I have a script that is running and uses
lspci -s 0a.00.1
This returns
0a.00.1 usb controller some text device 4dc9
I want to get those last 4 characters inline such that
lspci -s 0a.00.1 | some command to give me the last 4 characters.
How about tail, with the -c switch. For example, to get the last 4 characters of "hello":
echo "hello" | tail -c 5
ello
Note that I used 5 (4+1) because a newline character is added by echo. As suggested by Brad Koch below, use echo -n to prevent the newline character from being added.
Do you really want the last four characters? It looks like you want the last "word" on the line:
awk '{ print $NF }'
This will work if the ID is 3 characters, or 5, as well.
Using sed:
lspci -s 0a.00.1 | sed 's/^.*\(.\{4\}\)$/\1/'
Output:
4dc9
Try this, say if the string is stored in the variable foo.
foo=`lspci -s 0a.00.1` # the foo value should be "0a.00.1 usb controller some text device 4dc9"
echo ${foo:(-4)} # which should output 4dc9
I usually use
echo 0a.00.1 usb controller some text device 4dc9 | rev | cut -b1-4 | rev
4dc9
If the real request is to copy the last space-separated string regardless of its length, then the best solution seems to be using ... | awk '{print $NF}' as given by #Johnsyweb. But if this is indeed about copying a fixed number of characters from the end of a string, then there is a bash-specific solution without the need to invoke any further subprocess by piping:
$ test="1234567890"; echo "${test: -4}"
7890
$
Please note that the space between colon and minus character is essential, as without it the full string will be delivered:
$ test="1234567890"; echo "${test:-4}"
1234567890
$
Try using grep:
lspci -s 0a.00.1 | grep -o ....$
This will print last 4 characters of every line.
However if you'd like to have last 4 characters of the whole output, use tail -c4 instead.
One more way to approach this is to use <<< notation:
tail -c 5 <<< '0a.00.1 usb controller some text device 4dc9'
instead of using named variables, develop the practice of using the positional parameters, like this:
set -- $( lspci -s 0a.00.1 ); # then the bash string usage:
echo ${1:(-4)} # has the advantage of allowing N PP's to be set, eg:
set -- $(ls *.txt)
echo $4 # prints the 4th txt file.
Related
Without using sed or awk, only cut, how do I get the last field when the number of fields are unknown or change with every line?
You could try something like this:
echo 'maps.google.com' | rev | cut -d'.' -f 1 | rev
Explanation
rev reverses "maps.google.com" to be moc.elgoog.spam
cut uses dot (ie '.') as the delimiter, and chooses the first field, which is moc
lastly, we reverse it again to get com
Use a parameter expansion. This is much more efficient than any kind of external command, cut (or grep) included.
data=foo,bar,baz,qux
last=${data##*,}
See BashFAQ #100 for an introduction to native string manipulation in bash.
It is not possible using just cut. Here is a way using grep:
grep -o '[^,]*$'
Replace the comma for other delimiters.
Explanation:
-o (--only-matching) only outputs the part of the input that matches the pattern (the default is to print the entire line if it contains a match).
[^,] is a character class that matches any character other than a comma.
* matches the preceding pattern zero or more time, so [^,]* matches zero or more non‑comma characters.
$ matches the end of the string.
Putting this together, the pattern matches zero or more non-comma characters at the end of the string.
When there are multiple possible matches, grep prefers the one that starts earliest. So the entire last field will be matched.
Full example:
If we have a file called data.csv containing
one,two,three
foo,bar
then grep -o '[^,]*$' < data.csv will output
three
bar
Without awk ?...
But it's so simple with awk:
echo 'maps.google.com' | awk -F. '{print $NF}'
AWK is a way more powerful tool to have in your pocket.
-F if for field separator
NF is the number of fields (also stands for the index of the last)
There are multiple ways. You may use this too.
echo "Your string here"| tr ' ' '\n' | tail -n1
> here
Obviously, the blank space input for tr command should be replaced with the delimiter you need.
