "0" returned, which is not what you want.

expr "A%aB%bC%cX" : ".*\(%[^auUlLcCsS%]\)"

returns "%b", though. So your mileage may vary.

Best regards

Axel

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On 2022-01-15, Janis Papanagnou <[email protected]> wrote:

I want to do some regexp matches on strings with expr, a tool I rarely
used, and I'm puzzled about its correct usage for the intended purpose,

There are two properties to be aware of:
* expr(1)'s match operator uses a _basic_ regular expression.
* The regular expression is anchored to the beginning of the string
with an implicit '^'.

If possible I'd want either the substring or the index of the match in
the string and I thought that expr might serve as a light-weight tool
to avoid grep or awk.

You can extract a substring with \(...\).

--
Christian "naddy" Weisgerber [email protected]

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