I have a value in a variable that may be absolute or relative url, and I need to check which one it is.
I have found that there's a =~ operator in [[, but I can't get it to work. What am I doing wrong?
url="http://test"
if [[ "$url" =~ "^http://" ]];
then echo "absolute.";
fi;
You need to use regex without quote:
url="http://test"
if [[ "$url" =~ ^http:// ]]; then
echo "absolute."
fi
This outputs `absolute. as regex needs to be without quote in newer BASH (after BASH v3.1)
Or avoid regex and use glob matching:
if [[ "$url" == "http://"* ]]; then
echo "absolute."
fi
Related
From a variable how to extract integers that will be in format *\d+.\d+.\d+* (4.12.3123) using bash.
filename="xzxzxzxz4.12.3123fsfsfsfsfsfs"
I have tried:
filename="xzxzxzxz4.12.3123fsfsfsfsfsfs"
if [[ "$filename" =~ (.*)(\d+.\d+.\d+)(.*) ]]; then
echo ${BASH_REMATCH}
echo ${BASH_REMATCH[1]}
echo ${BASH_REMATCH[2]}
echo ${BASH_REMATCH[3]}
else
echo 'nej'
fi
which does not work.
The easiest way to work with regexes in Bash, in terms of consistency between Bash versions and escaping, is to put the regex into a single-quoted variable and then use it unquoted, as below:
re='[0-9]+\.[0-9]+\.[0-9]+'
[[ $filename =~ $re ]] && printf '%s\n' "${BASH_REMATCH[#]}"
The main issue with your approach were that you were using the "Perl-style" \d, so in fact you could make your code work with:
if [[ "$filename" =~ (.*)([0-9]+\.[0-9]+\.[0-9]+)(.*) ]]; then
echo "${BASH_REMATCH[2]}"
fi
But this unnecessarily creates 3 capture groups, when you don't even need one. Note that I also changed . (any character) to \. (a literal .).
one way to extract:
grep -oP '\d\.\d+\.\d+' <<<$xfilename
There is one more way
$ filename="xzxzxzxz4.12.3123fsfsfsfsfsfs"
$ awk '{ if (match($0, /[0-9].[0-9]+.[0-9]+/, m)) print m[0] }' <<< "$filename"
4.12.3123
I have a file (file.txt) that contains some text like:
000000000+000+0+00
000000001+000+0+00
000000002+000+0+00
and I am trying to check each line to make sure that it follows the format:
character*9, "+", character*3, "+", etc
so far I have:
#!/bin/ksh
file=file.txt
line_number=1
for line in $(cat $file)
do
if [[ "$line" != "[[.]]{9}+[[.]]{3}+[[.]]{1}+[[.]]{2} ]" ]]
then
echo "Invalid number ($line) check line $line_number"
exit 1
fi
let "line_number++"
done
however this does not evaluate correctly, no matter what I put in the lines the program terminates.
When you want line numbers of the mismatches, you can use grep -vn. Be careful with writing a correct regular expression, and you will have
grep -Evn "^.{9}[+].{3}[+].[+].{2}$" file.txt
This is not in the layout that you want, so change the layout with sed:
grep -Evn "^.{9}[+].{3}[+].[+].{2}$" file.txt |
sed -r 's/([^:]*):(.*)/Invalid number (\2) check line number \1./'
EDIT:
I changed .{1} into ..
The sed is also over the top. When you need spme explanation, you can start with echo "Linenr:Invalid line"
I'm having funny results putting the regex in the condition directly:
$ line='000000000+000+0+00'
$ [[ $line =~ ^.{9}\+.{3}\+.\+..$ ]] && echo ok
ksh: syntax error: `~(E)^.{9}\+.{3}\+.\+..$ ]] && echo ok
' unexpected
But if I save the regex in a variable:
$ re="^.{9}\+.{3}\+.\+..$"
$ [[ $line =~ $re ]] && echo ok
ok
So you can do
#!/bin/ksh
file=file.txt
line_number=1
re="^.{9}\+.{3}\+.\+..$"
while IFS= read -r line; do
if [[ ! $line =~ $re ]]; then
echo "Invalid number ($line) check line $line_number"
exit 1
fi
let "line_number++"
done < "$file"
You can also use a plain glob pattern:
if [[ $line != ?????????+???+?+?? ]]; then echo error; fi
ksh glob patterns have some regex-like syntax. If there's an optional space in there, you can handle that with the ?(sub-pattern) syntax
pattern="?????????+???+?( )?+??"
