There's a string example
"j2sdk/1.8.0_25-static j2sdk/1.8.0_45 j2sdk/1.8.0_p120 j2sdk/1.8.0_40 j2sdk/1.8.0_51"
I want to find the ones matched with format j2sdk/1.8.0_xxx, but xxx only with digits, here, I want below strings be matched
j2sdk/1.8.0_45
j2sdk/1.8.0_40
j2sdk/1.8.0_51
I wrote below code, but when run, it only get the first matched j2sdk/1.8.0_45, anything wrong with my code?
avail_versions="j2sdk/1.8.0_25-static j2sdk/1.8.0_45 j2sdk/1.8.0_p120 j2sdk/1.8.0_40 j2sdk/1.8.0_51"
patern='j2sdk\/1\.8\.0_[0-9]+\s+'
if [[ $avail_versions =~ $patern ]];then
echo matched
echo ${BASH_REMATCH[0]}
echo ${BASH_REMATCH[1]}
echo ${BASH_REMATCH[2]}
fi
The results is that BASH_REMATCH[0] is j2sdk/1.8.0_45, BASH_REMATCH[1] and [2] are empty
I expected I can get them in BASH_REMATH[1],BASH_REMATH[2],BASH_REMATH[3].
Is there other way in Bash I can get expected matches.
Thanks
I split the input at spaces and add back the space after each word.
for s in $avail_versions ; do
s="$s "
if [[ $s =~ $patern ]];then
echo ${BASH_REMATCH[0]}
fi
done
j2sdk/1.8.0_45
j2sdk/1.8.0_40
j2sdk/1.8.0_51
Related
I have a string like this
foo:collection:indexation [options] [--] <text> <text_1> <text_2> <text_3> <text_4>
And i want to use bash regex to get an array or string that I can split to get this in order to check if the syntax is correct
["text", "text_1", "text_2", "text_3", "text_4"]
I have tried to do this :
COMMAND_OUTPUT=$($COMMAND_HELP)
# get the output of the help
# regex
ARGUMENT_REGEX="<([^>]+)>"
GOOD_REGEX="[a-z-]"
# get all the arguments
while [[ $COMMAND_OUTPUT =~ $ARGUMENT_REGEX ]]; do
ARGUMENT="${BASH_REMATCH[1]}"
# bad syntax
if [[ ! $ARGUMENT =~ $GOOD_REGEX ]]; then
echo "Invalid argument '$ARGUMENT' for the command $FILE"
echo "Must only use characters [a-z:-]"
exit 5
fi
done
But the while does not seem to be appropriate since I always get the first match.
How can I get all the matches for this regex ?
Thanks !
The loop doesn't work because every time you're just testing the same input string against the regexp. It doesn't know that it should start scanning after the match from the previous iteration. You'd need to remove the part of the string up to and including the previous match before doing the next test.
A simpler way is to use grep -o to get all the matches.
$COMMAND_HELP | grep -o "$ARGUMENT_REGEX" | while read ARGUMENT; do
if [[ ! $ARGUMENT =~ $GOOD_REGEX ]]; then
echo "Invalid argument '$ARGUMENT' for the command $FILE"
echo "Must only use characters [a-z:-]"
exit 5
fi
done
Bash doesn't have this directly, but you can achieve a similar effect with a slight modification.
string='foo...'
re='<([^>]+)>'
while [[ $string =~ $re(.*) ]]; do
string=${BASH_REMATCH[2]}
# process as before
done
This matches the regex we want and also everything in the string after the regex. We keep shortening $string by assigning only the after-our-regex portion to it on every iteration. On the last iteration, ${BASH_REMATCH[2]} will be empty so the loop will terminate.
i have some regex that is behaving oddly in my shell script i have variables, and i have tried every what way to get them to behave, and they dont seem to do any regex, and i know my regex quite well thanks to regex101, here is what a sample looks like
fname="direcheck"
FIND="*"
if [[ $fname =~ $FIND ]]; then
echo "no quotes"
fi
if [[ "$fname" =~ "$FIND" ]]; then
echo "with quotes"
fi
right now it will display nothing
if i change find to
FIND="[9]*"
then it prints no quotes
if i say
FIND="[a-z]*"
then it prints no quotes
if i say
FIND="dircheck"
then nothing prints
if i say
FIND="*ck"
then nothing prints
I don't get how this regex is working
how do i use these variables, and what is the proper syntax?
* and *ck are invalid regular expressions. It would work (with no quotes) if you were comparing with ==, not =~. If you want to use the same functionality that you get in == for them, the equivalent regexps are .* and .*ck.
[9]* is any number (including zero) of characters that are 9. There is zero characters 9 in your direcheck, so it matches. (Edited from brainfart, thanks chepner)
dircheck is not found in direcheck, so not printing anything is hardly surprising.
[a-z]* is any number of characters that are between a and z (i.e. any number of lowercase letters). This will match, assuming it's not quoted.
I finally figured it out, and why it was working so oddly
[a-z]* and [9]* and [anythinghere]* they all match because it matches zero or more times. so "direcheck" has [9] zero or more times.
so
if [[ "$fname" =~ $FIND ]]; then
or
if [[ $fname =~ $FIND ]]; then
are both correct, and
if [[ "$fname" =~ "$FIND" ]]; then
matches only when the string matches exactly because $FIND is matched as a literal string not regex
I need to match a string $str that contains any of
foo{77..93}
and capture the above substring in a variable.
