I've got two test files, namely, ttt.txt and ttt2.txt, the Content of which is shown as below:
#ttt.txt
(132) 123-2131
543-732-3123
238-3102-312
#ttt2.txt
1
2
3
I've already tried the following commands in bash and it works fine:
if grep -oE "(\(\d{3}\)[ ]?\d{3}-\d{4})|(\d{3}-\d{3}-\d{4})" ttt1.txt ; then echo "found"; fi
# with output 'found'
if grep -oE "(\(\d{3}\)[ ]?\d{3}-\d{4})|(\d{3}-\d{3}-\d{4})" ttt2.txt ; then echo "found"; fi
But when I combine the above command with xargs, it complains error '-bash: syntax error near unexpected token `then''. Could anyone give me some explanation? Thanks in advance!
ll | awk '{print $9}' | grep ttt | xargs -I $ if grep --quiet -oE "(\(\d{3}\)[ ]?\d{3}-\d{4})|(\d{3}-\d{3}-\d{4})" $; then echo "found"; fi
$ is a special character in bash (it marks variables) so don't use it as your xargs marker, you'll only get confused.
The real problem here though is that you are passing if grep --quiet -oE "(\(\d{3}\)[ ]?\d{3}-\d{4})|(\d{3}-\d{3}-\d{4})" $ as the argument to xargs, and then the remainder of the line is being treated as a new command, because it breaks at the ;.
You can wrap the whole thing in a sub-invocation of bash, so that xargs sees the whole command:
$ ll | awk '{print $9}' | grep ttt | xargs -I xx bash -c 'if grep --quiet -oE "(\(\d{3}\)[ ]?\d{3}-\d{4})|(\d{3}-\d{3}-\d{4})" xx; then echo "found"; fi'
found
Finally, ll | awk '{print $9}' | grep ttt is a needlessly complicated way of listing the files that you're looking for. You actually you don't need any of the code above, just do this:
$ if grep --quiet -oE "(\(\d{3}\)[ ]?\d{3}-\d{4})|(\d{3}-\d{3}-\d{4})" ttt*; then echo "found"; fi
found
Alternatively, if you want to process each file in turn (which you don't need here, but you might want when this gets more complicated):
for file in ttt*
do
if grep --quiet -oE "(\(\d{3}\)[ ]?\d{3}-\d{4})|(\d{3}-\d{3}-\d{4})" "$file"
then
echo "found"
fi
done
Related
I have to print number of folders in my directory, so i use
ls -l $1| grep "^d" | wc -l
after that, I would liked to add a text in the same line.
any ideas?
If you don’t want to use a variable to hold the output you can use echo and put your command in $( ) on that echo line.
echo $(ls -l $1| grep "^d" | wc -l ) more text to follow here
Assign the result to a variable, then print the variable on the same line as the directory name.
folders=$(ls -l "$1" | grep "^d" | wc -l)
printf "%s %d\n" "$1" "$folders"
Also, remember to quote your variables, otherwise your script won't work when filenames contain whitespace.
I want to make a bash script to use grep to search for lines which have multiple patterns (case-insensitive). I want to create a bash script which I can use as follows:
myscript file.txt pattern1 pattern2 pattern3
and it should get traslated to:
grep -i --color=always pattern1 file.txt | grep -i pattern2 | grep -i pattern3
I tried following bash script, but it is not working:
#!/bin/bash
grep -i --color=always $2 $1 | grep -i $3 | grep -i $4 | grep -i $5 | grep -i $6 | grep -i $7
The error is:
Usage: grep [OPTION]... PATTERN [FILE]...
Try 'grep --help' for more information.
Usage: grep [OPTION]... PATTERN [FILE]...
Try 'grep --help' for more information.
Usage: grep [OPTION]... PATTERN [FILE]...
Try 'grep --help' for more information.
Usage: grep [OPTION]... PATTERN [FILE]...
Try 'grep --help' for more information.
I think you can do a recursive function:
search() {
if [ $# -gt 0 ]; then
local pat=$1
shift
grep "$pat" | search "$#"
else
cat
fi
}
In your script you would call this function and pass the search patterns as arguments. Say that $1 is the file and the rest of the arguments are patterns then you would do
file=$1
shift
cat "$file" | search "$#"
When you have GNU awk, you can use
awk 'BEGIN {IGNORECASE=1} /pattern1/ && /pattern2/ && /pattern3/' file.txt
EDIT:
You can use this in a script like this:
inputfile="$1"
shift
awk -f <(echo "BEGIN {IGNORECASE=1}"; printf " /%s/ &&" $* | sed 's/&&$//') "${inputfile}"
If you omit one argument or other at the end, so that $3 etc. will be missing, then some grep command will not receive an argument and will whine.
I want to check if multiple lines in a file exist in bash.
so for that I use grep -q which works with only one line:
if grep -q string1 "/path/to/file";then
echo 'exists'
else
echo 'does not exist'
fi
I tried many things in various combinations, for example:
if grep -q [ string1 ] && grep -q [ string2 ] "path/to/file";then
I also tried it with -E:
grep -E 'pattern1' filename | grep -E 'pattern2'
but nothing seems to work. Any ideas?
