I have file that looks like:
This is a RESTRICTED site.
All connections are monitored and recorded.
Disconnect IMMEDIATELY if you are not an authorized user!
sftp> cd outbox
sftp> ls -ltr
-rw------- 1 0 0 1911 Jun 12 20:40 61N0584832_EDIP000749728818_MFC_20190612203409.txt
-rw------- 1 0 0 1878 Jun 13 06:01 613577165_EDIP000750181517_MFC_20190613055207.txt
I want to print only the .txt file names, ideally in one command.
I can do:
grep -e '^-' outfile.log > outfile.log2
..which gives only the lines that start with '-'.
-rw------- 1 0 0 1911 Jun 12 20:40 61N0584832_EDIP000749728818_MFC_20190612203409.txt
-rw------- 1 0 0 1878 Jun 13 06:01 613577165_EDIP000750181517_MFC_20190613055207.txt
And then:
awk '{print $9}' outfile.log2 > outfile.log3
..which gives the desired output:
61N0584832_EDIP000749728818_MFC_20190612203409.txt
613577165_EDIP000750181517_MFC_20190613055207.txt
And so the question is, can these 2 commands be combined into 1?
You may use a single awk:
awk '/^-/{ print $9 }' file > outputfile
Or
awk '/^-/{ print $9 }' file > tmp && mv tmp file
It works like this:
/^-/ - finds each line starting with -
{ print $9 } - prints Field 9 of the matching lines only.
Seems like matching the leading - is not really want you want. If you want to just get the .txt files as output, filter on the file name:
awk '$9 ~ /\.txt$/{print $9}' input-file
Using grep with PCRE enabled (-P) flag:
grep -oP '^-.* \K.*' outfile.log
61N0584832_EDIP000749728818_MFC_20190612203409.txt
613577165_EDIP000750181517_MFC_20190613055207.txt
'^-.* \K.*' : Line starting with - till last white space are matched but ignored (anything left of \K will be matched and ignored) and matched part right of \K will be printed.
Since he clearly writes I want to print only the .txt file names, we should test for txt file and since file name are always the latest column we make it more portable by only test the latest filed line this:
awk '$NF ~ /\.txt$/{print $NF}' outfile.log > outfile.log2
61N0584832_EDIP000749728818_MFC_20190612203409.txt
613577165_EDIP000750181517_MFC_20190613055207.txt
Related
I'm running this on an AIX 6.1.
The intended purpose of this command is to display the following information in the following format:
GetUsedRAM:GetUsedSwap:CPU_0_System:CPU_0_User:…CPU_N_System:CPU_N_User
The command is composed of several sub commands:
echo `vmstat 1 2 | tr -s ' ' ':' | cut -d':' -f4,5,14-15 | tail -1 | sed 's/\([0-9]*:[0-9]*:\)\([0-9]*:[0-9]*\)/\1/'``mpstat -a 1 1 | tr -s ' ' '|' | head -8 | tail -4 | cut -d'|' -f 25,27 | awk -F "|" '{printf "%.0f:%.0f:",$2,$1}' | sed '$s/.$//'| sed -e "s/ \{1,\}$//"| awk '{int a[10];split($1, a,":");printf("%d:%d:%d:%d:%d:%d:%d:%d",a[0],a[1],a[2],a[3],a[4],a[5],a[6],a[7])}'`
Which I'll re format for clarity:
echo \
`vmstat 1 2 |
tr -s ' ' ':' |
cut -d':' -f4,5,14-15 |
tail -1 |
sed 's/\([0-9]*:[0-9]*:\)\([0-9]*:[0-9]*\)/\1/' \
` \
`mpstat -a 1 1 |
tr -s ' ' '|' |
head -8 |
tail -4 |
cut -d'|' -f 25,27 |
awk -F "|" '{printf "%.0f:%.0f:",$2,$1}' |
sed '$s/.$//' |
sed -e "s/ \{1,\}$//" |
awk '{int a[10];split($1, a,":");printf("%d:%d:%d:%d:%d:%d:%d:%d",a[0],a[1],a[2],a[3],a[4],a[5],a[6],a[7])}' \
`
I understand all of the tr, cut, head tail, and (roughly) vmstat/mpstat commands. The first sed is where I get lost, I've tried running the command in smaller segments and not quite sure why it seems to work as a whole but not when I truncate the command before the next tr.
