The first command does work as expected but not the second one.
I want to add some text either at the beginning or the end.
# grep `date +'%y%m%d'` /var/log/mysqld.log
100101 10:56:00 mysqld started
100101 10:56:02 InnoDB: Started; log sequence number 1 2052750649
# sed 's/^/computer /g' < grep `date +'%y%m%d'` /var/log/mysqld.log
bash: grep: No such file or directory
# expected output
computer 100101 10:56:00 mysqld started
computer 100101 10:56:02 InnoDB: Started; log sequence number 1 2052750649
As you wrote it with input redirection, it looks for a file called grep in the current directory and tries to read its contents, not execute it. You need to use a pipe instead:
grep `date +'%y%m%d'` /var/log/mysqld.log | sed 's/^/computer /'
I also removed the 'g' modifier from your sed as it is completely unnecessary.
Just one awk command
awk -vd=$(date +'%y%m%d') '$0~d{ print "computer "$0 }' /var/log/mysqld.log
Related
Given a file, for example:
potato: 1234
apple: 5678
potato: 5432
grape: 4567
banana: 5432
sushi: 56789
I'd like to grep for all lines that start with potato: but only pipe the numbers that follow potato:. So in the above example, the output would be:
1234
5432
How can I do that?
grep 'potato:' file.txt | sed 's/^.*: //'
grep looks for any line that contains the string potato:, then, for each of these lines, sed replaces (s/// - substitute) any character (.*) from the beginning of the line (^) until the last occurrence of the sequence : (colon followed by space) with the empty string (s/...// - substitute the first part with the second part, which is empty).
or
grep 'potato:' file.txt | cut -d\ -f2
For each line that contains potato:, cut will split the line into multiple fields delimited by space (-d\ - d = delimiter, \ = escaped space character, something like -d" " would have also worked) and print the second field of each such line (-f2).
or
grep 'potato:' file.txt | awk '{print $2}'
For each line that contains potato:, awk will print the second field (print $2) which is delimited by default by spaces.
or
grep 'potato:' file.txt | perl -e 'for(<>){s/^.*: //;print}'
All lines that contain potato: are sent to an inline (-e) Perl script that takes all lines from stdin, then, for each of these lines, does the same substitution as in the first example above, then prints it.
or
awk '{if(/potato:/) print $2}' < file.txt
The file is sent via stdin (< file.txt sends the contents of the file via stdin to the command on the left) to an awk script that, for each line that contains potato: (if(/potato:/) returns true if the regular expression /potato:/ matches the current line), prints the second field, as described above.
or
perl -e 'for(<>){/potato:/ && s/^.*: // && print}' < file.txt
The file is sent via stdin (< file.txt, see above) to a Perl script that works similarly to the one above, but this time it also makes sure each line contains the string potato: (/potato:/ is a regular expression that matches if the current line contains potato:, and, if it does (&&), then proceeds to apply the regular expression described above and prints the result).
Or use regex assertions: grep -oP '(?<=potato: ).*' file.txt
grep -Po 'potato:\s\K.*' file
-P to use Perl regular expression
-o to output only the match
\s to match the space after potato:
\K to omit the match
.* to match rest of the string(s)
sed -n 's/^potato:[[:space:]]*//p' file.txt
One can think of Grep as a restricted Sed, or of Sed as a generalized Grep. In this case, Sed is one good, lightweight tool that does what you want -- though, of course, there exist several other reasonable ways to do it, too.
This will print everything after each match, on that same line only:
perl -lne 'print $1 if /^potato:\s*(.*)/' file.txt
This will do the same, except it will also print all subsequent lines:
perl -lne 'if ($found){print} elsif (/^potato:\s*(.*)/){print $1; $found++}' file.txt
These command-line options are used:
-n loop around each line of the input file
-l removes newlines before processing, and adds them back in afterwards
-e execute the perl code
You can use grep, as the other answers state. But you don't need grep, awk, sed, perl, cut, or any external tool. You can do it with pure bash.
