Regex word boundaries in double bracket version of test - regex

It seems that I can't get word boundaries to work in [[:
$ echo foo | md5sum
d3b07384d113edec49eaa6238ad5ff00 -
$ [[ "$(echo foo | md5sum)" =~ ^d3b07384d113edec49eaa6238ad5ff00 ]] && echo ok
ok
$ [[ "$(echo foo | md5sum)" =~ ^d3b07384d113edec49eaa6238ad5ff00\b ]] && echo ok
$ ## no output
Are word boundaries not accepted in [[? Or am I missing something?

This seems to work, although it's a bit verbose:
[[ "$(echo foo | md5sum)" =~ $(echo '^d3b07384d113edec49eaa6238ad5ff00\b') ]] && echo ok

The problem seems related to the backslash losing the meaning you intend when interpreted by the shell. There's probably some incantation of quoting that would eliminate the issue, but for me it's sometimes just easier to dump the output of a construct into Perl for further processing.
If you can accept a solution that invokes Perl on your system, this works:
echo foo | md5sum | perl -nE 'say "ok" if m/^\bd3b07384d113edec49eaa6238ad5ff00\b/'
If you're stuck with a Perl that predates v5.10, then this:
echo foo | md5sum | perl -lne 'print "ok" if m/^\bd3b07384d113edec49eaa6238ad5ff00\b/'
The solution is fairly self-explanatory if you read through perlrun, which explains what the various command line switches do. We're using -n to cause Perl to process some input, -E to tell Perl to evaluate some code using modern (5.10+) features (say), and the rest just reads as you would expect.
For older Perl versions (pre-5.10), say wasn't available, so the command line switches change to -l, -n, and -e: The first strips newlines from input (not useful), and adds them to output (useful, because print doesn't do that, where the newer say does). And the -e to evaluate some code using pre-5.10 semantics.

Related

Find regular expression in a file matching a given value

I have some basic knowledge on using regular expressions with grep (bash).
But I want to use regular expressions the other way around.
For example I have a file containing the following entries:
line_one=[0-3]
line_two=[4-6]
line_three=[7-9]
Now I want to use bash to figure out to which line a particular number matches.
For example:
grep 8 file
should return:
line_three=[7-9]
Note: I am aware that the example of "grep 8 file" doesn't make sense, but I hope it helps to understand what I am trying to achieve.
Thanks for you help,
Marcel
As others haven pointed out, awk is the right tool for this:
awk -F'=' '8~$2{print $0;}' file
... and if you want this tool to feel more like grep, a quick bash wrapper:
#!/bin/bash
awk -F'=' -v seek_value="$1" 'seek_value~$2{print $0;}' "$2"
Which would run like:
./not_exactly_grep.sh 8 file
line_three=[7-9]
My first impression is that this is not a task for grep, maybe for awk.
Trying to do things with grep I only see this:
for line in $(cat file); do echo 8 | grep "${line#*=}" && echo "${line%=*}" ; done
Using while for file reading (following comments):
while IFS= read -r line; do echo 8 | grep "${line#*=}" && echo "${line%=*}" ; done < file
This can be done in native bash using the syntax [[ $value =~ $regex ]] to test:
find_regex_matching() {
local value=$1
while IFS= read -r line; do # read from input line-by-line
[[ $line = *=* ]] || continue # skip lines not containing an =
regex=${line#*=} # prune everything before the = for the regex
if [[ $value =~ $regex ]]; then # test whether we match...
printf '%s\n' "$line" # ...and print if we do.
fi
done
}
...used as:
find_regex_matching 8 <file
...or, to test it with your sample input inline:
find_regex_matching 8 <<'EOF'
line_one=[0-3]
line_two=[4-6]
line_three=[7-9]
EOF
...which properly emits:
line_three=[7-9]
You could replace printf '%s\n' "$line" with printf '%s\n' "${line%%=*}" to print only the key (contents before the =), if so inclined. See the bash-hackers page on parameter expansion for a rundown on the syntax involved.
This is not built-in functionality of grep, but it's easy to do with awk, with a change in syntax:
/[0-3]/ { print "line one" }
/[4-6]/ { print "line two" }
/[7-9]/ { print "line three" }
If you really need to, you could programmatically change your input file to this syntax, if it doesn't contain any characters that need escaping (mainly / in the regex or " in the string):
sed -e 's#\(.*\)=\(.*\)#/\2/ { print "\1" }#'
As I understand it, you are looking for a range that includes some value.
You can do this in gawk:
$ cat /tmp/file
line_one=[0-3]
line_two=[4-6]
line_three=[7-9]
$ awk -v n=8 'match($0, /([0-9]+)-([0-9]+)/, a){ if (a[1]<n && a[2]>n) print $0 }' /tmp/file
line_three=[7-9]
Since the digits are being treated as numbers (vs a regex) it supports larger ranges:
$ cat /tmp/file
line_one=[0-3]
line_two=[4-6]
line_three=[75-95]
line_four=[55-105]
$ awk -v n=92 'match($0, /([0-9]+)-([0-9]+)/, a){ if (a[1]<n && a[2]>n) print $0 }' /tmp/file
line_three=[75-95]
line_four=[55-105]
If you are just looking to interpret the right hand side of the = as a regex, you can do:
$ awk -F= -v tgt=8 'tgt~$2' /tmp/file
You would like to do something like
grep -Ef <(cut -d= -f2 file) <(echo 8)
This wil grep what you want but will not display where.
With grep you can show some message:
echo "8" | sed -n '/[7-9]/ s/.*/Found it in line_three/p'
Now you would like to transfer your regexp file into such commands:
sed 's#\(.*\)=\(.*\)#/\2/ s/.*/Found at \1/p#' file
Store these commands in a virtual command file and you will have
echo "8" | sed -nf <(sed 's#\(.*\)=\(.*\)#/\2/ s/.*/Found at \1/p#' file)

