34
This is what the cut command is for.
cut -d';' -f2-
19
With date, shuf and xargs:
Convert start and end date to "seconds since 1970-01-01 00:00:00 UTC" and use shuf to print six random values in this range. Pipe this result to xargs and convert the values to the desired date format.
Edit: If you want dates of the year 2017 to be included in the output, you have add one year -1s (2017-12-31 23:59:59) to the end ...
14
Another sed variant:
sed 's/^[^;]*;//' file
Replace any non-semicolon characters ([^;]*) at the beginning of the line (the first ^) followed by a semicolon with an empty string.
10
You can turn the problem into generating a random number between a number representing the first possible date and a number representing the last possible date (actually the one right after the last possible), in unix epoch format. Everything else is handled by standard date conversions. gawk has a better random number resolution than bash (float vs 15 bits ...
10
Perl:
perl -MTime::Piece -sE '
my $t1 = Time::Piece->strptime("$y1-01-01 00:00:00", "%Y-%m-%d %H:%M:%S");
my $t2 = Time::Piece->strptime("$y2-12-31 23:59:59", "%Y-%m-%d %H:%M:%S");
do {
my $time = $t1 + int(rand() * ($t2 - $t1));
say $time->dmy;
} for 1..6;
' -- -y1=1987 -y2=2017
Sample output:
10-01-1989
30-04-...
9
Inside the command substitution you have "$FILE" | sed -e 's/\\/\//gp', which the shell expands to (the equivalent of) '\\edi3\welshch\test' | sed -e 's/\\/\//gp'. Since it's a command, the shell goes looking for a file called \\edi3\welshch\test to run.
You probably meant to use echo "$FILE" | sed ... to pass the contents of FILE to sed via the pipe.
...
9
Here's one way, mostly in awk:
#!/bin/sh
[ "$end" -ge "$start" ] || exit 1
awk -v start=$1 -v end=$2 '
BEGIN {
srand();
for(i=1; i <= 6; i++)
printf "%02d/%02d/%d\n", 1 + rand() * 28, 1 + rand() * 12, start + rand() * (end-start);
}
' < /dev/null
The shell script takes two parameters and passes them as variables to awk, who reads ...
8
Using variable name prefix matching and variable indirection in bash:
proj_env_repo_db_username=username
proj_env_repo_db_host=host
proj_env_repo_db_port=port
for variable in "${!proj_env_repo@}"; do
export "TF_ENV${variable#proj_env_repo}"="${!variable}"
done
The loop uses "${!proj_env_repo@}" to generate a list of variable names that share the name ...
8
In awk:
awk 'BEGIN { FS = ";" } ; { print $2 }'
or, even simpler
awk -F";" '{print $2}'
(thank you, ilkkachu, for this simplified version)
6
With GNU sed:
sed -E 's/^.*;(.*)/\1/' file
With GNU grep:
grep -Po '(?!.*;).*' file
6
Run this from your terminal to get a random 11-digit number generated each time:
$ tr -cd '[:digit:]' < /dev/urandom | fold -w 11 | head -n 1
6
< inputfile egrep ';' | cut -f 2 -d ';'
-This command and you will get this!
6
You simply have an extra / at the beginning, which confused sed into thinking that you want to perform a command (called <) on a line that matches the pattern /s/. Instead, use:
sed -i 's/<ADD ANSWER HERE>/answer/' config.yaml
6
To replace single quotes (') it's easiest to put the sed command within double quotes and escape the double quote in the replacement:
$ cat quotes.txt
I'm Alice
$ sed -e "s/'/\"/g" quotes.txt
I"m Alice
Note that the single quote is not special within double quotes, so it must not be escaped.
If, instead one wants to replace backticks (`), as the ...
5
There are, as the existing answers already show, many ways to do this. Here's another GNU grep approach:
$ grep -oP ';\K.*' file
1563101282.M178569P409830.de122.abteilung.com,S=1258,W=1287:2,
1563102004.M49870P436474.de122.abteilung.com,S=1258,W=1287:2,
1563102961.M195946P457876.de122.abteilung.com,S=1258,W=1287:2,
1563103921.M334168P463856.de122.abteilung....