Rather than running multiple grep commands you can use this gnu-awk command to assert presence of multiple strings in a file:
awk -v RS='\\Z' '/string1/ && /string2/ && /string3/{e=1} END{exit !e}' file &&
echo 'exists' || echo 'does not exist'
RS=\Z will make awk read all the input in a single record separator
Using && between multiple search terms will make sure all the search words exist in input file
This will print exists only if all 3 search terms exists in the input file.
since #iruvar hasn't posted his comment as answer, i'll put it here:
grep -q string_1 file && grep -q string_2 file
now, here is my contribution. is #anubhava's more computationally complex awk answer, which reads the file only once, any faster than #iruvar's simpler answer, which reads the file three times?
awk 11.730 s
grep && grep 0.258 s
no.
this surely will depend on the speed of the filesystem vs the cpu, and on how much caching goes on, but on my system, which is probably a typical B+/A- workstation, grep kw1 file && grep kw2 file && grep kw3 file is ~50x as fast as #anubhava's awk solution. this held true both on ssd and spindle raid. (details: test file was 5,000,000 lines, 160M, and had kw1 on the first line, kw2 on the 2.5 millionth, and kw3 on the 5 millionth.)
some easy optimization is possible, for example, if you can solve your problem by matching whole lines, do so (with grep -x); it's twice as fast in this case.
for many (e.g., >1,000) files, it is faster to use grep -l and xargs:
grep -l kw1 *.txt | xargs grep -l kw2 | xargs grep -q kw3
as opposed to a loop:
for f in *.txt; do
grep -q kw1 $f && grep -q kw2 $f && grep -q kw3 $f
done
with the same test file, grep -l | xargs grep took 0.258 s, just like grep && grep. with two test files, it was still no faster than grep && grep. with 2000 test files of 5,000 lines each, none of which contained any matches, grep -l | xargs grep was ~10x as faster as grep && grep.
There are a couple ambiguities in your question, but assuming you want pattern_1 and pattern_2 to exist in a file (not on the same line) then you can do this.
for file in *; do
egrep -q pattern_1 $file && egrep -q pattern_2 $file && echo $file
done
With grep -p you can match multiply patterns in the same line:
grep -P '(?=.*string1)(?=.*string2)' file
The above will print lines that matches string1 and string2.
(?=...) is a positive lookaheads which matches a pattern without making it a part of the match.
And -z will slurp the whole file:
% seq 1 100 | grep -qzP '(?=.*1)(?=.*5)'; echo $?
0
% seq 1 100 | grep -qzP '(?=.*1)(?=.*a)'; echo $?
1
You can do it like this:
if grep -q 'string1' /path/to/file; then
if grep -q 'string2' /path/to/file; then
echo exists
else
echo 'does not exist'
else
echo 'does not exist'
fi
Or:
grep -q 'string1' /path/to/file &&
grep -q 'string2' /path/to/file &&
echo exists ||
echo 'does not exist'
you can use "-q" to search using grep
if grep -q string1 "/path/to/file" && grep -q string2 "/path/to/file";then
echo 'exists'
else
echo 'does not exist'
fi
I want to pass a variable in my grep command in Linux bash script. Variable is a text file from Internet and i want to find some words in it.
I have tried the following command in my bash:
cat "$var" | grep -Po '(?<=\d[a-zA-Z]).*\..*(?=[a-zA-Z]\d)'
echo "$var" | grep -Po '(?<=\d[a-zA-Z]).*\..*(?=[a-zA-Z]\d)'
grep -Po '(?<=\d[a-zA-Z]).*\..*(?=[a-zA-Z]\d)' <<< "$var"
but i dont get a right Result.
How can i do it?
Here is my bash:
#!/bin/sh
urltext=$(curl -s https://example.com)
string=$(grep -Po '(?<=\d[a-zA-Z]).*\..*(?=[a-zA-Z]\d)' "$urltext" | tr '.' '\n' )
cat $string
What's supposed to be contained in the variable ?
It's a file ?
grep <options> <expression> "$var"
It's a string ?
echo "$var"|grep <options> <expression>
grep <options> <expression> <(echo "$var")
NB : try -e option if there are several lines in $var
I have a command that should be executed by a shell script.
Actually the command does not matter the only thing that is important the further command execution and the right escaping of the critical parts.
The command that usually is executed normally in putty is something like this(maybe some additional flags for ls)
rm -r `ls /test/parse_first/ | awk '{print $2}' | grep trash`
but now I have a batch of such command so I would like to execute them in a loop
like
for i in {0..100}
do
str=str$i
${!str}
done
where str is :
str0="rm -r `ls /test/parse_first/ | awk '{print $2}' | grep trash`"
str1="rm -r `ls /test/parse_second/ | awk '{print $2}' | grep trash`"
and that gives me a lot of headache cause the execution done by ${!str} brakes the quotations and inline shell between `...` marks
my_rm() { rm -r `ls /test/$1 | awk ... | grep ... `; }
for i in `whatevr`; do
my_rm $i
done;
Getting this right is surprisingly tricky, but it can be done:
for i in $(seq 0 100)
do
str=str$i
eval "eval \"\$$str\""
done
You can also do:
for i in {0..10}
do
<whatevercommand>
done
It's actually simpler to place them on arrays and use glob patterns:
#!/bin/bash
shopt -s nullglob
DIRS=("/test/parse_first/" "/test/parse_second/")
for D in "${DIRS[#]}"; do
for T in "$D"/*trash*; do
rm -r -- "$T"
done
done
And if rm could accept multiple arguments, you don't need to have an extra loop:
for D in "${DIRS[#]}"; do
rm -r -- "$D"/*trash*
done
UPDATE:
#!/bin/bash
readarray -t COMMANDS <<'EOF'
rm -r `ls /test/parse_first/ | awk '{print $2}' | grep trash
rm -r `ls /test/parse_second/ | awk '{print $2}' | grep trash
EOF
for C in "${COMMANDS[#]}"; do
eval "$C"
done
Or you could just read commands from another file:
readarray -t COMMANDS < somefile.txt