Shell: Checking if argument exists and matches expression - regex

I'm new to shell scripting and trying to write the ability to check if an argument exists and if it matches an expression. I'm not sure how to write expressions, so this is what I have so far:
#!/bin/bash
if [[ -n "$1"] && [${1#*.} -eq "tar.gz"]]; then
echo "Passed";
else
echo "Missing valid argument"
fi
To run the script, I would type this command:
# script.sh YYYY-MM.tar.gz
I believe what I have is
if the YYYY-MM.tar.gz is not after script.sh it will echo "Missing valid argument" and
if the file does not end in .tar.gz it echo's the same error.
However, I want to also check if the full file name is in YYYY-MM.tar.gz format.

if [[ -n "$1" ]] && [[ "${1#*.}" == "tar.gz" ]]; then
-eq: (equal) for arithmetic tests
==: to compare strings
See: help test

You can also use:
case "$1" in
*.tar.gz) ;; #passed
*) echo "wrong/missing argument $1"; exit 1;;
esac
echo "ok arg: $1"

As long as the file is in the correct YYYY-MM.tar.gz format, it obviously is non-empty and ends in .tar.gz as well. Check with a regular expression:
if ! [[ $1 =~ [0-9]{4}-[0-9]{1,2}.tar.gz ]]; then
echo "Argument 1 not in correct YYYY-MM.tar.gz format"
exit 1
fi
Obviously, the regular expression above is too general, allowing names like 0193-67.tar.gz. You can adjust it to be as specific as you need it to be for your application, though. I might recommend
[1-9][0-9]{3}-([1-9]|10|11|12).tar.gz
to allow only 4-digit years starting with 1000 (support for the first millennium ACE seems unnecessary) and only months 1-12 (no leading zero).

Related

Bash script with regex and capturing group

I'm working on a bash script to rename automatically files on my Synology NAS.
I have a loop for the statement of the files and everything is ok until I want to make my script more efficient with regex.
I have several bits of code which are working like as expected:
filename="${filename//[-_.,\']/ }"
filename="${filename//[éèēěëê]/e}"
But I have this:
filename="${filename//t0/0}"
filename="${filename//t1/1}"
filename="${filename//t2/2}"
filename="${filename//t3/3}"
filename="${filename//t4/4}"
filename="${filename//t5/5}"
filename="${filename//t6/6}"
filename="${filename//t7/7}"
filename="${filename//t8/8}"
filename="${filename//t9/9}"
And, I would like to use captured group to have something like this:
filename="${filename//t([0-9]{1,2})/\1}"
filename="${filename//t([0-9]{1,2})/${BASH_REMATCH[1]}}"
I've been looking for a working syntax without success...
The shell's parameter expansion facility does not support regular expressions. But you can approximate it with something like
filename=$(sed 's/t\([0-9]\)/\1/g' <<<"$filename")
This will work regardless of whether the first digit is followed by additional digits or not, so dropping that requirement simplifies the code.
If you want the last or all t[0-9]{1,2}s replaced:
$ filename='abt1cdt2eft3gh'; [[ "$filename" =~ (.*)t([0-9]{1,2}.*) ]] && filename="${BASH_REMATCH[1]}${BASH_REMATCH[2]}"; echo "$filename"
abt1cdt2ef3gh
$ filename='abt1cdt2eft3gh'; while [[ "$filename" =~ (.*)t([0-9]{1,2}.*) ]]; do filename="${BASH_REMATCH[1]}${BASH_REMATCH[2]}"; done; echo "$filename"
ab1cd2ef3gh
Note that the "replace all" case above would keep iterating until all t[0-9]{1,2}s are changed, even ones that didn't exist in the original input but were being created by the loop, e.g.:
$ filename='abtt123de'; while [[ "$filename" =~ (.*)t([0-9]{1,2}.*) ]]; do filename="${BASH_REMATCH[1]}${BASH_REMATCH[2]}"; echo "$filename"; done
abt123de
ab123de
whereas the sed script in #tripleee's answer would not do that:
$ filename='abtt123de'; filename=$(sed 's/t\([0-9]\)/\1/g' <<<"$filename"); echo "$filename"
abt123de

Regex pattern that recognises file extension in Bash script not accurate to capture compressed files