Try this (semicolons are there to allow you to put it all on one line):
$ while read line;
do
if [[ "${line%%:\ *}" == "potato" ]];
then
echo ${line##*:\ };
fi;
done< file.txt
## tells bash to delete the longest match of ": " in $line from the front.
$ while read line; do echo ${line##*:\ }; done< file.txt
1234
5678
5432
4567
5432
56789
or if you wanted the key rather than the value, %% tells bash to delete the longest match of ": " in $line from the end.
$ while read line; do echo ${line%%:\ *}; done< file.txt
potato
apple
potato
grape
banana
sushi
The substring to split on is ":\ " because the space character must be escaped with the backslash.
You can find more like these at the linux documentation project.
Modern BASH has support for regular expressions:
while read -r line; do
if [[ $line =~ ^potato:\ ([0-9]+) ]]; then
echo "${BASH_REMATCH[1]}"
fi
done
grep potato file | grep -o "[0-9].*"
I have the following log line
2021-10-28 10:26:19,624 INFO [Native-Thread-7f9e6effd700] idol.nifi.connector.GetWeb GetWeb[id=c59c415c-017c-1000-195c-e85818b0a032] Processing: [depth:0] https://qed.qld.gov.au/about-us/rti/disclosure-log/disclosure-log-2015
I want to extract the date and everything that comes after the word Processing: so that is looks like
2021-10-28 10:26:19 [depth:0] https://qed.qld.gov.au/about-us/rti/disclosure-log/disclosure-log-2015
I am not sure how to achieve this with grep or sed?
#!/bin/bash cat ./logs/*.log | grep Processing: | grep -E "(\d\d\d\d-\d\d-\d\d \d\d:\d\d:\d\d)"
is as far as I got.
sed 's/,.*Processing://' file
Output:
2021-10-28 10:26:19 [depth:0] https://qed.qld.gov.au/about-us/rti/disclosure-log/disclosure-log-2015
See: man sed and The Stack Overflow Regular Expressions FAQ
I have a CSV. I want to edit the 35th field of the CSV and write the change back to the 35th field. This is what I am doing on bash:
awk -F "," '{print $35}' test.csv | sed -i 's/^0/+91/g'
so, I am pulling the 35th entry using awk and then replacing the "0" in the starting position in the string with "+91". This one works perfet and I get desired output on the console.
Now I want this new entry to get written in the file. I am thinking of sed's "in -place" replacement feature but this fetuare needs and input file. In above command, I cannot provide input file because my primary command is awk and sed is taking the input from awk.
Thanks.
You should choose one of the two tools. As for sed, it can be done as follows:
sed -ri 's/^(([^,]*,){34})0([^,]*)/\1+91\3/' test.csv
Not sure about awk, but #shellter's comment might help with that.
The in-place feature of sed is misnamed, as it does not edit the file in place. Instead, it creates a new file with the same name. eg:
$ echo foo > foo
$ ln -f foo bar
$ ls -i foo bar # These are the same file
797325 bar 797325 foo
$ echo new-text > foo # Changes bar
$ cat bar
new-text
$ printf '/new/s//newer\nw\nq\n' | ed foo # Edit foo "in-place"; changes bar
9
newer-text
11
$ cat bar
newer-text
$ ls -i foo bar # Still the same file
797325 bar 797325 foo
$ sed -i s/new/newer/ foo # Does not edit in-place; creates a new file
$ ls -i foo bar
797325 bar 792722 foo
Since sed is not actually editing the file in place, but writing a new file and then renaming it to the old file, you might as well do the same.
awk ... test.csv | sed ... > test.csv.1 && mv test.csv.1 test.csv
There is the misperception that using sed -i somehow avoids the creation of the temporary file. It does not. It just hides the fact from you. Sometimes abstraction is a good thing, but other times it is unnecessary obfuscation. In the case of sed -i, it is the latter. The shell is really good at file manipulation. Use it as intended. If you do need to edit a file in place, don't use the streaming version of ed; just use ed
So, it turned out there are numerous ways to do it. I got it working with sed as below:
sed -i 's/0\([0-9]\{10\}\)/\+91\1/g' test.csv
But this is little tricky as it will edit any entry which matches the criteria. however in my case, It is working fine.
