if [[ "$len" -lt "$MINLEN" && "$line" =~ \[*\.\] ]]
This is from Advanced bash scripting guide "Example 10-1. Inserting a blank line between paragraphs in a text file"
As I understand this matches "any string or a dot character". Right ?
It matches zero or more open bracket characters (\[*), followed by a period and a close square bracket (\.\]). Note that it only requires that a match exist somewhere in "$line", not that the whole string match. Here's a demo:
$ showmatch() { [[ "$1" =~ \[*\.\] ]] && echo "matched: '${BASH_REMATCH[0]}'" || echo "no match"; }
$ showmatch "abc[.]def"
matched: '[.]'
$ showmatch "abc.]def"
matched: '.]'
$ showmatch "abc[[[[[[[.]def"
matched: '[[[[[[[.]'
$ showmatch "abc[[[[[[[xyz.]def"
matched: '.]'
$ showmatch "abc[[[[[[[.xyz]def"
no match
...and I'm pretty sure that's not what it's supposed to be doing in that example script.
It means any string ended with dot inside bracers, for example: [.]
[abc.]
Update: +1 to Gordon Davisson, who has summed it up pretty well... so I've redacted my original post
In brief: You can test the result of a bash regex match like this:
[[ "[*.]" =~ \[*\.\] ]] ; echo ${BASH_REMATCH[0]}
Related
I'm trying to find the word "PASS_MAX_DAYS" in a file using grep
grep "^PASS_MAX_DAYS" /etc/login.defs
then I save it in a variable and compare it to a regular expression that has the value 90 or less.
regex = "PASS_MAX_DAYS\s*([0-9]|[1-8][0-9]|90)"
grep output is: PASS_MAX_DAYS 120
so my function should print a fail, however it matches:
function audit_Control () {
if [[ $cmd =~ $regex ]]; then
echo match
else
echo fail
fi
}
cmd=`grep "^PASS_MAX_DAYS" /etc/login.defs`
regex="PASS_MAX_DAYS\s*([0-9]|[1-8][0-9]|90)"
audit_Control "$cmd" "$regex"
The problem is that the bash test [[ using the regex match operator =~ does not support the common escapes such as \s for whitespace or \W for non-word-characters.
It does support posix predefined character classes, so you can use [[:space:]] in place of \s
Your regex would then be:
regex="PASS_MAX_DAYS[[:space:]]*([0-9]|[1-8][0-9]|90)"
You may want to add anchors ^ and $ to ensure a whole-line match, then the regex is
regex="^PASS_MAX_DAYS[[:space:]]*([0-9]|[1-8][0-9]|90)$"
Without the end-of-line anchor you could match lines that have trailing numbers after the match, so PASS_MAX_DAYS 9077 would match PASS_MAX_DAYS 90 and the trailing "77" would not prevent the match.
This answer also has some very useful information about bash's [[ ]] construction with the =~ operator.
I believe you have a problem with your regex, please try that version:
PASS_MAX_DAYS\s*([0-9]|[1-8][0-9]|90)$
[lucas#lucasmachine ~]$ cat test.sh
#!/bin/bash
function audit_Control () {
if [[ $cmd =~ $regex ]]; then
echo match
else
echo fail
fi
}
regex="PASS_MAX_DAYS\s*([0-9]|[1-8][0-9]|90)$"
audit_Control "$cmd" "$regex"
[lucas#lucasmachine ~]$ export cmd="PASS_MAX_DAYS 123"
[lucas#lucasmachine ~]$ ./test.sh
fail
[lucas#lucasmachine ~]$ export cmd="PASS_MAX_DAYS 1"
[lucas#lucasmachine ~]$ ./test.sh
match
I can explain the problem was the other regex was not checking the end of line, so, your were matching "PASS_MAX_DAYS 1"23 so 23 were not being "counted" to your regex. Your regex was really matching part of the text.. Now with the end of line it should match exactly 1 digit find a end of line, or [1-8][0-9] end of line or 90 end of line.
For the same regex applied to the same string, why does grep -E match, but the Bash =~ operator in [[ ]] does not?