This is the only solution possible for using nothing but cut:
echo "s.t.r.i.n.g." | cut -d'.' -f2-
[repeat_following_part_forever_or_until_out_of_memory:] | cut -d'.' -f2-
Using this solution, the number of fields can indeed be unknown and vary from time to time. However as line length must not exceed LINE_MAX characters or fields, including the new-line character, then an arbitrary number of fields can never be part as a real condition of this solution.
Yes, a very silly solution but the only one that meets the criterias I think.
If your input string doesn't contain forward slashes then you can use basename and a subshell:
$ basename "$(echo 'maps.google.com' | tr '.' '/')"
This doesn't use sed or awk but it also doesn't use cut either, so I'm not quite sure if it qualifies as an answer to the question as its worded.
This doesn't work well if processing input strings that can contain forward slashes. A workaround for that situation would be to replace forward slash with some other character that you know isn't part of a valid input string. For example, the pipe (|) character is also not allowed in filenames, so this would work:
$ basename "$(echo 'maps.google.com/some/url/things' | tr '/' '|' | tr '.' '/')" | tr '|' '/'
the following implements A friend's suggestion
#!/bin/bash
rcut(){
nu="$( echo $1 | cut -d"$DELIM" -f 2- )"
if [ "$nu" != "$1" ]
then
rcut "$nu"
else
echo "$nu"
fi
}
$ export DELIM=.
$ rcut a.b.c.d
d
An alternative using perl would be:
perl -pe 's/(.*) (.*)$/$2/' file
where you may change \t for whichever the delimiter of file is
It is better to use awk while working with tabular data. You don't have to master on command. If it can be achieved by awk, why not use that? I suggest you do not waste your precious time, and use a handful of commands to get the job done.
Example:
# $NF refers to the last column in awk
ll | awk '{print $NF}'
If you have a file named filelist.txt that is a list paths such as the following:
c:/dir1/dir2/file1.h
c:/dir1/dir2/dir3/file2.h
then you can do this:
rev filelist.txt | cut -d"/" -f1 | rev
Adding an approach to this old question just for the fun of it:
$ cat input.file # file containing input that needs to be processed
a;b;c;d;e
1;2;3;4;5
no delimiter here
124;adsf;15454
foo;bar;is;null;info
$ cat tmp.sh # showing off the script to do the job
#!/bin/bash
delim=';'
while read -r line; do
while [[ "$line" =~ "$delim" ]]; do
line=$(cut -d"$delim" -f 2- <<<"$line")
done
echo "$line"
done < input.file
$ ./tmp.sh # output of above script/processed input file
e
5
no delimiter here
15454
info
Besides bash, only cut is used.
Well, and echo, I guess.
choose -1
choose supports negative indexing (the syntax is similar to Python's slices).
I realized if we just ensure a trailing delimiter exists, it works. So in my case I have comma and whitespace delimiters. I add a space at the end;
$ ans="a, b"
$ ans+=" "; echo ${ans} | tr ',' ' ' | tr -s ' ' | cut -d' ' -f2
b
I need to remove any letters that occur after the first comma in a line
some.file
JAN,334X,333B,337A,338D,332Q,335H,331U
Expected Result:
JAN,334,333,337,338,332,335,331
Code:
sed -i 's/\[0-9][0-9][0-9].*,/[0-9][0-9][0-9],/g' some.file
What am I doing wrong?
You could also use a small loop (this is GNU sed);
sed ':;s/[A-Z],/,/2;t;s/[A-Z]$//'
It only deletes the second letter preceding a comma, and loops. Finally, it deletes the letter at the line's end, if there is one.
Try this
$ sed 's/,\([0-9]*\)[^,]*/,\1/g' <<<'JAN,334X,333B,337A,338D,332Q,335H,331U'
JAN,334,333,337,338,332,335,331
You need to capture the digits with round parenthesis in order to use the captured string in the replacement. The option g does this for every occurrence.