line1="000000000+000+0+00"
line2="000000000+000+ 0+00"
[[ $line1 == $pattern ]] && echo match || echo no match # => match
[[ $line2 == $pattern ]] && echo match || echo no match # => match
Read the "File Name Generation" section of the ksh man page.
Your regex looks bad - using sites like https://regex101.com/ is very helpful. From your description, I suspect it should look more like one of these;
^.{9}\+.{3}\+.{1}\+.{2}$
^[^\+]{9}\+[^\+]{3}\+[^\+]{1}\+[^\+]{2}$
^[0-9]{9}\+[0-9]{3}\+[0-9]{1}\+[0-9]{2}$
From the ksh manpage section on [[ - you would probably want to be using =~.
string =~ ere
True if string matches the pattern ~(E)ere where ere is an extended regular expression.
Note: As far as I know, ksh regex doesn't follow the normal syntax
You may have better luck with using grep:
# X="000000000+000+0+00"
# grep -qE "^[^\+]{9}\+[^\+]{3}\+[^\+]{1}\+[^\+]{2}$" <<<"${X}" && echo true
true
Or:
if grep -qE "^[^\+]{9}\+[^\+]{3}\+[^\+]{1}\+[^\+]{2}$" <<<"${line}"
then
exit 1
fi
You may also prefer to use a construct like below for handling files:
while read line; do
echo "${line}";
done < "${file}"
In a list of files:
javascript-custom-rules-plugin-1.0-SNAPSHOT.jar
README.txt
sonar-build-breaker-plugin-2.0.jar
sonar-javascript-plugin-2.11.jar
tmo-custom-rules-1.0.jar
I am attempting to match these filenames by regex.
My Script
#!/usr/bin/env bash
install_location=/usr/local/sonar/extensions/plugins
for f in $(ls -1 $install_location)
do
# remove any previous versions of this plugin
if [[ "$f" =~ ".*tmo-custom-rules-(.+)\.jar" ]]
then
echo "found $f. will remove"
else
echo "$f doesn't match"
fi
done
I've tried if [[ "$f" =~ ".*tmo-custom-rules-(.+)\.jar" ]] and if [[ "$f" == *"tmo-custom-rules" ]] to no avail.
I'm getting
javascript-custom-rules-plugin-1.0-SNAPSHOT.jar doesn't match
README.txt doesn't match
sonar-build-breaker-plugin-2.0.jar doesn't match
sonar-javascript-plugin-2.11.jar doesn't match
tmo-custom-rules-1.0.jar doesn't match
when I expect found tmo-custom-rules-1.0.jar. will remove
I've run my regular expression through many regular expression testers with the data above, and they all return the correct matches, but I can't get it to work here in my script.
How can I loop through, and check to see if any files matches this regular expression?
In BASH regex must be unquoted so this should work:
[[ $f =~ .*tmo-custom-rules-(.+)\.jar ]]
Or better:
re=".*tmo-custom-rules-(.+)\.jar"
[[ $f =~ $re ]]
However you don't even need regex and can use shell glob matching:
#!/usr/bin/env bash
install_location=/usr/local/sonar/extensions/plugins
for f in "$install_location"/*
do
# remove any previous versions of this plugin
if [[ $f == *tmo-custom-rules-*.jar ]]
then
echo "found $f. will remove"
else
echo "$f doesn't match"
fi
done
Note that you can avoid using output of ls which is not always fit for scripting.