I created this little Bash script that has one argument (a filename) and the script is supposed to respond according to the extension of the file:
#!/bin/bash
fileFormat=${1}
if [[ ${fileFormat} =~ [Ff][Aa]?[Ss]?[Tt]?[Qq]\.?[[:alnum:]]+$ ]]; then
echo "its a FASTQ file";
elif [[ ${fileFormat} =~ [Ss][Aa][Mm] ]]; then
echo "its a SAM file";
else
echo "its not fasta nor sam";
fi
It's ran like this:
sh script.sh filename.sam
If it's a fastq (or FASTQ, or fq, or FQ, or fastq.gz (compressed)) I want the script to tell me "it's a fastq". If it's a sam, I want it to tell me it's a sam, and if not, I want to tell me it's neither sam or fastq.
THE PROBLEM: when I didn't consider the .gz (compressed) scenario, the script ran well and gave the result I expected, but something is happening when I try to add that last part to account for that situation (see third line, the part where it says .?[[:alnum:]]+ ). This part is meant to say "in the filename, after the extension (fastq in this case), there might be a dot plus some word afterwards".
My input is this:
sh script.sh filename.fastq.gz
And it works. But if I put:
sh script.sh filename.fastq
It says it's not fastq. I wanted to put that last part as optional, but if I add a "?" at the end it doesn't work. Any thoughts? Thanks! My question would be to fix that part in order to work for both cases.
You may use this regex:
fileFormat="$1"
if [[ $fileFormat =~ [Ff]([Aa][Ss][Tt])?[Qq](\.[[:alnum:]]+)?$ ]]; then
echo "its a FASTQ file"
elif [[ $fileFormat =~ [Ss][Aa][Mm]$ ]]; then
echo "its a SAM file"
else
echo "its not fasta nor sam"
fi
Here (\.[[:alnum:]]+)? makes last group optional which is dot followed by 1+ alphanumeric characters.
When you run it as:
./script.sh filename.fastq
its a FASTQ file
./script.sh fq
its a FASTQ file
./script.sh filename.fastq.gz
its a FASTQ file
./script.sh filename.sam
its a SAM file
./script.sh filename.txt
its not fasta nor sam
The immediate problem is that you are requiring at least one [[:alnum:]] character after .fastq. This is easy to fix per se with * instead of +.
Regex is not a particularly happy solution to this problem, though.
case $fileFormat in
*.[Ff][Aa][Ss][Tt][Qq] | *.[Ff][Aa][Ss][Tt][Qq].*)
echo "$0: $fileFormat is a FASTQ file" >&2 ;;
*.[Ss][Aa][Mm] )
echo "$0: $fileFormat is a SAM file" >%2 ;;
esac
is portable all the way back to the original Bourne sh. In Bash 4.x you could lowercase the filename before the comparison so as to simplify the glob patterns.
Notice also how the diagnostics contain the name of the script and print to standard error instead of standard output.