Similar implementation of above logic in perl:
perl -p -i -e 's/\b0(\d{10})\b/\+91$1/g;' test.csv
Again, same caveat as mentioned above.
More precise way of doing it as shown by Lev Levitsky because it will operate specifically on the 35th field
sed -ri 's/^(([^,]*,){34})0([^,]*)/\1+91\3/g' test.csv
For more complex situations, I will have to consider using any of the csv modules of perl.
Thanks everyone for your time and input. I surely know more about sed/awk after reading your replies.
This might work for you:
sed -i 's/[^,]*/+91/35' test.csv
EDIT:
To replace the leading zero in the 35th field:
sed 'h;s/[^,]*/\n&/35;/\n0/!{x;b};s//+91/' test.csv
or more simply:
|sed 's/^\(\([^,]*,\)\{34\}\)0/\1+91/' test.csv
If you have moreutils installed, you can simply use the sponge tool:
awk -F "," '{print $35}' test.csv | sed -i 's/^0/+91/g' | sponge test.csv
sponge soaks up the input, closes the input pipe (stdin) and, only then, opens and writes to the test.csv file.
As of 2015, moreutils is available in package repositories of several major Linux distributions, such as Arch Linux, Debian and Ubuntu.
Another perl solution to edit the 35th field in-place:
perl -i -F, -lane '$F[34] =~ s/^0/+91/; print join ",",#F' test.csv
These command-line options are used:
-i edit the file in-place
-n loop around every line of the input file
-l removes newlines before processing, and adds them back in afterwards
-a autosplit mode – split input lines into the #F array. Defaults to splitting on whitespace.
-e execute the perl code
-F autosplit modifier, in this case splits on ,
#F is the array of words in each line, indexed starting with 0
$F[34] is the 35 element of the array
s/^0/+91/ does the substitution
I would like the option of extracting the following string/data:
/work/foo/processed/25
/work/foo/processed/myproxy
/work/foo/processed/sample
=or=
25
myproxy
sample
But it would help if I see both.
From this output using cut or perl or anything else that would work:
Found 3 items
drwxr-xr-x - foo_hd foo_users 0 2011-03-16 18:46 /work/foo/processed/25
drwxr-xr-x - foo_hd foo_users 0 2011-04-05 07:10 /work/foo/processed/myproxy
drwxr-x--- - foo_hd testcont 0 2011-04-08 07:19 /work/foo/processed/sample
Doing a cut -d" " -f6 will get me foo_users, testcont. I tried increasing the field to higher values and I'm just not able to get what I want.
I'm not sure if cut is good for this or something like perl?
The base directories will remain static /work/foo/processed.
Also, I need the first line Found Xn items removed. Thanks.
You can do a substitution from beginning to the first occurrence of / , (non greedily)
$ your_command | ruby -ne 'print $_.sub(/.*?\/(.*)/,"/\\1") if /\//'
/work/foo/processed/25
/work/foo/processed/myproxy
/work/foo/processed/sample
Or you can find a unique separator (field delimiter) to split on. for example, the time portion is unique , so you can split on that and get the last element. (2nd element)
$ ruby -ne 'print $_.split(/\s+\d+:\d+\s+/)[-1] if /\//' file
/work/foo/processed/25
/work/foo/processed/myproxy
/work/foo/processed/sample
With awk,
$ awk -F"[0-9][0-9]:[0-9][0-9]" '/\//{print $NF}' file
/work/foo/processed/25
/work/foo/processed/myproxy
/work/foo/processed/sample
perl -lanF"\s+" -e 'print #F[-1] unless /^Found/' file
Here is an explanation of the command-line switches used:
-l: remove line break from each line of input, then add one back on print
-a: auto-split each line of input into an #F array
-n: loop through each line of input
-F: the regexp pattern to use for the auto-split (with -a)
-e: the perl code to execute (for each line of input if using -n or -p)
If you want to just output the last portion of your directory path, and the basedir is always '/work/foo/processed', I would do this:
perl -nle 'print $1 if m|/work/foo/processed/(\S+)|' file
Try this out :
<Your Command> | grep -P -o '[\/\.\w]+$'
OR if the directory '/work/foo/processed' is always static then:
<Your Command>| grep -P -o '\/work\/foo\/processed\/.+$'
-o : Show only the part of a matching line that matches PATTERN.
