For a list of files, I'd like to match the ones not ending with .txt. I am currently using this expression:
.*(txt$)|(html\.txt$)
This expression will match everything ending in .txt, but I'd like it to do the opposite.
Should match:
happiness.html
joy.png
fear.src
Should not match:
madness.html.txt
excitement.txt
I'd like to get this so I can use it in pair with fswatch:
fswatch -0 -e 'regex here' . | xargs -0 -n 1 -I {} echo "{} has been changed"
The problem is it doesn't seem to work.
PS: I use the tag bash instead of fswatch because I don't have enough reputation points to create it. Sorry!
Try using a lookbehind, like this:
.*$(?<!\.txt)
Demonstration
Basically, this matches any line of text so long as the last 4 characters are not ".txt".
You can use Negative Lookahead for this purpose.
^(?!.*\.txt).+$
Live Demo
You can use this expression with grep using option -P:
grep -Po '^(?!.*\.txt).+$' file
Since question has been tagged as bash, lookaheads may not be supported (except grep -P), here is one grep solution that doesn't need lookaheads:
grep -v '\.txt$' file
happiness.html
joy.png
fear.src
EDIT: You can use this xargs command to avoid matching *.txt files:
xargs -0 -n 1 -I {} bash -c '[[ "{}" == *".txt" ]] && echo "{} has been changed"'
It really depends what regular expression tool you are using. Many tools provide a way to invert the sense of a regex. For example:
bash
# succeeds if filename ends with .txt
[[ $filename =~ "."txt$ ]]
# succeeds if filename does not end with .txt
! [[ $filename =~ "."txt$ ]]
# another way of writing the negative
[[ ! $filename =~ "."txt$ ]]
grep
# succeeds if filename ends with .txt
egrep -q "\.txt$" <<<"$filename"
# succeeds if filename does not end with .txt
egrep -qv "\.txt$" <<<"$filename"
awk
/\.txt$/ { print "line ends with .txt" }
! /\.txt$/ { print "line doesn't end with .txt" }
$1 ~ /\.txt$/ { print "first field ends with .txt" }
$1 !~ /\.txt$/ { print "first field doesn't end with .txt" }
For the adventurous, a posix ERE which will work in any posix compatible regex engine
/[^t]$|[^x]t$|[^t]xt$|[^.]txt$/
Related
Given a file, for example:
potato: 1234
apple: 5678
potato: 5432
grape: 4567
banana: 5432
sushi: 56789
I'd like to grep for all lines that start with potato: but only pipe the numbers that follow potato:. So in the above example, the output would be:
1234
5432
How can I do that?
grep 'potato:' file.txt | sed 's/^.*: //'
grep looks for any line that contains the string potato:, then, for each of these lines, sed replaces (s/// - substitute) any character (.*) from the beginning of the line (^) until the last occurrence of the sequence : (colon followed by space) with the empty string (s/...// - substitute the first part with the second part, which is empty).
or
grep 'potato:' file.txt | cut -d\ -f2
For each line that contains potato:, cut will split the line into multiple fields delimited by space (-d\ - d = delimiter, \ = escaped space character, something like -d" " would have also worked) and print the second field of each such line (-f2).
or
grep 'potato:' file.txt | awk '{print $2}'
For each line that contains potato:, awk will print the second field (print $2) which is delimited by default by spaces.
or
grep 'potato:' file.txt | perl -e 'for(<>){s/^.*: //;print}'
All lines that contain potato: are sent to an inline (-e) Perl script that takes all lines from stdin, then, for each of these lines, does the same substitution as in the first example above, then prints it.
or
awk '{if(/potato:/) print $2}' < file.txt
The file is sent via stdin (< file.txt sends the contents of the file via stdin to the command on the left) to an awk script that, for each line that contains potato: (if(/potato:/) returns true if the regular expression /potato:/ matches the current line), prints the second field, as described above.
or
perl -e 'for(<>){/potato:/ && s/^.*: // && print}' < file.txt
The file is sent via stdin (< file.txt, see above) to a Perl script that works similarly to the one above, but this time it also makes sure each line contains the string potato: (/potato:/ is a regular expression that matches if the current line contains potato:, and, if it does (&&), then proceeds to apply the regular expression described above and prints the result).