Comparison of the different answers
Test data:
$ > data; for ((x=1000000;x>0;x--)); do echo 'JAN,334X,333B,337A,338D,332Q,335H,331U' >> data; done
My answer is the slowest:
$ time sed 's/,\([0-9]*\)[^,]*/,\1/g' < data >/dev/null
real 0m16.368s
user 0m16.296s
sys 0m0.024s
Michael is a bit faster:
$ time sed ':;s/[A-Z],/,/2;t;s/[A-Z]$//' < data >/dev/null
real 0m9.669s
user 0m9.624s
sys 0m0.012s
But Sundeep is the fastet:
$ time sed 's/[A-Z]//4g' < data >/dev/null
real 0m4.905s
user 0m4.856s
sys 0m0.028s
Some issues are:
No need to escape [.
Your replace value is wrong. Ex: s/regex/replace/g
Use this:
sed -e 's/\([0-9]\+\)[a-zA-Z],/\1,/g' -e 's/\([0-9]\+\)[a-zA-Z]$/\1/g' file
You should omit the * and the first \ looks like a mistake i.e.
sed -i 's/[0-9][0-9][0-9].,/[0-9][0-9][0-9],/g' some.file
but I think you also want to capture the number ...
sed -i 's/\([0-9][0-9][0-9]\).,/\1,/g' some.file
Would be helpful if you posted your actual output as well ...
Since question is tagged linux, this GNU sed option comes in handy
$ echo 'JAN,334X,333B,337A,338D,332Q,335H,331U' | sed -E 's/[A-Z](,|$)/\1/2g'
JAN,334,333,337,338,332,335,331
2g means replace from 2nd match onwards till end of line
If number of letters is known for first column, this can be simplified to
$ echo 'JAN,334X,333B,337A,338D,332Q,335H,331U' | sed 's/[A-Z]//4g'
JAN,334,333,337,338,332,335,331
No need for sed, coreutils will do:
paste -d, <(cut -d, -f1 data) <(cut -d, -f2- data | tr -d 'A-Z')
This takes .3 seconds on my computer when run on the data file generated in ceving's answer.
In a very large file I need to find the position (line number) of a string, then extract the 2 lines above and below that string.
To do this right now - I launch vi, find the string, note it's line number, exit vi, then use sed to extract the lines surrounding that string.
Is there a way to streamline this process... ideally without having to run vi at all.
Maybe using grep like this:
grep -n -2 your_searched_for_string your_large_text_file
Will give you almost what you expect
-n : tells grep to print the line number
-2 : print 2 additional lines (and the wanted string, of course)
You can do
grep -C 2 yourSearch yourFile
To send it in a file, do
grep -C 2 yourSearch yourFile > result.txt
Use grep -n string file to find the line number without opening the file.
you can use cat -n to display the line numbers and then use awk to get the line number after a grep in order to extract line number:
cat -n FILE | grep WORD | awk '{print $1;}'
although grep already does what you mention if you give -C 2 (above/below 2 lines):
grep -C 2 WORD FILE
You can do it with grep -A and -B options, like this:
grep -B 2 -A 2 "searchstring" | sed 3d
grep will find the line and show two lines of context before and after, later remove the third one with sed.
If you want to automate this, simple you can do a Shell Script. You may try the following:
#!/bin/bash
VAL="your_search_keyword"
NUM1=`grep -n "$VAL" file.txt | cut -f1 -d ':'`
echo $NUM1 #show the line number of the matched keyword
MYNUMUP=$["NUM1"-1] #get above keyword
MYNUMDOWN=$["NUM1"+1] #get below keyword
sed -n "$MYNUMUP"p file.txt #display above keyword
sed -n "$MYNUMDOWN"p file.txt #display below keyword
The plus point of the script is you can change the keyword in VAL variable as you like and execute to get the needed output.
I want to do this:
run a command
capture the output
select a line
select a column of that line
Just as an example, let's say I want to get the command name from a $PID (please note this is just an example, I'm not suggesting this is the easiest way to get a command name from a process id - my real problem is with another command whose output format I can't control).