bash regular expression test: if vs grep

I need to scan each line of a file looking for any characters above hex \x7E. The file has several million rows, so improving efficiency would be great. So far, reading each line in a while loop, this works and finds lines with invalid characters:
echo "$line" | grep -P "[\x7F-\xFF]" > /dev/null 2>&1
if [ $? -eq 0 ]; then...
But this doesn't:
if [[ "$line" =~ [\x7F-\xFF] ]]; then...
I'm assuming it would be more efficient the second way, if I could get it to work. What am I missing?
If you're interested in efficiency, you shouldn't write your loop in bash. You should rethink your program in terms of pipes and use efficient tools.
That said, you can do this with
LC_CTYPE=C LC_COLLATE=C
if [[ "$line" =~ [$'\x7f'-$'\xff'] ]]
then
echo "It contains bytes \x7F or up"
fi
I basically have to split the file. Valid records go to one file, invalid records go to another.
sed -n '/[^\x0-\x7e]/w badrecords
//! w goodrecords'
If you're already using Perl regular expressions, you might as well use perl for the task:
perl -ne '
if (/[\x7F-\xFF]/) {print STDERR $_} else {print}
' file > valid 2> invalid
I'd bet that's faster than a bash loop.
I suspect this would be more efficient, even though it processes the file twice:
grep -P "[\x7F-\xFF]" file > invalid
grep -vP "[\x7F-\xFF]" file > valid
You'd want to write your grep code as
if grep -qP "[\x7F-\xFF]" <<< "$line"; then...

What is the regular expression to get two numbers after a pattern in Solaris with a script?

I want a pair of numbers after a pattern in a line, the pattern is 'mt=' and its position is variable in the line, so if I have mt=83, I only want 83.
I have the next code with an example of the line:
LINE=(10)un=5518666915/(34)ni=vvaummas03.me/ty=12/it=4/et=13/(8)id=ead57a5c/pt=7/to=20130408T155959Z/ot=2/(15)ed=ChangedGreeting/(16)ms=session_11218671/mt=81/
echo "$(expr substr $LINE $(($(echo $LINE | grep -b -o 'mt=' | cut -d: -f1)+4)) 2)"
I know the instruction can be improved, but it worked for me in CentOS, Cygwin and OpenSUSE, but when I run it in Solaris, the terminal show me that grep option -o and the -f1 instruction are invalid. I'm also having trouble with the instruction awk and another options like -c with cut; its very limited so I am searching for common commands on Solaris. I am not allowed to change or install anything on the system.
Any suggestions?
Since you have only a limited set of options it would be possible to do this using only Bash 3 or newer versions with regex matching:
if [[ "$LINE" =~ ^.*mt=([0-9]+).*$ ]]; then
printf "%s\n" "${BASH_REMATCH[1]}"
fi
This will capture any number with a least one digit, following the pattern mt= in the string.
How about this:
~]$ LINE="(10)un=5518666915/(34)ni=vvaummas03.me/ty=12/it=4/et=13/(8)id=ead57a5c/pt=7/to=20130408T155959Z/ot=2/(15)ed=ChangedGreeting/(16)ms=session_11218671/mt=81/"
~]$ echo $LINE | sed -e "s/.*mt=\([0-9]\+\).*/\1/"
81

