I'd like to run:
grep nfs /etc/fstab | awk '{print $2}'
[root#nyproxy5 ~]# grep nfs /etc/fstab | awk '{print $2}'
/proxy_logs
/proxy_dump
/sync_logs
[root#nyproxy5 ~]#
And to get the output in one line delimited by space.
How can I do that?
If you don't mind a space (and no newline) at the end, you could do use this awk script:
awk '/nfs/{printf "%s ", $2}' /etc/fstab
For lines that match the pattern /nfs/, the second column is printed followed by a space. As a general rule, piping grep into awk is unnecessary as awk can do the pattern matching itself.
If you would like a newline at the end, you could use the END block:
awk '/nfs/{printf "%s ", $2}END{print ""}' /etc/fstab
This prints an empty string, followed by the output record separator (which is a newline). This will mean that you always have a newline in the output even if no matching records were found. If that's a problem, you could use a flag:
awk '/nfs/{f=1;printf "%s ", $2}END{if(f)print ""}' /etc/fstab
The flag f is set to true if the pattern is ever matched, causing the newline to be printed.
Newlines or any other character could be removed or replaced with tr command:
grep nfs /etc/fstab | awk '{print $2}' | tr -c '\n' ' '
If you want to get rid of tabs also:
| tr -c '\t' ' '
Related
Need help on awk
awk to ignore leading and trailing space and blank lines and commented lines if any from a file
Here you go,
grep "MyText" FromMyLog.log |awk -F " " '{print $2}'|awk -F "#" '{print $1}'
Here MyText is the key to grep from file FromMyLog.log
-F is used to avoid the following value, here space between quotes.
'{print $2}' will print the 2nd argument from the output, you can use $1, $2 as your requirement.
awk -F "#" This will ignore the commented lines.
This is just a hint for you, Modify the code with your requirements. This works for me while grep.
grep -v '^$\|^\s*\#' <filename> or egrep -v '^[[:space:]]*$|^ *#' <file_name> (if white spaces)
I think this is what you were asking for:
$> echo -e ' abc \t
\t efg
# alskdjfl
#
awk
# askdfh
' |
awk '
# match if first none space character is not a hash sign
/^[[:space:]]*[^#]/ {
# delete any spaces from start and end of line
sub(/^[[:space:]]*/, "");
sub(/[[:space:]]*$/, "", NF); # `NF` is Number of Fields
print
}'
abc
efg
awk
This can be folded onto a single line if so needed. Any problems, an actual example of the input (in a code block in your question) would be helpful.
Here's one way to extract required content ignoring spaces
FILE CONTENT
Server: 192.168.XX.XX
Address 1: 192.168.YY.YY
Name: central.google.com
Now to extract the server's address without spaces.
COMMAND
awk -F':' '/Server/ '{print $2}' YOURFILENAME | tr -s " "
option -s for squeezing the repetition of spaces.
which gives,
192.168.XX.XX
Here, notice that there is one leading space in the address.
To completely ignore spaces you can change that to,
awk -F':' '/Server/ '{print $2}' YOURFILENAME | tr -d [:space:]
option -d for removing particular characters, which is [:space:] here.
which gives,
192.168.YY.YY
without any leading or trailing spaces.
tr is an UNIX utility for translating, or deleting, or squeezing repeated characters. tr refers to translate here.
Examples:
tr [:lower:] [:upper:]
gives,
YOUAREAWESOME
for
youareawesome
Hope that helps.
I have the following line. I can grep one part but struggling with also grepping the second portion.
Line:
html:<TR><TD>PICK_1</TD><TD>36.0000</TD><TD>1000000</TD><TD>26965</TD><TD>100000000</TD><TD>97074000</TD><TD>2926000</TD><TD>2.926%</TD><TD>97.074%</TD></TR>
I want to have the following results after grepping this line.
PICK_1 97.074%
Currently just grepping first portion via following command.
grep -Po "<TR><TD>[A-Z0-9_]+" test.txt
Appreciate any help on how I can go about doing this. Thanks.
Use awk with a custom field separator:
awk -F'[<>TDR/]+' '{ print $2, $(NF-1) }' file
This splits the line on things that look like one or more opening or closing <TD> or <TR> tags, and prints the second and second-last field.
Warning: this will break on almost every input except the one that you've shown, since awk, grep and friends are designed for processing text, not HTML.
If you always have the same number of fields delimited by "TD" tags, you can try with this (dirty) awk:
awk -F'[<TD>|</TD>]' '{print $8 " " $80}'
Or this combination of column and awk:
column -t -s "</TD>" | awk -F' ' '{print $3 " " $11}'
Or with sed instead of column:
sed -e 's/<TD>/ /g' | awk -F' ' '{print $3 " " $11}'
try provide each patter after "-e" option
grep -e PICK_1 -e "<TR><TD>[A-Z0-9_]+" test.txt
awk -F'[<>]' '{print $5,$(NF-4)}' file
PICK_1 97.074%
I am validating few columns in a pipe delimited file. My second column is defaulted with '*'.