I'm also not so sure on the awk command although I understand the premise vaguely, as a function allowing formatted output.
Similarly, I have a vague understanding of sed being a command allowing certain strings/characters being replaced in some file.
I'm not able to make out what this specific implementation in the above case is.
Could anyone provide some clarity or direction as to exactly what is happening at each sed and awk step within the context of the entire command?
Thanks for your help.
Simplification
This two simpler commands will get the exact same output:
# GetUsedRAM:GetUsedSwap:CPU_0_System:CPU_0_User:…CPU_N_System:CPU_N_User
# Select fields 4,5 of last line, and format with :
comm1=`vmstat 1 2 |
awk '$4~/[0-9]/{avm=$4;fre=$5} END{printf "%s:%s",avm,fre}'
`
# Select fields 27 (sy) and 25 (us) for four cpu, print as decimal.
comm2=`mpstat -A 1 1 |
awk -v firstline=6 -v cpus=4 '
BEGIN{start=firstline-1; end=firstline+cpus;}
NR>start && NR<end {printf( ":%d:%d", $27,$25)}'
`
echo "${comm1}${comm2}"
Description.
Description of original commands
The whole command is the concatenation of two commands.
The first command:
The output of the vmstat is shown in this link.
The columns 4 and 5 are 'avm' and 'fre'. The output in columns 14 and 15,
seem to be 'us' (user) and 'sy' (system). And I say seem as no output
from the user is available to confirm.
The first command
`vmstat 1 2 | # Execute the command vmstat.
tr -s ' ' ':' | # convert all spaces to colon (:).
cut -d':' -f4,5,14-15 | # select fields 4,5,14,and 15
tail -1 | # select last line.
sed 's/\([0-9]*:[0-9]*:\)\([0-9]*:[0-9]*\)/\1/' \ # See below.
`
The sed command selects inside braces all digits [0-9]* before a colon
repeated twice. And then again (without the last colon). That's the whole
string in two parts: « (dd:dd:)(dd:dd) » (d means digit).
And finally, it replaces such whole string by what was selected inside
the first braces /\1/.
All this complexity just removes fields 14 and 15 as selected by cut.
A simpler command with exactly the same output is:
Select fields 4,5 of last line, and format with (:).
`vmstat 1 2 | awk '
$4~/[0-9]/{avm=$4;fre=$5} END{printf "%s:%s:",avm,fre}'
`
The second command:
The output of mpstat -A is similar to this one from Linux.
And also similar to this AIX mpstat -d output.
However, the exact output of AIX 6.1 for mpstat -a (ALL) on the computer
used could have several variations. Anyway, guided by the intended final
output desired: CPU_0_System:CPU_0_User:…CPU_N_System:CPU_N_User.
It seems that the columns to be selected should be us (user) and sy
(sys) percent of time that used the cpu for all cpu in use,
which seem to be four on the computer measured.
The manual for AIX 6.1 mpstat is here.
It has a list of all the 40 columns that are presented when the option
-a ALL is used:
CPU min maj mpcs mpcr dev soft dec ph cs ics bound rq push
S3pull S3grd S0rd S1rd S2rd S3rd S4rd S5rd S3hrd S4hrd S5hrd
sysc us sy wa id pc %ec ilcs vlcs lcs %idon %bdon %istol %bstol %nsp
us and sy are listed as the fields 27 and 28, however the command presented
by the user selects fields number 25 and 27. Close but not the same. The
only way to confirm would be to receive the output of the command from the user.
For testing I will be using the output of mpstat 5 1 from here.