You can do this with expr using the colon operator:
if expr "$f" : '.*tmo-custom-rules-.*\.jar' > /dev/null; then
echo matches
fi
Note that the regular expressions in this context are assumed to be anchored to the start of the line.
I have a regular expression:
^(.+?)(\.[^.]+$|$)
which separates a file name and the file extension (if there is one)
http://movingtofreedom.org/2008/04/01/regex-match-filename-base-and-extension/
Works perfectly fine in Perl
Say $FILE ='.myfile.form.txt'
$1 is '.myfile.form' and
$2 is '.txt', as they should be
I know Bash regex and Perl regex aren't the same, but I've never had a problem with Bash Rematching until now
But when I try to use in in a Bash script as, say...
FILE='.myfile.form.txt'
[[ $FILE =~ ^(.+?)(\.[^.]+$|$) ]]
${BASH_REMATCH[1]} will just have the entire file name (.myfile.form.txt), and nothing in ${BASH_REMATCH[2]}
I'm wondering what's wrong/going on here
Thanks for any help!
regex(7) which is referenced by regex(3) which is referenced by bash(1) makes no mention of greediness modifiers. Your pattern cannot be implemented in bash regex.
This doesn't mean you can't achieve what you want, though.
[[ $FILE =~ ^(.+)(\.[^.]*)$ ]] || [[ $FILE =~ ^(.*)()$ ]]
file="${BASH_REMATCH[1]}"
ext="${BASH_REMATCH[2]}"
Or something more straightforward like
if [[ $FILE =~ ^(.+)(\.[^.]*)$ ]]; then
file="${BASH_REMATCH[1]}"
ext="${BASH_REMATCH[2]}"
else
file="$FILE"
ext=""
fi
What is the correct way to escape a dollar sign in a bash regex? I am trying to test whether a string begins with a dollar sign. Here is my code, in which I double escape the dollar within my double quotes expression:
echo -e "AB1\nAB2\n\$EXTERNAL_REF\nAB3" | while read value;
do
if [[ ! $value =~ "^\\$" ]];
then
echo $value
else
echo "Variable found: $value"
fi
done
This does what I want for one box which has:
GNU bash, version 3.2.25(1)-release (x86_64-redhat-linux-gnu)
And the verbose output shows
+ [[ ! $EXTERNAL_REF =~ ^\$ ]]
+ echo 'Variable found: $EXTERNAL_REF'
However, on another box which uses
GNU bash, version 4.1.2(1)-release (x86_64-redhat-linux-gnu)
The comparison is expanded as follows
+ [[ ! $EXTERNAL_REF =~ \^\\\$ ]]
+ echo '$EXTERNAL_REF'
Is there a standard/better way to do this that will work across all implementations?
Many thanks
Why do you use a regular expression here? A glob is enough:
#!/bin/bash
while read value; do
if [[ "$value" != \$* ]]; then
echo "$value"
else
echo "Variable found: $value"
fi
done < <(printf "%s\n" "AB1" "AB2" '$EXTERNAL_REF' "AB3")
Works here with shopt -s compat32.
The regex doesn't need any quotes at all. This should work:
if [[ ! $value =~ ^\$ ]];
I would replace the double quotes with single quotes and remove a single \ and have the changes as below
$value =~ "^\\$"
can also be used as
$value =~ '^\$'
I never found the solution either, but for my purposes, I settled on the following workaround:
if [[ "$value" =~ ^(.)[[:alpha:]_][[:alnum:]_]+\\b && ${BASH_REMATCH[1]} == '$' ]]; then
echo "Variable found: $value"
else
echo "$value"
fi
Rather than trying to "quote" the dollar-sign, I instead match everything around it and I capture the character where the dollar-sign should be to do a direct-string comparison on. A bit of a kludge, but it works.
Alternatively, I've taken to using variables, but just for the backslash character (I don't like storing the entire regex in a variable because I find it confusing for the regex to not appear in the context where it's used):
bs="\\"
string="test\$test"
if [[ "$string" =~ $bs$ ]]; then
echo "output \"$BASH_REMATCH\""
fi