I would like to pipe log file with tail -f but only the last line. My command is
tail -f logfile.log | while read line; do /usr/bin/gammu --sendsms TEXT ****** -text "$line"; done
I saw that I can use grep --line-buffered but my new line can't have any char and with what I have tried it doesn't work.
The error shows the problem is other place, which you don't show to us.
Below test code is fine for me.
tail -1f logfile.log | while read line
do
echo /usr/bin/gammu --sendsms TEXT ****** -text "$line"
done
Related
I am running this command "ls -lah -lF | grep /$ | awk '{print $2,$5,$9}' >> test.txt"
when it runs and sends the results to the text file the output is all in one row. How do I get it to formate each directory in a row?
It isn't in one line. Just checked. Check, how your editor handles Unix line endings.
Team,
I want to grep for a substring container - and then only output that string and not whole line. how can i? I know i can awk on space and pull using $ but want to know how to do in grep?
echo $test_pods_info | grep -F 'test-'
output
test-78ac951e-89a6-4199-87a4-db8a1b8b054f export-9b55f0d5-071d-431-1d2ux0-avexport-xavierisp-sjc4--a4dd85-102 1/1 Running 0 19h
expected output
test-78ac951e-89a6-4199-87a4-db8a1b8b054f
awk is more suitable for this as you want to get first field in a matching line:
awk '/test-/{print $1}' <<< "$taxIncluded"
test-78ac951e-89a6-4199-87a4-db8a1b8b054f
If you really want to use grep then this might be what you're looking for:
grep -o 'test-\S*' <<< "$taxIncluded"
or:
grep -o 'test-[^[:space:]]*' <<< "$taxIncluded"
Try
echo $test_pods_info | grep -o 'test-'
the -o option is:
show[ing] only the part of a line matching PATTERN
according to grep --help. Of course, this will only print test-, so you'll need to rework your regex:
grep -oE '(test).*[[:space:]]\b'
Figured it out..
echo $test_pods_info | grep -o "\test-\w*-\w*\-\w*\-\w*\-\w*"
outoput
test-78ac951e-89a6-4199-87a4-db8a1b8b054f
but i wish there is simple way. like \test-*\
I want the output of the sed file edit to go into my log file name d_selinuxlog.txt. Currently, grep outputs the specified string as well as 3 other strings above and below in the edited file.
#!/bin/bash
{ getenforce;
sed -i s/SELINUX=enforcing/SELINUX=disabled /etc/selinux/config;
grep "SELINUX=*" /etc/selinux/config > /home/neb/scropts/logs/d_selinuxlog.txt;
setenforce 0;
getenforce; }
I want to be seeing just SELINUX=disabled in the log file
All the lines with the lines SELINUX are going to match, even the commented ones, so, you need to omit that ones, and the * from the match.
grep "SELINUX=" /etc/selinux/config | grep -v "#"
This is my output
17:52:07 alvaro#lykan /home/alvaro
$ grep "SELINUX=" /etc/selinux/config | grep -v "#"
SELINUX=disabled
17:52:22 alvaro#lykan /home/alvaro
I have a space-separated file that looks like this:
$ cat in_file
GCF_000046845.1_ASM4684v1_protein.faa WP_004920342.1 Chal_sti_synt_C
GCF_000046845.1_ASM4684v1_protein.faa WP_004927566.1 Chal_sti_synt_C
GCF_000046845.1_ASM4684v1_protein.faa WP_004919950.1 FAD_binding_3
GCF_000046845.1_ASM4684v1_protein.faa WP_004920342.1 FAD_binding_3
I am using the following shell script utilizing grep to search for strings:
$ cat search_script.sh
grep "GCF_000046845.1_ASM4684v1_protein.faa WP_004920342.1" Pfam_anntn_temp.txt
grep "GCF_000046845.1_ASM4684v1_protein.faa WP_004920342.1" Pfam_anntn_temp.txt
The problem is that I want each grep command to return only the first instance of the string it finds exclusive of the previous identical grep command's output.
I need an output which would look like this:
$ cat out_file
GCF_000046845.1_ASM4684v1_protein.faa WP_004920342.1 Chal_sti_synt_C
GCF_000046845.1_ASM4684v1_protein.faa WP_004920342.1 FAD_binding_3
in which line 1 is exclusively the output of the first grep command and line 2 is exclusively the output of the second grep command. How do I do it?
P.S. I am running this on a big file (>125,000 lines). So, search_script.sh is mostly composed of unique grep commands. It is the identical commands' execution that is messing up my downstream analysis.
I'm assuming you are generating search_script.sh automatically from the contents of in_file. If you can count how many times you'll repeat the same grep command you can just use grep once and use head, for example if you know you'll be using it 2 times:
grep "foo" bar.txt | head -2
Will output the first 2 occurrences of "foo" in bar.txt.