$ D=Dw4EWRwer
$ echo $D|grep -qE '^[A-Z][A-Za-z0-9]{1,2}[[:alnum:]_-\ ]{1,22}$' || echo wrong pattern
$ [[ "${D}" =~ ^[A-Z][A-Za-z0-9]{1,2}[[:alnum:]_-\ ]{1,22}$ ]] || echo wrong pattern
wrong pattern
Update: I confirm this worked:
[[ "${D}" =~ ^[A-Z][A-Za-z0-9]{1,2}[[:alnum:]\ _-]{1,22}$ ]] || echo wrong pattern
The problem (for both versions of the code) is on this character class:
[[:alnum:]_-\ ]
In the grep version, because the regex is enclosed in single quotes, the backslash doesn't escape anything and the character range received by grep is exactly how it is represented above.
In the bash version, the backslash (\) escapes the space that follows it and the actual character class used by [[ ]] to test is [[:alnum:]_- ].
Because in ASCII table the underscore (_) comes after both space () and backslash (\), neither of these character classes is correct.
For the bash version you can use:
[[ "${D}" =~ ^[A-Z][A-Za-z0-9]{1,2}[[:alnum:]_-\ ]{1,22}$ ]]; echo $?
to verify its outcome. If the regex is incorrect, the exit code is 2.
If you want to put a dash (-) into a character class you have to put it either as the first character in the class (just after [ or [^ if it is a negating class) or as the last character in the class (right before the closing]`).
The grep version of the code should be (there is no need to escape anything inside a string enclosed in single quotes):
$ echo $D | grep -qE '^[A-Z][A-Za-z0-9]{1,2}[[:alnum:]_ -]{1,22}$' || echo wrong pattern
The bash version of your code should be:
[[ "${D}" =~ ^[A-Z][A-Za-z0-9]{1,2}[[:alnum:]_\ -]{1,22}$ ]] || echo wrong pattern
Based on your comment, you want the bracket expression to contain alphanumeric characters, spaces, underscores and dashes, so the dash is not supposed to indicate a range. To add a hyphen to a bracket expression, it has to be the first or last character in it. Additionally, you don't have to escape things in bracket expressions, so you can drop the backslash. Your grep regex includes a literal \ in the bracket expression:
$ grep -q '[\]' <<< '\' && echo "Match"
Match
In the Bash regex, the space has to be escaped because the string is first read by the shell, but see below how to avoid that.
First, fixing your regex:
^[A-Z][A-Za-z0-9]{1,2}[[:alnum:]_ -]{1,22}$
The backslash is gone, and the hyphen is moved to the end. Using this with grep works fine:
$ D=Dw4EWRwer
$ grep -E '^[A-Z][A-Za-z0-9]{1,2}[[:alnum:]_ -]{1,22}$' <<< "$D"
Dw4EWRwer
To use the regex within [[ ]] directly, the space has to be escaped:
$ [[ $D =~ ^[A-Z][A-Za-z0-9]{1,2}[[:alnum:]_\ -]{1,22}$ ]] && echo "Match"
Match
I would make the following changes:
Use character classes where possible: [A-Z] is [[:upper:]], [A-Za-z0-9] is [[:alnum:]]
Store the regex in a variable for usage in [[ ]]; this has two advantages: no escaping characters special to the shell, and compatibility with older Bash versions, as the quoting requirements changed between 3.1 and 3.2 (see the Patterns article in the BashGuide).
The regex would then become this for grep:
$ grep -E '^[[:upper:]][[:alnum:]][[:alnum:]_ -]{1,22}$' <<< "$D"
Dw4EWRwer
and this in Bash:
$ re='^[[:upper:]][[:alnum:]][[:alnum:]_ -]{1,22}$'
$ [[ $D =~ $re ]] && echo "Match"
Match
I'm trying to remove colouring codes from a string; e.g. from: \033[36;1mDISK\033[0m to: DISK
my regex looks like this: \033.*?m so match '\033' followed by any number of chars, terminated by 'm'
when I search for the pattern, it finds a match; [[ "$var" =~ $regex ]] evaluates to true
however when I try to replace matches, nothing happens and the same string is returned.