If I run ps I get:
PID TTY TIME CMD
11383 pts/1 00:00:00 bash
11771 pts/1 00:00:00 ps
Now I do ps | egrep 11383 and get
11383 pts/1 00:00:00 bash
Next step: ps | egrep 11383 | cut -d" " -f 4. Output is:
<absolutely nothing/>
The problem is that cut cuts the output by single spaces, and as ps adds some spaces between the 2nd and 3rd columns to keep some resemblance of a table, cut picks an empty string. Of course, I could use cut to select the 7th and not the 4th field, but how can I know, specially when the output is variable and unknown on beforehand.
One easy way is to add a pass of tr to squeeze any repeated field separators out:
$ ps | egrep 11383 | tr -s ' ' | cut -d ' ' -f 4
I think the simplest way is to use awk. Example:
$ echo "11383 pts/1 00:00:00 bash" | awk '{ print $4; }'
bash
Please note that the tr -s ' ' option will not remove any single leading spaces. If your column is right-aligned (as with ps pid)...
$ ps h -o pid,user -C ssh,sshd | tr -s " "
1543 root
19645 root
19731 root
Then cutting will result in a blank line for some of those fields if it is the first column:
$ <previous command> | cut -d ' ' -f1
19645
19731
Unless you precede it with a space, obviously
$ <command> | sed -e "s/.*/ &/" | tr -s " "
Now, for this particular case of pid numbers (not names), there is a function called pgrep:
$ pgrep ssh
Shell functions
However, in general it is actually still possible to use shell functions in a concise manner, because there is a neat thing about the read command:
$ <command> | while read a b; do echo $a; done
The first parameter to read, a, selects the first column, and if there is more, everything else will be put in b. As a result, you never need more variables than the number of your column +1.
So,
while read a b c d; do echo $c; done
will then output the 3rd column. As indicated in my comment...
A piped read will be executed in an environment that does not pass variables to the calling script.
out=$(ps whatever | { read a b c d; echo $c; })
arr=($(ps whatever | { read a b c d; echo $c $b; }))
echo ${arr[1]} # will output 'b'`
The Array Solution
So we then end up with the answer by #frayser which is to use the shell variable IFS which defaults to a space, to split the string into an array. It only works in Bash though. Dash and Ash do not support it. I have had a really hard time splitting a string into components in a Busybox thing. It is easy enough to get a single component (e.g. using awk) and then to repeat that for every parameter you need. But then you end up repeatedly calling awk on the same line, or repeatedly using a read block with echo on the same line. Which is not efficient or pretty. So you end up splitting using ${name%% *} and so on. Makes you yearn for some Python skills because in fact shell scripting is not a lot of fun anymore if half or more of the features you are accustomed to, are gone. But you can assume that even python would not be installed on such a system, and it wasn't ;-).
try
ps |&
while read -p first second third fourth etc ; do
if [[ $first == '11383' ]]
then
echo got: $fourth
fi
done
Your command
ps | egrep 11383 | cut -d" " -f 4
misses a tr -s to squeeze spaces, as unwind explains in his answer.
However, you maybe want to use awk, since it handles all of these actions in a single command:
ps | awk '/11383/ {print $4}'
This prints the 4th column in those lines containing 11383. If you want this to match 11383 if it appears in the beginning of the line, then you can say ps | awk '/^11383/ {print $4}'.
Using array variables
set $(ps | egrep "^11383 "); echo $4
or
A=( $(ps | egrep "^11383 ") ) ; echo ${A[3]}
Similar to brianegge's awk solution, here is the Perl equivalent:
ps | egrep 11383 | perl -lane 'print $F[3]'
-a enables autosplit mode, which populates the #F array with the column data.
Use -F, if your data is comma-delimited, rather than space-delimited.
