Extracting numbers between square brackets

[TideMan]

I'm feeling my way with bash scripting.......

The first line of my file is:
prmslmsl, [61][1][1]
and I want to extract the first no in the square brackets.

head -n 1 GrADS.out | cut -d\, -f 2 | cut -d\] -f 1 | cut -d\[ -f 2

gives me the answer, but it's very ugly.

Is there a better way?

[ghostdog74]

if you have Python

# python -c "first=open('file').readline();print first.split('[')[1][:-1]"
61

awk

awk -F"[" 'NR==1{sub("]","",$2);print $2}'  file

[andrewr]

Depends on what "neat" is... using bash alone, I would do:
x=`head -n 1 GrADS.out`; x=${x#*[}; x=${x%%]*}; echo $x

Or, there is always sed --- available on just about all UNIX's.
Assuming there are no numbers before the first one, you can do:
head -n 1 GrADS.out | sed -ne "s%[^0-9]*\([0-9]*\).*%\1%p"

What's your preference?

[ghostdog74]

Depends on what "neat" is... using bash alone, I would do:
x=`head -n 1 GrADS.out`; x=${x#*[}; x=${x%%]*}; echo $x

well if you use head, that's not considered bash alone

read x < "GrADS.out"

[andrewr]

aye, I wasn't specific -- I meant matching by bash alone....
And you see a workaround .... too.

or how about....

read -d ] x < GrADS.out; echo ${x#*[}

OOOOHHHHH.... or, do an

IFS="[]"; read -a x < GrADS.out; echo ${x[1]}

(could be executed in a subshell to save saving IFS and restoring... recurse.. curse...
....

Too bad sed doesn't allow positive matches .... it would make it so much more readable.



Oh, and since I cant sleep -- & my thread question isn't getting answers -- how about one that avoids your solution (read) but is more FLEXIBLE!!! USING TRULY ONLY BaSh!!!
(Pick a line, any line.... !1 = line 1, !2=line 2, etc....

x=( `bash -c 'history -cr GrADS.out; history -p "!1:gs/[/ /:gs/]//"' )
echo ${x[1]}

oddly, bash -c is required, as merely executing in a subshell does NOT reset the current history line to 0 -- a parenthetical inside the `( history -cr ... )` inherits the present history line number.
Unfortunately, I don't see a way to reset BASH's notion of the present history line number without calling bash, directly.... so, it is slower than a mere fork() -- challenge Q: does anyone know how to make that work without the bash -c ??? :shrug:

[TideMan]

Thanks you blokes.

For me, the neatest is:

read -d ] x < GrADS.out; echo ${x#*[}

Now, I'm off to find out how it works exactly...............

[TideMan]

I'm trying to understand this:
read -d ] x < GrADS.out; echo ${x#*[}

OK, I understand this bit:
read -d ] x < GrADS.out | echo $x
It gives:
prmslmsl, [61

First of all, why do we use ;, not | for the next bit? Either seems to work OK.

And I guess ${x#*[} means retain all the characters after the [
Where could I learn this stuff?

[andrewr]

I'm trying to understand this:
read -d ] x < GrADS.out; echo ${x#*[}

OK, I understand this bit:
read -d ] x < GrADS.out | echo $x
It gives:
prmslmsl, [61

First of all, why do we use ;, not | for the next bit? Either seems to work OK.

And I guess ${x#*[} means retain all the characters after the [
Where could I learn this stuff?

Actually, I didn't use | between the commands because echo does not use std input.
the semicolon ; simply means the same as an enter button. It allows multiple unrelated/unpiped commands on one line.
You can get away with | because echo ignores the standard input and isn't confused by it.
I like your use of | better, and wish I had thought of that twisted thinking for 1 reason only -- BASH has bugs in job control (see the end of the man file -- under BUGS). Type: man bash
the | would prevent that problem .... hmmm.... neat.

for non-bug learning; With the manual up type:
/Parameter Expansion
followed by enter.

That will search for the section explaining how to remove parts of variables.
Actually, ${x#*[} is negative thinking only -- remove everything up to and including the [

Cheers.

P.S.
Sheeesh.... found another BASH BUG. Anyone know why this *DOESNT* work?

read -d ] < GrADS.out | echo ${REPLY#*[}

What is wrong with it, logically?
If nothing, could someone write to the bug list of whoever maintains BASH.
It has lots of quirks....

[TideMan]

Thank you for your help, andrewr
${x#*[} seems like gobbledygook at first, but as I read the manual, it means take x and remove (#) everything (*) up to and including the [, which leaves 61.

Actually, I might revise my assessment of elegance and go for:
IFS="[]"; read -a x < GrADS.out; echo ${x[1]}
having just learned about IFS