# mpstat 5 1
System configuration: lcpu=4 ent=1.0 mode=Uncapped
cpu min maj mpc int cs ics rq mig lpa sysc us sy wt id pc %ec lcs
0 4940 0 1 632 685 268 0 320 100 263924 42 55 0 4 0.57 35.1 277
1 990 0 3 1387 2234 805 0 684 100 130290 28 47 0 25 0.27 16.6 649
2 3943 0 2 531 663 223 0 389 100 276520 44 54 0 3 0.57 34.9 270
3 1298 0 2 1856 2742 846 0 752 100 82141 31 40 0 29 0.22 13.4 650
ALL 11171 0 8 4406 6324 2142 0 2145 100 752875 39 51 0 10 1.63 163.1 1846
The second command
`mpstat -A 1 1 | # execute command
tr -s ' ' '|' | # replace all spaces with (|).
head -8 | # select 8 first lines.
tail -4 | # select last four lines.
cut -d'|' -f 25,27 | # select fields 25 and 27
awk -F "|" '{printf "%.0f:%.0f:",$2,$1}' | # print the fields as integers.
sed '$s/.$//' | # on the last line ($), substitute the last character (.$) by nothing.
sed -e "s/ \{1,\}$//" | # remove trailing space(s).
awk '{
int a[10];
split($1, a,":");
printf("%d:%d:%d:%d:%d:%d:%d:%d",a[0],a[1],a[2],a[3],a[4],a[5],a[6],a[7])
}' \
`
About the int: For older versions of awk, calling a function without the parentheses is equivalent to call the function on $0. int is equivalent to int($0), which is not printed, nor used. The same happens to the value of a[10].
The split sets each value of the command in a[i]. Then, all values of a[i] are printed as decimals.
The equivalent, and way simpler is:
Command #2
`mpstat -A 1 1 |
awk -v firstline=6 -v cpus=4 '
BEGIN{start=firstline-1; end=firstline+cpus;}
NR>start && NR<end {printf( ":%d:%d", $27,$25)}'
`
I needed to extract all hits from one list (list.txt) which can be found in one of the columns of another (here in Data.txt) into a third (output.txt).
Data.txt (tab delimited)
some_data more_data other_data here yet_more_data etc
A B 2 Gee;Whiz;Hello 13 12
A B 2 Gee;Whizz;Hi 56 32
E 4 Btm;Lol 16 2
T 3 Whizz 13 3
List.txt
Gee
Whiz
Lol
Ideally output.txt looks like
some_data more_data other_data here yet_more_data etc
A B 2 Gee;Whiz;Hello 13 12
A B 2 Gee;Whizz;Hi 56 32
E 4 Btm;Lol 16 2
So I tried a shell script
for ids in List.txt
do
grep $ids Data.txt >> output.txt
done
except I typed out everything (cut and paste actually) in List.txt in said script.
Unfortunately it gave me an output.txt including the last line, I assume as 'Whizz' contains 'Whiz'.
I also tried cat Data.txt | egrep -F "List.txt" and that resulted in grep: conflicting matchers specified -- I suppose that was too naive of me. The actual files: List.txt contains a sorted list of 985 words, Data.txt has 115576 rows with 17 columns.
Some help/guidance would be much appreciated thanks.
Try something like this:
for ids in List.txt
do
grep "[TAB;]$ids[TAB;]" Data.txt >> output.txt
done
But it has two drawbacks:
"Data.txt" is scanned multiple times
You can get one line multiple times.
If it is problem try two step version:
cat List.txt | sed -e "s/.*/[TAB;]\0[TAB;]/g" > List_mod.txt
grep -f List_mod.txt Data.txt > output.txt
Note:
TAB character can be inserted by combination Ctrl-V following by Tab key in command line, and Tab character in editor. You have to check if your edit does not change tab to series of spaces.
The UNIX tool for general text processing is "awk":
awk '
NR==FNR { list[$0]; next }
{
for (word in list) {
if ($0 ~ "[\t;]" word "[\t;]") {
print
next
}
}
}
' List.txt Data.txt > output.txt
I have two files like this:
abc.txt
a
b
z
1
10
and abcd.txt
a
b
c
d
1
10
100
1000
I would like:
a
b
1
10
I would like to use grep -fw abc.txt abcd.txt to search through every line of abc.txt and print lines which match the entire word. If I just use grep -f, I get lines 100 since the pattern '10' matches '100'. But grep -f -w abc.txt abcd.txt produces:
a
b
1
and doesn't print out the 10. So, I guess, what is the best way to match every line in abc.txt with the entire line of abcd.txt ?
When I do ls -l | grep ^d it lists only directories in the current directory.