Regular expressions don't work as expected in bash if-else block's condition

My pattern defined to match in if-else block is :
pat="17[0-1][0-9][0-9][0-9].AUG"
nln=""
In my script, I'm taking user input which needs to be matched against the pattern, which if doesn't match, appropriate error messages are to be shown. Pretty simple, but giving me a hard time though. My code block from the script is this:
echo "How many days' AUDIT Logs need to be searched?"
read days
echo "Enter file name(s)[For multiple files, one file per line]: "
for(( c = 0 ; c < $days ; c++))
do
read elements
if [[ $elements =~ $pat ]];
then
array[$c]="$elements"
elif [[ $elements =~ $nln ]];
then
echo "No file entered.Run script again. Exiting"
exit;
else
echo "Invalid filename entered: $elements.Run script again. Exiting"
exit;
fi
done
The format I want from the user for filenames to be entered is this:
170402.AUG
So basically yymmdd.AUG (where y-year,m-month,d-day), with trailing or leading spaces is fine. Anything other than that should throw "Invalid filename entered: $elements.Run script again. Exiting" message. Also I want to check if if it is a blank line with a "Enter" hit, it should give an error saying "No file entered.Run script again. Exiting"
However my code, even if I enter something like "xxx" as filename, which should be throwing "Invalid filename entered: $elements.Run script again. Exiting", is actually checking true against a blank line, and throwing "No file entered.Run script again. Exiting"
Need some help with handling the regular expressions' check with user input, as otherwise rest of my script works just fine.
I think as discussed in the comments you are confusing with the glob match and a regEx match, what you have defined as pat is a glob match which needs to be equated with the == operator as,
pat="17[0-1][0-9][0-9][0-9].AUG"
string="170402.AUG"
[[ $string == $pat ]] && printf "Match success\n"
The equivalent ~ match would be to something as
pat="17[[:digit:]]{4}\.AUG"
[[ $string =~ $pat ]] && printf "Match success\n"
As you can see the . in the regex syntax has been escaped to deprive of its special meaning ( to match any character) but just to use as a literal dot. The POSIX character class [[:digit:]] with a character count {4} allows you to match 4 digits followed by .AUG
And for the string empty check do as suggested by the comments from Cyrus, or by Benjamin.W
[[ $elements == "" ]]
(or)
[[ -z $elements ]]
I would not bug the user with how many days (who want count 15 days or like)? Also, why only one file per line? You should help the users, not bug them like microsoft...
For the start:
show_help() { cat <<'EOF'
bla bla....
EOF
}
show_files() { echo "${#files[#]} valid files entered: ${files[#]}"; }
while read -r -p 'files? (h-help)> ' line
do
case "$line" in
q) echo "quitting..." ; exit 0 ;;
h) show_help ; continue;;
'') (( ${#files} )) && show_files; continue ;;
l) show_files ; continue ;;
p) (( ${#files} )) && break || { echo "No files enterd.. quitting" ; exit 1; } ;; # go to processing
esac
# select (grep) the valid patterns from the entered line
# and append them into the array
# using the -P (if your grep know it) you can construct very complex regexes
files+=( $(grep -oP '17\d{4}.\w{3}' <<< "$line") )
done
echo "processing files ${files[#]}"
Using such logic you can build really powerful and user-friendly app. Also, you can use -e for the read enable the readline functions (cursor keys and like)...
But :) Consider just create a simple script, which accepts arguments. Without any dialogs and such. example:
myscript -h
same as above, or some longer help text
myscript 170402.AUG 170403.AUG 170404.AUG 170405.AUG
will do whatever it should do with the files. Main benefit, you could use globbing in the filenames, like
myscript 1704*
and so on...
And if you really want the dialog, it could show it when someone runs the script without any argument, e.g.:
myscript
will run in interactive mode...

bash regular expression format

My code have problem with compare var with regular expression.
The main problem is problem is here
if [[ “$alarm” =~ ^[0-2][0-9]\:[0-5][0-9]$ ]]
This "if" is never true i dont know why even if i pass to "$alarm" value like 13:00 or 08:19 its always false and write "invalid clock format".
When i try this ^[0-2][0-9]:[0-5][0-9]$ on site to test regular expressions its work for example i compered with 12:20.
I start my script whith command ./alarm 11:12
below is whole code
#!/bin/bash
masa="`date +%k:%M`"
mp3="$HOME/Desktop/alarm.mp3" #change this
echo "lol";
if [ $# != 1 ]; then
echo "please insert alarm time [24hours format]"
echo "example ./alarm 13:00 [will ring alarm at 1:00pm]"
exit;
fi
alarm=$1
echo "$alarm"
#fix me with better regex >_<
if [[ “$alarm” =~ ^[0-2][0-9]\:[0-5][0-9]$ ]]
then
echo "time now $masa"
echo "alarm set to $alarm"
echo "will play $mp3"
else
echo "invalid clock format"
exit;
fi
while [ $masa != $alarm ];do
masa="`date +%k:%M`" #update time
sleep 1 #dont overload the cpu cycle
done
echo $masa
if [ $masa = $alarm ];then
echo ringggggggg
play $mp3 > /dev/null 2> /dev/null &
fi
exit
I can see a couple of issues with your test.
Firstly, it looks like you may be using the wrong kind of double quotes around your variable (“ ”, rather than "). These "fancy quotes" are being concatenated with your variable, which I assume is what causes your pattern to fail to match. You could change them but within bash's extended tests (i.e. [[ instead of [), there's no need to quote your variables anyway, so I would suggest removing them entirely.
Secondly, your regular expression allows some invalid dates at the moment. I would suggest using something like this:
re='^([01][0-9]|2[0-3]):[0-5][0-9]$'
if [[ $alarm =~ $re ]]
I have deliberately chosen to use a separate variable to store the pattern, as this is the most widely compatible way of working with bash regexes.