Field 3 is printed since Perl starts counting from 0 rather than 1
Getting the correct line (example for line no. 6) is done with head and tail and the correct word (word no. 4) can be captured with awk:
command|head -n 6|tail -n 1|awk '{print $4}'
Instead of doing all these greps and stuff, I'd advise you to use ps capabilities of changing output format.
ps -o cmd= -p 12345
You get the cmmand line of a process with the pid specified and nothing else.
This is POSIX-conformant and may be thus considered portable.
Bash's set will parse all output into position parameters.
For instance, with set $(free -h) command, echo $7 will show "Mem:"
This question already has answers here:
How can I remove the first line of a text file using bash/sed script?
(19 answers)
Closed 3 years ago.
I have a very long file which I want to print, skipping the first 1,000,000 lines, for example.
I looked into the cat man page, but I did not see any option to do this. I am looking for a command to do this or a simple Bash program.
You'll need tail. Some examples:
$ tail great-big-file.log
< Last 10 lines of great-big-file.log >
If you really need to SKIP a particular number of "first" lines, use
$ tail -n +<N+1> <filename>
< filename, excluding first N lines. >
That is, if you want to skip N lines, you start printing line N+1. Example:
$ tail -n +11 /tmp/myfile
< /tmp/myfile, starting at line 11, or skipping the first 10 lines. >
If you want to just see the last so many lines, omit the "+":
$ tail -n <N> <filename>
< last N lines of file. >
Easiest way I found to remove the first ten lines of a file:
$ sed 1,10d file.txt
In the general case where X is the number of initial lines to delete, credit to commenters and editors for this:
$ sed 1,Xd file.txt
If you have GNU tail available on your system, you can do the following:
tail -n +1000001 huge-file.log
It's the + character that does what you want. To quote from the man page:
If the first character of K (the number of bytes or lines) is a
`+', print beginning with the Kth item from the start of each file.
Thus, as noted in the comment, putting +1000001 starts printing with the first item after the first 1,000,000 lines.
If you want to skip first two line:
tail -n +3 <filename>
If you want to skip first x line:
tail -n +$((x+1)) <filename>
A less verbose version with AWK:
awk 'NR > 1e6' myfile.txt
But I would recommend using integer numbers.
Use the sed delete command with a range address. For example:
sed 1,100d file.txt # Print file.txt omitting lines 1-100.
Alternatively, if you want to only print a known range, use the print command with the -n flag:
sed -n 201,300p file.txt # Print lines 201-300 from file.txt
This solution should work reliably on all Unix systems, regardless of the presence of GNU utilities.
Use:
sed -n '1d;p'
This command will delete the first line and print the rest.
If you want to see the first 10 lines you can use sed as below:
sed -n '1,10 p' myFile.txt
Or if you want to see lines from 20 to 30 you can use:
sed -n '20,30 p' myFile.txt
Just to propose a sed alternative. :) To skip first one million lines, try |sed '1,1000000d'.
Example:
$ perl -wle 'print for (1..1_000_005)'|sed '1,1000000d'
1000001
1000002
1000003
1000004
1000005
You can do this using the head and tail commands:
head -n <num> | tail -n <lines to print>
where num is 1e6 + the number of lines you want to print.
This shell script works fine for me:
#!/bin/bash
awk -v initial_line=$1 -v end_line=$2 '{
if (NR >= initial_line && NR <= end_line)
print $0
}' $3
Used with this sample file (file.txt):
one
two
three
four
five
six
The command (it will extract from second to fourth line in the file):
edu#debian5:~$./script.sh 2 4 file.txt
Output of this command:
two
three
four
Of course, you can improve it, for example by testing that all argument values are the expected :-)
cat < File > | awk '{if(NR > 6) print $0}'
I needed to do the same and found this thread.
I tried "tail -n +, but it just printed everything.
The more +lines worked nicely on the prompt, but it turned out it behaved totally different when run in headless mode (cronjob).
I finally wrote this myself:
skip=5
FILE="/tmp/filetoprint"
tail -n$((`cat "${FILE}" | wc -l` - skip)) "${FILE}"