What I'd like to know is what does the character ^ in ^d mean?
The caret ^ and the dollar sign $ are meta-characters that respectively match the empty string at the beginning and end of a line.The grep is matching only lines that start with "d".
To complement the good answer by The New Idiot, I want to point out that this:
ls -l | grep ^d
Shows all directories in the current directory. That's because the ls -l adds a d in the beginning of the directories info.
The format of ls -l is like:
-rwxr-xr-x 1 user group 0 Jun 12 12:25 exec_file
-rw-rw-r-- 1 user group 0 Jun 12 12:25 normal_file
drwxr-xr-x 16 user group 4096 May 24 12:46 dir
^
|___ see the "d"
To make it more clear, you can ls -lF to include a / to the end of the directories info:
-rwxr-xr-x 1 user group 0 Jun 12 12:25 exec_file*
-rw-rw-r-- 1 user group 0 Jun 12 12:25 normal_file
drwxr-xr-x 16 user group 4096 May 24 12:46 dir/
So ls -lF | grep /$ will do the same as ls -l | grep ^d.
It has two meanings. One as 'The New Idiot' above pointed out. The other, equally useful, is within character class expression, where it means negation: grep -E '[^[:digit:]]' accepts any character except a digit. The^` must be the first character within [].
How do I tell grep to only print out lines if the "filename" matches when I'm piping through ls? I want it to ignore everything on each line until after the timestamp. There must be some easy way to do this on a single command.
As you can see, without it, if I searched for the file "rwx", it would return not only the line with rwx.c, but also the first three lines because of permissions. I was going to use AWK but I want it to display the whole last line if I search for "rwx".
Any ideas?
EDIT: Thanks for the hacks below. However, it would be great to have a more bug-free method. For example, if I had a file named "rob rob", I wouldn't be able to use the stated solutions.
drwxrwxr-x 2 rob rob 4096 2012-03-04 18:03 .
drwxrwxr-x 4 rob rob 4096 2012-03-04 12:38 ..
-rwxrwxr-x 1 rob rob 13783 2012-03-04 18:03 a.out
-rw-rw-r-- 1 rob rob 4294 2012-03-04 18:02 function1.c
-rw-rw-r-- 1 rob rob 273 2012-03-04 12:54 function1.c~
-rw-rw-r-- 1 rob rob 16 2012-03-04 18:02 rwx.c
-rw-rw-r-- 1 rob rob 16 2012-03-04 18:02 rob rob
The following will list only file name, and one file in each row.
$ ls -1
To include . files
$ ls -1a
Please note that the argument is number "1", not letter "l".
Why don't you use grep and match the file name following the timestamp?
grep -P "[0-9]{2}:[0-9]{2} $FILENAME(\.[a-zA-Z0-9]+)?$"
The [0-9]{2}:[0-9]{2} is for the time, the $FILENAME is where you'd put rob rob or rwx, and the trailing (\.[a-zA-Z0-9]+)? is to allow for an optional extension.
Edit: #JonathanLeffler below points out that when files are older than bout 6 months the time column gets replaced by a year - this is what happens on my computer anyhow. You could do ([0-9]{2}:[0-9]{2}|(19|20)[0-9]{2}) to allow time OR year, but you may be best of using awk (?).
[foo#bar ~/tmp]$ls -al
total 8
drwxrwxr-x 2 foo foo 4096 Mar 5 09:30 .
drwxr-xr-- 83 foo foo 4096 Mar 5 09:30 ..
-rw-rw-r-- 1 foo foo 0 Mar 5 09:30 foo foo
-rw-rw-r-- 1 foo foo 0 Mar 5 09:29 rwx.c
-rw-rw-r-- 1 foo foo 0 Mar 5 09:29 tmp
[foo#bar ~/tmp]$export filename='foo foo'
[foo#bar ~/tmp]$echo $filename
foo foo
[foo#bar ~/tmp]$ls -al | grep -P "[0-9]{2}:[0-9]{2} $filename(\.[a-zA-Z0-9]+)?$"
-rw-rw-r-- 1 cha66i cha66i 0 Mar 5 09:30 foo foo
(You could additionally extend to matching the whole line if you wanted:
^ # start of line
[d-]([r-][w-][x-]){3} + # permissions & space (note: is there a 't' or 's'
# sometimes where the 'd' can be??)
[0-9]+ # whatever that number is
[\w-]+ [\w-]+ + # user/group (are spaces allowed in these?)