Or use regex assertions: grep -oP '(?<=potato: ).*' file.txt
grep -Po 'potato:\s\K.*' file
-P to use Perl regular expression
-o to output only the match
\s to match the space after potato:
\K to omit the match
.* to match rest of the string(s)
sed -n 's/^potato:[[:space:]]*//p' file.txt
One can think of Grep as a restricted Sed, or of Sed as a generalized Grep. In this case, Sed is one good, lightweight tool that does what you want -- though, of course, there exist several other reasonable ways to do it, too.
This will print everything after each match, on that same line only:
perl -lne 'print $1 if /^potato:\s*(.*)/' file.txt
This will do the same, except it will also print all subsequent lines:
perl -lne 'if ($found){print} elsif (/^potato:\s*(.*)/){print $1; $found++}' file.txt
These command-line options are used:
-n loop around each line of the input file
-l removes newlines before processing, and adds them back in afterwards
-e execute the perl code
You can use grep, as the other answers state. But you don't need grep, awk, sed, perl, cut, or any external tool. You can do it with pure bash.
Try this (semicolons are there to allow you to put it all on one line):
$ while read line;
do
if [[ "${line%%:\ *}" == "potato" ]];
then
echo ${line##*:\ };
fi;
done< file.txt
## tells bash to delete the longest match of ": " in $line from the front.
$ while read line; do echo ${line##*:\ }; done< file.txt
1234
5678
5432
4567
5432
56789
or if you wanted the key rather than the value, %% tells bash to delete the longest match of ": " in $line from the end.
$ while read line; do echo ${line%%:\ *}; done< file.txt
potato
apple
potato
grape
banana
sushi
The substring to split on is ":\ " because the space character must be escaped with the backslash.
You can find more like these at the linux documentation project.
Modern BASH has support for regular expressions:
while read -r line; do
if [[ $line =~ ^potato:\ ([0-9]+) ]]; then
echo "${BASH_REMATCH[1]}"
fi
done
grep potato file | grep -o "[0-9].*"
I'm trying to count the words with at least two vowels in all the .txt files in the directory. Here's my code so far:
#!/bin/bash
wordcount=0
for i in $HOME/*.txt
do
cat $i |
while read line
do
for w in $line
do
if [[ $w == .*[aeiouAEIOU].*[AEIOUaeiou].* ]]
then
wordcount=`expr $wordcount + 1`
echo $w ':' $wordcount
else
echo "In else"
fi
done
done
echo $i ':' $wordcount
wordcount=0
done
Here is my sample from a txt file
Last modified: Sun Aug 20 18:18:27 IST 2017
To remove PPAs
sudo apt-get install ppa-purge
sudo ppa-purge ppa:
The problem is it doesn't match the pattern in the if statement for all the words in the text file. It goes directly to the else statement. And secondly, the wordcount in echo $i ':' $wordcount is equal to 0 which should be some value.
Immediate Issue: Glob vs Regex
[[ $string = $pattern ]] doesn't perform regex matching; instead, it's a glob-style pattern match. While . means "any character" in regex, it matches only itself in glob.
You have a few options here:
Use =~ instead to perform regular expression matching:
[[ $w =~ .*[aeiouAEIOU].*[AEIOUaeiou].* ]]
Use a glob-style expression instead of a regex:
[[ $w = *[aeiouAEIOU]*[aeiouAEIOU]* ]]
Note the use of = rather than == here; while either is technically valid, the former avoids building finger memory that would lead to bugs when writing code for a POSIX implementation of test / [, as = is the only valid string comparison operator there.