Here's my complete script:
regex="\033.*?m"
var="\033[36;1mDISK\033[0m"
if [[ "$var" =~ $regex ]]
then
echo "matches"
echo ${var//$regex}
else
echo "doesn't match!"
fi
The problem appears to be with the match any number of any character part of the regex. I can successfully replace DISK but if I change that to D.*K or D.*?K it fails.
Note in all above cases the pattern claims to match the string but fails when replacing. Not too sure where to go with this now, any help appreciated.
Thanks
The following should do it:
$ var="\033[36;1mDISK\033[0m"
$ newvar=$(printf ${var} | sed -r "s/\x1B\[([0-9]{1,2}(;[0-9]{1,2})?)?[m|K]//g")
$ echo ${newvar}
returns:
DISK
Now verify!
$ echo $var | od
0000000 030134 031463 031533 035466 066461 044504 045523 030134
0000020 031463 030133 005155
0000026
$ echo $newvar | od
0000000 044504 045523 000012
0000005
To use the parameter expansion substitution operator, you need to use an extended glob.
shopt -s extglob
newvar=${var//\\033\[*([0-9;])m}
To break it down:
\\033\[ - match the encoded escape character and [.
*([0-9;]) - match zero or more digits or semicolons. You could use +([0-9;]) to (more correctly?) match one or more digits or semicolons
m - the trailing m.
I'm sure this is a simple oversight, but I don't see it, and I'm not sure why this regex is matching more than it should:
#!/bin/bash
if [[ $1 =~ ([0-9]+,)+[0-9]+ ]]; then
{
echo "found list of jobs"
}
fi
This is with input that looks like "02,48,109,309,183". Matching that is fine
However, it is also matching input that has no final number and is instead "09,28,34,"
Should the [0-9]+ at the end dictate the final character be at least 1+ numbers?
You have to add markers for beginning (^) and end ($) of input:
#!/bin/bash
if [[ $1 =~ ^([0-9]+,)+[0-9]+$ ]]; then
echo "found list of jobs"
fi
Otherwise it matches 09,28,34, because it matches from 0 until 4, ignoring everything that follows.
Your regex only has to match somewhere in the string, not from start to end. To make it match the whole string, use the ^ and $ meta-characters:
#!/bin/bash
if [[ $1 =~ ^([0-9]+,)+[0-9]+$ ]]; then
echo "found list of jobs"
fi
(Incidentally, you don't need { and } to define a block in Bash, that's the job of then and fi)
In bash, I am trying to match valid attributes that are present in an array. Attributes may be 'disabled' by preceding them with a bang (exclamation mark, !), in which case they must not be matched. I have this:
[[ ${TESTS[#]} =~ [^\!]match ]]
which will return true if the word 'match' is in TESTS and not preceded by a !.
It works, except when the word match is in the first position in the array. The problem is the regexp is saying 'match preceded by something that isn't a !'. When it's the first item it is preceded by nothing and therefore does not match.
How do I modify the above to say 'match not preceded by !' ?
From reading answers to other questions I have tried (?<!!)match but this does not work.
Use this re:
([^\!]|^)match
Example of usage:
$ [[ match =~ (^|[^\!])match ]] && echo matches || echo "doesn't match"
matches
$ [[ xmatch =~ (^|[^\!])match ]] && echo matches || echo "doesn't match"
matches
$ [[ '!match' =~ (^|[^\!])match ]] && echo match || echo "doesn't match"
doesn't match
In general, it would be also correct to use assertions here, but bash uses POSIX regular expressions and they know nothing about assertions. But with grep (GNU grep), or perl, or anything that supports PCRE you can do it:
$ echo match | grep -qP '(?<!!)match' && echo matches || echo "doesn't match"
matches
$ echo xmatch | grep -qP '(?<!!)match' && echo matches || echo "doesn't match"
matches
$ echo '!match' | grep -qP '(?<!!)match' && echo matches || echo "doesn't match"
doesn't match