[0-9]+ + # file size (modify for -h switch??)
(19|20)[0-9]{2}- # yyyy (modify if you want to allow <1900)
(1[012]|0[1-9])- # mm
(0[1-9]|[12][0-9]|3[012]) + # dd
([01][0-9]|2[0-3]):[0-6][0-9] +# HH:MM (24hr)
$filename(\.[a-zA-Z0-9]+)? # filename & optional extension
$ # end of line
. You get the point, tailor to your needs.)
Assuming that you aren't prepared to do:
ls -ld $(ls -a | grep rwx)
then you need to exploit the fact that there are 8 columns with space separation before the file name starts. Using egrep (or grep -E), you could do:
ls -al | egrep "^([^ ]+ +){8}.*rwx"
This looks for 'rwx' after the 8th column. If you want the name to start with rwx, omit the .*. If you want the name to end with rwx, add a $ at the end. Note that I used double quotes so you could interpolate a variable in place of the literal rwx.
This was tested on Mac OS X 10.7.3; the ls -l command consistently gives three columns for the date field:
-r--r--r-- 1 jleffler staff 6510 Mar 17 2003 README,v
-r--r--r-- 1 jleffler staff 26676 Mar 3 21:44 ccs.nmd
Your ls -l seems to be giving just two columns, so you'd need to change the {8} to {7} for your machine - and beware migrating between systems.
Well, if you're working with filenames that don't have spaces in them, you could do something like this:
grep 'rwx\S*$'
Aside frrm the fact that you can use pattern matching with ls, exaple ksh and bash,
which is probably what you should do, you can use the fact that filename occur in a
fixed position. awk (gawk, nawk or whaever you have) is a better choice for this.
If you have to use grep it smells like homework to me. Please tag it that way.
Assume the filename starting position is based on this output from ls -l in linux: 56
-rwxr-xr-x 1 Administrators None 2052 Feb 28 20:29 vote2012.txt
ls -l | awk ' substr($0,56) ~/your pattern even with spaces goes here/'
e.g.,
ls -l | awk ' substr($0,56) ~/^val/'
will find files starting with "val"
As a simple hack, just add a space before your filename so you don't match the beginning of the output:
ls -al | grep '\srwx'
Edit: OK, this is not as robust as it should be. Here's awk:
ls -l | awk ' $9 ~ /rwx/ { print $0 }'
This works for me, unlike ls -l & others as some folks pointed out. I like this because its really generic & gives me the base file name, which removes the path names before the file.
ls -1 /path_name |awk -F/ '{print $NF}'
Only one command you needed for this --
ls -al | gawk '{print $9}'
You can use this:
ls -p | grep -v /
this is super old, but i needed the answer and had a hard time finding it. i didn't really care about the one-liner part; i just needed it done. this is down and dirty and requires that you count the columns. i'm not looking for an upvote here, just leaving some options for future searcher-ers.
the helpful awk trick is here -- Using awk to print all columns from the nth to the last
if
YOUR_FILENAME="rob rob"
and
WHERE_FILENAMES_START=8
ls -al | while read x; do
y=$(echo "$x" | awk '{for(i=$WHERE_FILENAMES_START; i<=NF; ++i) printf $i""FS; print ""}')
[[ "$YOUR_FILENAME " = "$y" ]] && echo "$x"
done
if you save it as a bash script and swap out the vars with $2 and $1, throw the script in your usr bin... then you'll have your clean simple one-liner ;)
output will be:
> -rw-rw-r-- 1 rob rob 16 2012-03-04 18:02 rob rob
the question was for a one-liner so...
ls -al | while read x; do [[ "$YOUR_FILENAME " = "$(echo "$x" | awk '{for(i=WHERE_FILENAMES_START; i<=NF; ++i) printf $i""FS; print ""}')" ]] && echo "$x" ; done
(lol ;P)
on another note: mathematical.coffee your answer was rad. it didn't solve my version of this problem, so i didn't upvote, but i liked your regex breakdown :D