-P : Interpret PATTERN as a Perl regular expression.
In this example, the last word in the input will be matched .
(The word can also contain dot(s)),so file names like 'text_file1.txt', can be matched).
Ofcourse, you can change the pattern, as per your requirement.
If you know the columns will be the same, and you always list the full path name, you could try something like:
ls -l | cut -c79-
which would cut out the 79th character until the end. That might work in this exact case, but I think it would be better to find the basename of the last field. You could easily do this in awk or perl. Respond if this is not what you want and I'll add the awk and perl versions.
take the output of your ls command and pipe it to awk
your command|awk -F'/' '{print $NF}'
your_command | perl -pe 's#.*/##'
There are often times I will grep -n whatever file to find what I am looking for. Say the output is:
1234: whatev 1
5555: whatev 2
6643: whatev 3
If I want to then just extract the lines between 1234 and 5555, is there a tool to do that? For static files I have a script that does wc -l of the file and then does the math to split it out with tail & head but that doesn't work out so well with log files that are constantly being written to.
Try using sed as mentioned on
http://linuxcommando.blogspot.com/2008/03/using-sed-to-extract-lines-in-text-file.html. For example use
sed '2,4!d' somefile.txt
to print from the second line to the fourth line of somefile.txt. (And don't forget to check http://www.grymoire.com/Unix/Sed.html, sed is a wonderful tool.)
The following command will do what you asked for "extract the lines between 1234 and 5555" in someFile.
sed -n '1234,5555p' someFile
If I understand correctly, you want to find a pattern between two line numbers. The awk one-liner could be
awk '/whatev/ && NR >= 1234 && NR <= 5555' file
You don't need to run grep followed by sed.
Perl one-liner:
perl -ne 'if (/whatev/ && $. >= 1234 && $. <= 5555) {print}' file
Line numbers are OK if you can guarantee the position of what you want. Over the years, my favorite flavor of this has been something like this:
sed "/First Line of Text/,/Last Line of Text/d" filename
which deletes all lines from the first matched line to the last match, including those lines.
Use sed -n with "p" instead of "d" to print those lines instead. Way more useful for me, as I usually don't know where those lines are.
Put this in a file and make it executable:
#!/usr/bin/env bash
start=`grep -n $1 < $3 | head -n1 | cut -d: -f1; exit ${PIPESTATUS[0]}`
if [ ${PIPESTATUS[0]} -ne 0 ]; then
echo "couldn't find start pattern!" 1>&2
exit 1
fi
stop=`tail -n +$start < $3 | grep -n $2 | head -n1 | cut -d: -f1; exit ${PIPESTATUS[1]}`
if [ ${PIPESTATUS[0]} -ne 0 ]; then
echo "couldn't find end pattern!" 1>&2
exit 1
fi
stop=$(( $stop + $start - 1))
sed "$start,$stop!d" < $3
Execute the file with arguments (NOTE that the script does not handle spaces in arguments!):
Starting grep pattern
Stopping grep pattern
File path
To use with your example, use arguments: 1234 5555 myfile.txt
Includes lines with starting and stopping pattern.
If I want to then just extract the lines between 1234 and 5555, is
there a tool to do that?
There is also ugrep, a GNU/BSD grep compatible tool but one that offers a -K option (or --range) with a range of line numbers to do just that:
ugrep -K1234,5555 -n '' somefile.log
You can use the usual GNU/BSD grep options and regex patterns (but it also offers a lot more such as -K.)
If you want lines instead of line ranges, you can do it with perl: eg. if you want to get line 1, 3 and 5 from a file, say /etc/passwd:
perl -e 'while(<>){if(++$l~~[1,3,5]){print}}' < /etc/passwd