Sed substitute recursively

echo ddayaynightday | sed 's/day//g'
It ends up daynight
Is there anyway to make it substitute until no more match ?
My preferred form, for this case:
echo ddayaynightday | sed -e ':loop' -e 's/day//g' -e 't loop'
This is the same as everyone else's, except that it uses multiple -e commands to make the three lines and uses the t construct—which means "branch if you did a successful substitution"—to iterate.
This might work for you:
echo ddayaynightday | sed ':a;s/day//g;ta'
night
The g flag deliberately doesn't re-match against the substituted portion of the string. What you'll need to do is a bit different. Try this:
echo ddayaynightday | sed $':begin\n/day/{ s///; bbegin\n}'
Due to BSD Sed's quirkiness the embedded newlines are required. If you're using GNU Sed you may be able to get away with
sed ':begin;/day/{ s///; bbegin }'
with bash:
str=ddayaynightday
while true; do tmp=${str//day/}; [[ $tmp = $str ]] && break; str=$tmp; done
echo $str
The following works:
$ echo ddayaynightday | sed ':loop;/day/{s///g;b loop}'
night
Depending on your system, the ; may not work to separate commands, so you can use the following instead:
echo ddayaynightday | sed -e ':loop' -e '/day/{s///g
b loop}'
Explanation:
:loop # Create the label 'loop'
/day/{ # if the pattern space matches 'day'
s///g # remove all occurrence of 'day' from the pattern space
b loop # go back to the label 'loop'
}
If the b loop portion of the command is not executed, the current contents of the pattern space are printed and the next line is read.
Ok, here they're: while and strlen in bash.
Using them one may implement my idea:
Repeat until its length will stop changing.
There's neither way to set flag nor way to write such regex, to "substitute until no more match".

Return a regex match in a Bash script, instead of replacing it

I just want to match some text in a Bash script. I've tried using sed but I can't seem to make it just output the match instead of replacing it with something.
echo -E "TestT100String" | sed 's/[0-9]+/dontReplace/g'
Which will output TestTdontReplaceString.
Which isn't what I want, I want it to output 100.
Ideally, it would put all the matches in an array.
edit:
Text input is coming in as a string:
newName()
{
#Get input from function
newNameTXT="$1"
if [[ $newNameTXT ]]; then
#Use code that im working on now, using the $newNameTXT string.
fi
}
You could do this purely in bash using the double square bracket [[ ]] test operator, which stores results in an array called BASH_REMATCH:
[[ "TestT100String" =~ ([0-9]+) ]] && echo "${BASH_REMATCH[1]}"
echo "TestT100String" | sed 's/[^0-9]*\([0-9]\+\).*/\1/'
echo "TestT100String" | grep -o '[0-9]\+'
The method you use to put the results in an array depends somewhat on how the actual data is being retrieved. There's not enough information in your question to be able to guide you well. However, here is one method:
index=0
while read -r line
do
array[index++]=$(echo "$line" | grep -o '[0-9]\+')
done < filename
Here's another way:
array=($(grep -o '[0-9]\+' filename))
Pure Bash. Use parameter substitution (no external processes and pipes):
string="TestT100String"
echo ${string//[^[:digit:]]/}
Removes all non-digits.
I Know this is an old topic but I came her along same searches and found another great possibility apply a regex on a String/Variable using grep:
# Simple
$(echo "TestT100String" | grep -Po "[0-9]{3}")
# More complex using lookaround
$(echo "TestT100String" | grep -Po "(?i)TestT\K[0-9]{3}(?=String)")
With using lookaround capabilities search expressions can be extended for better matching. Where (?i) indicates the Pattern before the searched Pattern (lookahead),
\K indicates the actual search pattern and (?=) contains the pattern after the search (lookbehind).
https://www.regular-expressions.info/lookaround.html
The given example matches the same as the PCRE regex TestT([0-9]{3})String
Use grep. Sed is an editor. If you only want to match a regexp, grep is more than sufficient.
using awk
linux$ echo -E "TestT100String" | awk '{gsub(/[^0-9]/,"")}1'
100
I don't know why nobody ever uses expr: it's portable and easy.
newName()
{
#Get input from function
newNameTXT="$1"
if num=`expr "$newNameTXT" : '[^0-9]*\([0-9]\+\)'`; then
echo "contains $num"
fi
}
Well , the Sed with the s/"pattern1"/"pattern2"/g just replaces globally all the pattern1s to pattern 2.
Besides that, sed while by default print the entire line by default .
I suggest piping the instruction to a cut command and trying to extract the numbers u want :
If u are lookin only to use sed then use TRE:
sed -n 's/.*\(0-9\)\(0-9\)\(0-9\).*/\1,\2,\3/g'.
I dint try and execute the above command so just make sure the syntax is right.
Hope this helped.
using just the bash shell
declare -a array
i=0
while read -r line
do
case "$line" in
*TestT*String* )
while true
do
line=${line#*TestT}
array[$i]=${line%%String*}
line=${line#*String*}
i=$((i+1))
case "$line" in
*TestT*String* ) continue;;
*) break;;
esac
done
esac
done <"file"
echo ${array[#]}