If you have to do the grep commands separately, for example if you have other code in between the grep commands, you can mix head and tail:
grep "foo" bar.txt | head -1 | tail -1
Some other commands...
grep "foo" bar.txt | head -2 | tail -1
head -n displays the first n lines of the input
tail -n displays the last n lines of the input
If you really MUST always use the same command, but ensure that the outputs always differ, the only way I can think of to achieve this is using temporary files and a complex sequence of commands:
cat foo.bar.txt.tmp 2>&1 | xargs -I xx echo "| grep -v \\'xx\\' " | tr '\n' ' ' | xargs -I xx sh -c "grep 'foo' bar.txt xx | head -1 | tee -a foo.bar.txt.tmp"
So to explain this command, given foo as a search string and bar.txt as the filename, then foo.bar.txt.tmp is a unique name for a temporary file. The temporary file will hold the strings that have already been output:
cat foo.bar.txt.tmp 2>&1 : outputs the contents of the temporary file. If none is present, will output an error message to stdout, (important because if the output was empty the rest of the command wouldn't work.)
xargs -I xx echo "| grep -v \\'xx\\' " adds | grep -v to the start of each line in the temporary file, grep -v something excludes lines that include something.
tr '\n' ' ' replaces newlines with spaces, to have on a single string a sequence of grep -vs.
xargs -I xx sh -c "grep 'foo' bar.txt xx | head -1 | tee -a foo.bar.txt.tmp" runs a new command, grep 'foo' bar.txt xx | head -1 | tee -a foo.bar.txt.tmp, replacing xx with the previous output. xx should be the sequence of grep -vs that exclude previous outputs.
head -1 makes sure only one line is output at a time
tee -a foo.bar.txt.tmp appends the new output to the temporary file.
Just be sure to clear the temporary files, rm *.tmp, at the end of your script.
If I am getting question right and you want to remove duplicates based on last field of each line then try following(this should be easy task for awk).
awk '!a[$NF]++' Input_file
After reading the xargs man page, I am unable to understand the difference in exit codes from the following xargs invocations.
(The original purpose was to combine find and grep to check if an expressions exists in ALL the given files when I came across this behaviour)
To reproduce:
(use >>! if using zsh to force creation of file)
# Create the input files.
echo "a" >> 1.txt
echo "ab" >> 2.txt
# The end goal is to check for a pattern (in this case simply 'b') inside
# ALL the files returned by a find search.
find . -name "1.txt" -o -name "2.txt" | xargs -I {} grep -q "b" {}
echo $?
123 # Works as expected since 'b' is not present in 1.txt
find . -name "1.txt" -o -name "2.txt" | xargs grep -q "b"
echo $?
0 # Am more puzzled by why the behaviour is inconsistent
The EXIT_STATUS section on the man page says:
xargs exits with the following status:
0 if it succeeds
123 if any invocation of the command exited with status 1-125
124 if the command exited with status 255
125 if the command is killed by a signal
126 if the command cannot be run
127 if the command is not found
1 if some other error occurred.
I would have thought, that 123 if any invocation of the command exited with status 1-125 should apply irrespective of whether or not -I is used ?
Could you share any insights to explain this conundrum please?
Here is evidence of the effect of -I option with xargs with the help of a wrapper script which shows the number of invocations:
cat ./grep.sh
#/bin/bash
echo "I am being invoked at $(date +%Y%m%d_%H-%M-%S)"
grep $#
(the actual command being invoked, in this case grep doesn't really matter)
Now execute the same commands as in the question using the wrapper script instead:
❯ find . -name "1.txt" -o -name "2.txt" | xargs -I {} ./grep.sh -q "b" {}
I am being invoked at 20190410_09-46-29
I am being invoked at 20190410_09-46-30
❯ find . -name "1.txt" -o -name "2.txt" | xargs ./grep.sh -q "b"
I am being invoked at 20190410_09-46-53
I have just discovered a comment on the answer of a similar question that answers this question (complete credit to https://superuser.com/users/49184/daniel-andersson for his wisdom):
https://superuser.com/questions/557203/xargs-i-behaviour#comment678705_557230
Also, unquoted blanks do not terminate input items; instead the separator is the newline character. — this is central to understanding the behavior. Without -I, xargs only sees the input as a single field, since newline is not a field separator. With -I, suddenly newline is a field separator, and thus xargs sees three fields (that it iterates over). That is a real subtle point, but is explained in the man page quoted.
-I replace-str
Replace occurrences of replace-str in the initial-arguments
with names read from standard input. Also, unquoted blanks do
not terminate input items; instead the separator is the
newline character. Implies -x and -L 1.
Based on that,
find . -name "1.txt" -o -name "2.txt"
#returns
# ./1.txt
# ./2.txt
xargs -I {} grep -q "b" {}
# interprets the above as two separate lines since,
# with -I option the newline is now a *field separator*.
# So this results in TWO invocations of grep and since one of them fails,
# the overall output is 123 as documented in the EXIT_STATUS section
xargs grep -q "b"
# interprets the above as a single input field,
# so a single grep invocation which returns a successful exit code of 0 since the pattern was found in one of the files.