[ad_1] Simply use lsb_release: $ lsb_release -sr 12.04 See the man page for all available options. Note that not all platforms or Linux distributions have lsb_release. 3 [ad_2] solved Shell script to get the OS version [closed]
[ad_1] while IFS= read -r dir; do find “$dir” -type f done < <(envsubst < my_big_file) See https://unix.stackexchange.com/a/294400/133219 and the man page for more info on envsubst. 1 [ad_2] solved executing find command with a file having directory list
[ad_1] Assuming that your table is called df and the columns of your table are named X1, X2, X3 try the following: library(dplyr) library(tidyr) df %>% spread(X2, X3) 1 [ad_2] solved Conditional columns to row [duplicate]
[ad_1] You just need one command. sed ‘s/^/CREATE FROM /’ output.txt See also Counting lines or enumerating line numbers so I can loop over them – why is this an anti-pattern? 0 [ad_2] solved Shell command to print the statements with N number of words present in other file [closed]
[ad_1] You have: A program (python script) N (9) files to process with the program 1 argument (input) for the program Now, you want to run the script from command line for N (9) files with the same argument, right? My suggestion will be writing another script to run the script for that N files … Read more
[ad_1] I think this question is fine now, the input file is good enough after edit, I can fully understand what you ask for now. With awk, you need learn to use 2-d array, it will simplify the code. awk ‘BEGIN{print “Instance id Name Owner Cost.centre”} /TAG/{split($0,a,FS);a[4]=tolower(a[4]);$1=$2=$3=$4=””;b[a[3],a[4]]=$0;c[a[3]]} END{for (i in c) printf “%-18s%-26s%-14s%-20s\n”,i,b[i,”name”],b[i,”owner”],b[i,”cost.center”]}’ file Instance … Read more
[ad_1] You can do this: seq 100 | egrep -v “0$|5$” basically it looks for and removes (-v) any line ending in 0 or 5. 4 [ad_2] solved Printing numbers from one list, not found in another
[ad_1] #!/bin/bash is called the shebang (you missed the leading #). It tells which program will execute your script. clear is for clearing screen. echo outputs following argument to the standard output (your terminal by default). But you must not surround your string with parenthesis as it’s used for grouping command in a sub-shell. If … Read more
[ad_1] current_year=$(date +%Y) next_month=$(($(date +%m) + 1)) locale_next_month=$(date -d$current_year/$next_month/1 +%B) for day_of_month in $(seq 1 31) do if day_of_year=$(date -d$current_year/$next_month/$day_of_month +%j 2> /dev/null) then mkdir -p /home/applications/app_name/$current_year/$locale_next_month/$day_of_year fi done 1 [ad_2] solved Create directory structure in unix [closed]
[ad_1] sudo switches users and then executes bash, passing it the other arguments. bash running as the new user runs the command in the argument after -c. 3 [ad_2] solved What happens when I sudo bash -c? [closed]
[ad_1] Rather than trying to extend the syntax of the shell to do start: for …; do for …; do … done done if CONDITION; then goto start else SOMETHING fi you may do while true; do for …; do for …; do … done done if ! CONDITION; then SOMETHING break fi done or … Read more
[ad_1] I would do something like this: cat log | grep -o “number\=[0-9]\+” | cut -d= -f2 | sort | uniq -c The output for your example is: 7 1193046 6 466000000 5 [ad_2] solved search value and count number from log file
[ad_1] Try this: # wait 2 seconds or use “sleep 2″ SECONDS=0; while [ $SECONDS -ne 2 ]; do :; done n=”” for ((i=0;i<30;i++)); do n+=”$(($RANDOM%10))”; done; echo “$n” See: help while, help test and help for 4 [ad_2] solved Linux 30-digit random number generator [closed]
[ad_1] If you want to receive and pass it on to the shell, then use this: <?php if( !empty( $_POST ) ) { $min = $_POST[‘MIN’]; $max = $_POST[‘MAX’]; $norm = $_POST[‘NORM’]; echo shell_exec(“action.sh $min $max $norm”); } ?> <form method=”POST”> Min: <input type=”number” name=”MIN” value=”<?php echo $min;?>”> <br><br> Norm: <input type=”number” name=”NORM” value=”<?php echo … Read more
[ad_1] After grepping some data into a variable i want to print only the variable is having some data otherwise i dont wabt to print [ad_2] solved After grepping some data into a variable i want to print only the variable is having some data otherwise i dont wabt to print