Checking a string to see if it contains numeric character in UNIX

I'm new to UNIX, having only started it at work today, but experienced with Java, and have the following code:
#/bin/bash
echo "Please enter a word:"
read word
grep -i $word $1 | cut -d',' -f1,2 | tr "," "-"> output
This works fine, but what I now need to do is to check when word is read, that it contains nothing but letters and if it has numeric characters in print "Invalid input!" message and ask them to enter it again. I assumed regular expressions with an if statement would be the easy way to do this but I cannot get my head around how to use them in UNIX as I am used to the Java application of them. Any help with this would be greatly appreciated, as I couldn't find help when searching as all the solutions with regular expressions in linux I found only dealt with if it was either all numeric or not.
Yet another approach. Grep exits with 0 if a match is found, so you can test the exit code:
echo "${word}" | grep -q '[0-9]'
if [ $? = 0 ]; then
echo 'Invalid input'
fi
This is /bin/sh compatible.
Incorporating Daenyth and John's suggestions, this becomes
if echo "${word}" | grep '[0-9]' >/dev/null; then
echo 'Invalid input'
fi
The double bracket operator is an extended version of the test command which supports regexes via the =~ operator:
#!/bin/bash
while true; do
read -p "Please enter a word: " word
if [[ $word =~ [0-9] ]]; then
echo 'Invalid input!' >&2
else
break
fi
done
This is a bash-specific feature. Bash is a newer shell that is not available on all flavors of UNIX--though by "newer" I mean "only recently developed in the post-vacuum tube era" and by "not all flavors of UNIX" I mean relics like old versions of Solaris and HP-UX.
In my opinion this is the simplest option and bash is plenty portable these days, but if being portable to old UNIXes is in fact important then you'll need to use the other posters' sh-compatible answers. sh is the most common and most widely supported shell, but the price you pay for portability is losing things like =~.
If you're trying to write portable shell code, your options for string manipulation are limited. You can use shell globbing patterns (which are a lot less expressive than regexps) in the case construct:
export LC_COLLATE=C
read word
while
case "$word" in
*[!A-Za-z]*) echo >&2 "Invalid input, please enter letters only"; true;;
*) false;;
esac
do
read word
done
EDIT: setting LC_COLLATE is necessary because in most non-C locales, character ranges like A-Z don't have the “obvious” meaning. I assume you want only ASCII letters; if you also want letters with diacritics, don't change LC_COLLATE, and replace A-Za-z by [:alpha:] (so the whole pattern becomes *[![:alpha:]]*).
For full regexps, see the expr command. EDIT: Note that expr, like several other basic shell tools, has pitfalls with some special strings; the z characters below prevent $word from being interpreted as reserved words by expr.
export LC_COLLATE=C
read word
while expr "z$word" : 'z[A-Za-z]*$' >/dev/null; then
echo >&2 "Invalid input, please enter letters only"
read word
fi
If you only target recent enough versions of bash, there are other options, such as the =~ operator of [[ ... ]] conditional commands.
Note that your last line has a bug, the first command should be
grep -i "$word" "$1"
The quotes are because somewhat counter-intuitively, "$foo" means “the value of the variable called foo” whereas plain $foo means “take the value of foo, split it into separate words where it contains whitespace, and treat each word as a globbing pattern and try to expand it”. (In fact if you've already checked that $word contains only letters, leaving the quotes won't do any harm, but it takes more time to think of these special cases than to just put the quotes every times.)
Yet another (quite) portable way to do it ...
if test "$word" != "`printf "%s" "$word" | tr -dc '[[:alpha:]]'`"; then
echo invalid
fi
One portable (assuming bash >= 3) way to do this is to remove all numbers and test for length:
#!/bin/bash
read -p "Enter a number" var
if [[ -n ${var//[0-9]} ]]; then
echo "Contains non-numbers!"
else
echo "ok!"
fi
Coming from Java, it's important to note that bash has no real concept of objects or data types. Everything is a string, and complex data structures are painful at best.
For more info on what I did, and other related functions, google for bash string manipulation.
Playing around with Bash parameter expansion and character classes:
# cf. http://wiki.bash-hackers.org/syntax/pe
word="abc1def"
word="abc,def"
word=$'abc\177def'
# cf. http://mywiki.wooledge.org/BashFAQ/058 (no NUL byte in Bash variable)
word=$'abc\000def'
word="abcdef"
(
set -xv
[[ "${word}" != "${word/[[:digit:]]/}" ]] && echo invalid || echo valid
[[ -n "${word//[[:alpha:]]/}" ]] && echo invalid || echo valid
)
Everyone's answers seem to be based on the fact that the only invalid characters are numbers. The initial questions states that they need to check that the string contains "nothing but letters".
I think the best way to do it is
nonalpha=$(echo "$word" | sed 's/[[:alpha:]]//g')
if [[ ${#nonalpha} -gt 0 ]]; then
echo "Invalid character(s): $nonalpha"
fi
If you found this page looking for a way to detect non-numeric characters in your string (like I did!) replace [[:alpha:]] with [[:digit:]].