Larger Issue: Properly Reading Word-By-Word
Using for w in $line is innately unsafe. Use read -a to read a line into an array of words:
#!/usr/bin/env bash
wordcount=0
for i in "$HOME"/*.txt; do
while read -r -a words; do
for word in "${words[#]}"; do
if [[ $word = *[aeiouAEIOU]*[aeiouAEIOU]* ]]; then
(( ++wordcount ))
fi
done
done <"$i"
printf '%s: %s\n' "$i" "$wordcount"
wordcount=0
done
Try:
awk '/[aeiouAEIOU].*[AEIOUaeiou]/{n++} ENDFILE{print FILENAME":"n; n=0}' RS='[[:space:]]' *.txt
Sample output looks like:
$ awk '/[aeiouAEIOU].*[AEIOUaeiou]/{n++} ENDFILE{print FILENAME":"n; n=0}' RS='[[:space:]]' *.txt
one.txt:1
sample.txt:9
How it works:
/[aeiouAEIOU].*[AEIOUaeiou]/{n++}
Every time we find a word with two vowels, we increment variable n.
ENDFILE{print FILENAME":"n; n=0}
At the end of each file, we print the name of the file and the 2-vowel word count n. We then reset n to zero.
RS='[[:space:]]'
This tells awk to use any whitespace as a word separator. This makes each word into a record. Awk reads the input one record at a time.
Shell issues
The use of awk avoids a multitude of shell issues. For example, consider the line for w in $line. This will not work the way you hope. Consider a directory with these files:
$ ls
one.txt sample.txt
Now, let's take line='* Item One' and see what happens:
$ line='* Item One'
$ for w in $line; do echo "w=$w"; done
w=one.txt
w=sample.txt
w=Item
w=One
The shell treats the * in line as a wildcard and expands it into a list of files. Odds are you didn't want this. The awk solution avoids a variety of issues like this.
Using grep - this is pretty simple to do.
#!/bin/bash
wordcount=0
for file in ./*.txt
do
count=`cat $file | xargs -n1 | grep -ie "[aeiou].*[aeiou]" | wc -l`
wordcount=`expr $wordcount + $count`
done
echo $wordcount
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)
Given any string in bash, e.g flaccid, I want to match all characters in the string but the first 3 (in this case I want to exclude "fla" and match only "ccid"). The regex also needs to work in sed.
I have tried positive look behind and the following regex expressions (as well as various other unsuccessful ones):
^.{3}+([a-z,A-Z]+)
sed -r 's/(?<=^....)(.[A-Z]*)/,/g'
Google hasn't been very helpful as it only produce results like "get first 3 characters .."
Thanks in advance!
If you want to get all characters but the first 3 from a string, you can use cut:
str="flaccid"
cut -c 4- <<< "$str"
or bash variable subsitution:
str="flaccid"
echo "${str:3}"
That will strip the first 3 characters out of your string.
You may just use a capturing group within an expression like ^.{3}(.*) / ^.{3}([a-zA-Z]+) and grab the ${BASH_REMATCH[1]} contents:
#!/bin/bash
text="flaccid"
rx="^.{3}(.*)"
if [[ $text =~ $rx ]]; then
echo ${BASH_REMATCH[1]};
fi
See online Bash demo
In sed, you should also be using capturing groups / backreferences to get what you need. To just keep the first 3 chars, you may use a simple:
echo "flaccid" | sed 's/.\{3\}//'
See this regex demo. The .\{3\} matches exactly any 3 chars and will remove them from the beginning only, since g modifier is not used.
Now, both the solutions above will output ccid, returning the first 3 chars only.
Using sed, just remove them
echo string | sed 's/^...//g'
How is it that no-one has named the most simple and portable solution:
shell "Parameter expansions":
str="flacid"
echo "${str#???}
For a regex (bash):
$ str="flaccid"
$ regex='^.{3}(.*)$'
$ [[ $str =~ $regex ]] && echo "${BASH_REMATCH[1]}"
ccid
Same regex in sed:
$ echo "flaccid" | sed -E "s/$regex/\1/"
ccid
Or sed (Basic Regex):
$ echo "flaccid" | sed 's/^.\{3\}\(.*\)$/\1/'
ccid
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[#]}