So far I've got:
str=/random/string/containing/abc-foo78_efg/ # for example
if [[ $str =~ (foo[7-9][0-9]) ]]; then
id=${BASH_REMATCH[1]}
fi
echo $id # gives foo78
but this also captures ids outside of the target range (e.g. foo95).
Is there a way to restrict the regex to an exact integer range? (tried foo[77-93] but that doesn't work.
Thanks
If you want to use a regex, you're going to have to make it slightly more complex:
if [[ $str =~ foo(7[7-9]|8[0-9]|9[0-3]) ]]; then
id=${BASH_REMATCH[0]}
fi
Note that I have removed the capture group around the whole pattern and am now using the 0th element of the match array.
As an aside, for maximum compatibility with older versions of bash, I would recommend assigning the pattern to a variable and using in the test like this:
re='foo(7[7-9]|8[0-9]|9[0-3])'
if [[ $str =~ $re ]]; then
id=${BASH_REMATCH[0]}
fi
An alternative to using a regex would be to use an arithmetic context, like this:
if (( "${str#foo}" >= 77 && "${str#foo}" <= 93 )); then
id=$str
fi
This strips the "foo" part from the start of the variable so that the integer part can be compared numerically.
Sure is easy to do with Perl:
$ echo foo{1..100} | tr ' ' '\n' | perl -lne 'print $_ if m/foo(\d+)/ and $1>=77 and $1<=93'
foo77
foo78
foo79
foo80
foo81
foo82
foo83
foo84
foo85
foo86
foo87
foo88
foo89
foo90
foo91
foo92
foo93
Or awk even:
$ echo foo{1..100} | tr ' ' '\n' | awk -F 'foo' '$2>=77 && $2<=93
{print}'
foo77
foo78
foo79
foo80
foo81
foo82
foo83
foo84
foo85
foo86
foo87
foo88
foo89
foo90
foo91
foo92
foo93
I'm pretty new to bash scripting and regexp and have a question.
I want to check to see if my variable $name starts with a-d, e-h, i-l etc and do some stuff accordingly. If the string starts with "the." or "The." it should check the first letter after the period.
My problem is that if $name consists of "the.anchor" both the a-d0-9 and q-t will be true. Do you guys have any idea what's wrong?
if [[ $name =~ ^([tT]he\.)?[a-dA-D0-9]+ ]]; then
do some stuff
fi
if [[ $name =~ ^([tT]he\.)?[e-hE-H]+ ]]; then
do some stuff
fi
if [[ $name =~ ^([tT]he\.)?[i-lI-L]+ ]]; then
do some stuff
fi
if [[ $name =~ ^([tT]he\.)?[m-pM-P]+ ]]; then
do some stuff
fi
if [[ $name =~ ^([tT]he\.)?[q-tQ-T]+ ]]; then
do some stuff
fi
if [[ $name =~ ^([tT]he\.)?[u-wU-W]+ ]]; then
do some stuff
fi
if [[ $name =~ ^([tT]he\.)?[x-zX-Z]+ ]]; then
do some stuff
fi
Thanks in advance!
Your first part it optional:
([tT]he\.)?
So the.anchor matches the pattern ^([tT]he\.)?[a-dA-D0-9]+ because the the. matches `^([tT]he\.)? and the a matches [a-dA-D0-9]+. It matches ^([tT]he\.)?[q-tQ-T]+ because ^([tT]he\.)? is optional an t matches [q-tQ-T]+. Note not the whole input is consumed by the second pattern, in fact only the first character is grabbed.
You can verify this by having bash echo the match:
echo "${BASH_REMATCH[0]}"
Which should print the.anchor in the first case and t in the second.
You do not have an end anchor on the pattern so only part of the input needs to be matched. If you made the second pattern ^([tT]he\.)?[q-tQ-T]+$ then it would not match.
Alternatively you could make the the first part possessive - ^([tT]he\.)?+. This will mean that if the engine matches the first expression it will not be unmatched. In the latter case ^([tT]he\.)?+ will grab the the. and then not release it when [q-tQ-T]+ fails; this will cause the match to fail.
I figured out a way to fix my problem by using elif statements and putting the q-t part as the last one
I think the ? can be removed as the if statement is already doing the test. The + matches the preceding item at least once and would only be needed if you want to match more than one instance of the letters.
You can do it like this:
if [[ $name =~ ^[tT]he\.[a-dA-D0-9] ]]; then
do some stuff
fi
The condition will only return true if the first character after ^[tT]he\. is [a-dA-D0-9].
However, I tend to think case is a cleaner solution than if statements when matching lists of characters against variables.
case $name in
[tT]he\.[a-dA-D0-9]*)
do some stuff
;;
esac
I have tried stuff like =~ "\[[A-Za-z0-9]+\]" which I would expect would work but doesnt. I also tried "[[A-Za-z0-9]+]" and "\[[:alnum:]+\]". What am I doing wrong? Sample line I want to match: [RTNUT18] (I am iterating through a file, some lines are of this form)
This is my code snippet:
while read line;
do
if [[ $line =~ "^\[[A-Za-z0-9]+\]$" ]]; then
echo match
else
echo no match
fi
done < $1
This is a sample file:
[RBPAT7]
Whatever=foo,bla
Otherline
RRR
and I run:
./script.sh thefile.txt
I am not getting a hit on the [RBPAT7] line at all
Stuff like that isn't enough. You must use it in [[.
$ [[ [foo] =~ ^\[[A-Za-z0-9]+\]$ ]] ; echo $?
0
EDIT:
Unlike test, [[ does not need quotes around its arguments. Your code matches nothing, since you can't have " before the beginning of the line, nor " after the end. Remove the quotes.