5
Use jq to parse json data:
jq -r '._source.firstname'
With the input data from the question it shows the desired output.
4
If the difference in the number of years is less than about 90 you can use the $RANDOM variable in bash to give you an offset in number of days and use the limited ability of the date command to do the calculation.
#!/bin/bash
s=$(date +%s -d "1/1/$1") # start in seconds since 1 Jan 1970
e=$(date +%s -d "1/1/$(($2+1))") # start of end year +1 in ...
4
Considering that this input is Data01.txt, it may be done as:
awk -F";" '{ print $2 }' Data01.txt > Data01\ Ready.txt
Which will output Data01 Ready.txt file.
4
I'd use awk for that:
$ cat tst.awk
(!NF) { # blank line
b = ""; f = 1 # empty buffer, start buffering
}
/-->/ { # timestamp
f = 0 # stop buffering
if (p == $0) { # same timestamp
next # discard buffer, start over
}
p = $0 # save timestamp
printf "%s"...
3
Try this:
od -An -N8 -d /dev/random | sed -e 's| ||g' -e 's|\(.\{11\}\).*|\1|'
3
while IFS='=' read -r name value; do
new_name="TF_ENV${name#proj_env_repo}"
export "$new_name=$value"
unset "$name"
done < <( printenv | grep proj_env_repo )
The secret ingredients here:
Shell Parameter Expansion
IFS='=' read -r name value -- uses the = sign to split the incoming line into the component parts.
using a Process Substitution ...
2
Assuming that you have that list in list.txt, you can do the following:
grep -h L list.txt > list_L.txt
Explanation: grep L list.txt gives you all the lines containing an uppercase L. The option -h suppresses the output of the file name, but is only necessary if you are searching through multiple files (e.g. grep -h L lis*.txt). Adding > list_L.txt ...
2
Another grep example:
grep -E "[0-9]+ +L" file
+ matches one or more occurrence of the previous character, so in this case we match at least one numeric character, and at least one space, followed by a capital L.
-E is required to interpret the + as a special character instead of a literal '+'
output:
59 LOUIS 1202 +1
62 LEON 1137 +12
73 ...
2
With shuf from GNU coreutils:
shuf -n1 -i0-99999999999
Increase the value of the -n option if you need more than one number.
2
In "1,$1p", the $1 is expanded by the shell, and sed sees 1,actorsp. actorsp isn't a valid sed address, so it gives an error. The error text looks a bit weird to me though, it seems to recognize a as a command and then go back to wondering about the comma.
In '1,$1p', the $1 is not expanded by the shell, and sed sees 1,$1p. $ is a valid address, it's used ...
2
/etc/hosts entries can have one or more hostnames per IP address, so we need to check each hostname (i.e. all fields except the first field) to see if it has 2 or more . characters in it.
For example:
awk '{for (f=2;f<=NF;f++) {if (split($f,array,/\./)>2) {print;last}}}' /etc/hosts
or, with linefeeds and indentation added to improve readability:
...
2
To substitute individual characters, it's simpler to use tr(1):
FILE=$(echo "$FILE" | tr '\\' / )
2
A more general answer, since I can't comment on geekosaur's answer - You wouldn't put the ^ (start of line anchor). For example, if you want to replace the first occurence of A with B, you would have
sed 's/A/B/'
2
The variable has to be output (by echo) to sed. Using sed -n to suppress sed output. Edit - don't need sed -n if we omit the sed p flag.
#!/bin/bash
FILE='\\edi3\welshch\test'
FILEPATH="$(echo "$FILE" | sed 's/\\/\//g')"
echo $FILEPATH
2
sed -n '/Red Hat Satellite/,/Pool ID:/{ s/.*Pool ID:[[:blank:]]*\(.*\)/\1/p; }' file
This applies a substitution that picks out your string from after Pool ID:. It does this on any line between a line that contains the string Red Hat Satellite and the first line afterwards that contain Pool ID:. The command would print only the result of any successful ...
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