E.g. data of file to be validated:
abc|* |123
def|** |456
ghi|* |789
2nd record has 2 stars due to erroneous data.
I teied it as:
Value_to_match="*"
unmatch_count=cat <filename>| cut -d'|' -f2 | awk '{$1=$1};1' | grep -vw "$Value_to_match" | sort -n | uniq | wc -l
echo "unmatch_count"
This gives me count as 0 whereas I am expecting 1 (for **) as I have used -w with grep which is exact match and -v which is invert match.
How can I grep **?
The problem here is grep considering ** a regular expression. To prevent this, use -F to use fixed strings:
grep -F '**' file
However, you have an unnecessarily big set of piped operations, while awk alone can handle it quite well.
If you want to check lines containing ** in the second column, say:
$ awk -F"|" '$2 ~ /\*\*/' file
def|** |456
If you want to count how many of such lines you have, say:
$ awk -F"|" '$2 ~ /\*\*/ {sum++} END {print sum}' file
1
Note the usage of awk:
-F"|" to set the field separator to |.
$2 ~ /\*\*/ to say: hey, in every line check if the second field contains two asterisks (remember we sliced lines by |). We are escaping the * because it has a special meaning as a regular expression.
If you want to output those lines that have just one asterisk as second field, say:
$ awk -F"|" '$2 ~ /^*\s*$/' file
abc|* |123
ghi|* |789
Or check for those not matching this regex with !~:
$ awk -F"|" '$2 !~ /^*\s*$/' a
def|** |456
I would like to use xargs to list the contents of some files based on the output of command A. Xargs replace-str seem to be adding a space to the end and causing the command to fail. Any suggestions? I know this can be worked around using for loop. But curious to know how to do this using xargs.
lsscsi |awk -F\/ '/ATA/ {print $NF}' | xargs -L 1 -I % cat /sys/block/%/queue/scheduler
cat: /sys/block/sda /queue/scheduler: No such file or directory
The problem is not with xargs -I, which does not append a space to each argument, which can be verified as follows:
$ echo 'sda' | xargs -I % echo '[%]'
[sda]
Incidentally, specifying -L 1 in addition to -I is pointless: -I implies line-by-line processing.
Therefore, it must be the output from the command that provides input to xargs that contains the trailing space.
You can adapt your awk command to fix that:
lsscsi |
awk -F/ '/ATA/ {sub(/ $/,"", $NF); print $NF}' |
xargs -I % cat '/sys/block/%/queue/scheduler'
sub(/ $/,"", $NF) replaces a trailing space in field $NF with the empty string, thereby effectively removing it.
Note how I've (single-)quoted cat's argument so as to make it work even with filenames with spaces.
lsscsi |awk -F\/ '/ATA/ {print $NF}'| awk '{print $NF}' | xargs -L 1 -I % cat /sys/block/%/queue/scheduler
The first awk stmt splits by "/" so anything else is considered as field. In this is case "sda " becomes whole field including a space at the end. But by default, awk removes space . So after the pipe, the second awk prints $NF (which is last word of the line) and leaves out " " space as delimiter. awk { print $1 } will do the same because we have only one word, "sda" which is both first and last.
The string format is
Executed: variable_name
What is the simplest way to get the *variable_name* sub-string?
foo="Executed: variable_name"
echo ${foo##* } # Strip everything up to and including the rightmost space.
or
set -- $foo
echo $2
Unlike other solutions using awk, sed, and whatnot, these don't fork other programs and save thousands of CPU cycles since they execute completely in your shell. They are also more portable (unlike ${i/* /} which is a bashism).
With sed:
echo "Executed: variable_name" | sed 's/[^:]*: //'
Using awk:
echo 'Executed: variable_name' | awk -F' *: *' '{print $2}'
variable_name
If you have the string in a variable:
$ i="Executed: variable_name"
$ echo ${i/* /}
variable_name
If you have the string as output of a command
$ cmd
Executed: variable_name
$ cmd | awk '{print $NF}'
variable_name
Note that 'NF' means "number of fields", so $NF is always the last field on the line. Fields are assumed to be separated by spaces (unless -F is specified)
If your variable_name could have spaces in it, the
-F' *: *'
mentioned previously ensures that only the ": " is used as a field separator. However, this will preserve spaces at the end of the line if there are any.
If the line is mixed in with other output, you might need to filter.
Either grep..
$ cmd | grep '^Executed: ' | awk '{print $NF}'
or more clever awk..
$ cmd | awk '/^Executed: /{print $NF}'