How do I create a wage calculator? - bash

I want to be able to print an amount that goes up every second by a specific amount until I stop the script. Ive been trying to print an update of how much I'm being paid by the second (or hour) live whilst I work, how would I go about doing that in bash?

Here's a simple one:
#!/usr/bin/env bash
total=0
increment=50
while :
do
total=$((total + increment))
printf '$%.2f\n' $(bc <<< "scale=2; $total / 100")
sleep 1
done

Related

Bash date subtraction +%H:%M:%S

I looked all over online and can't seem to find anything that addresses what I am trying to do. I am using Bash on an Unbuntu VM.
I created the following script
start_code=$(date +%H:%M:%S)
end_code=$(date +%H:%M:%S)
echo $start_code
for i in {1..1000};
do echo $RANDOM >> filename.txt;
done
echo $end_code
The code works fine, but is there any way that I can subtract the variables start_code from end_code??
I have tried this many different ways one being total_code=$(($start_code - $end_code))
but I get errors or nothing returned with everything that I have tried. As I'm brand new to Bash and I'm not even sure if I can do this. Any help would be greatly appreciated.
There are better ways to compute time lapsed. You can do the subtraction with something like:
$ cat a.sh
#!/bin/sh
foo() {
# This function does stuff
sleep ${1-2}
}
start=$(date +%s)
foo 3
end=$(date +%s)
echo "time lapsed: $((end - start))"
time foo 4
$ ./a.sh
time lapsed: 3
real 0m4.006s
user 0m0.001s
sys 0m0.001s
In the above, we first store a time stamp before calling the function foo that just sleeps for 3 seconds. Then we compute a new time stamp and subtract. Then we do the same using the time builtin and let the function sleep for 4 seconds.

bash loop taking extremely long time

I have a list of times that I am looping through in the format HH:MM:SS to find the nearest but not past time. The code that I have is:
for i in ${times[#]}; do
hours=$(echo $i | sed 's/\([0-9]*\):.*/\1/g')
minutes=$(echo $i | sed 's/.*:\([0-9]*\):.*/\1/g')
currentHours=$(date +"%H")
currentMinutes=$(date +"%M")
if [[ hours -ge currentHours ]]; then
if [[ minutes -ge currentMinutes ]]; then
break
fi
fi
done
The variable times is an array of all the times that I am sorting through (its about 20-40 lines). I'd expect this to take less than 1 second however it is taking upwards of 5 seconds. Any suggestions for decreasing the time of the regular expression would be appreciated.
times=($(cat file.txt))
Here is a list of the times that are stored in a text file and are imported into the times variable using the above line of code.
6:05:00
6:35:00
7:05:00
7:36:00
8:08:00
8:40:00
9:10:00
9:40:00
10:11:00
10:41:00
11:11:00
11:41:00
12:11:00
12:41:00
13:11:00
13:41:00
14:11:00
14:41:00
15:11:00
15:41:00
15:56:00
16:11:00
16:26:00
16:41:00
16:58:00
17:11:00
17:26:00
17:41:00
18:11:00
18:41:00
19:10:00
19:40:00
20:10:00
20:40:00
21:15:00
21:45:00
One of the key things to understand in looking at bash scripts from a performance perspective is that while the bash interpreter is somewhat slow, the act of spawning an external process is extremely slow. Thus, while it can often speed up your scripts to use a single invocation of awk or sed to process a large stream of input, starting those invocations inside a tight loop will greatly outweigh the performance of those tools once they're running.
Any command substitution -- $() -- causes a second copy of the interpreter to be fork()ed off as a subshell. Invoking any command not built into bash -- date, sed, etc -- then causes a subprocess to be fork()ed off for that process, and then the executable associated with that process to be exec()'d -- something involves a great deal of OS-level overhead (the binary needs to be linked, loaded, etc).
This loop would be better written as:
IFS=: read -r currentHours currentMinutes < <(date +"%H:%M")
while IFS=: read -r hours minutes _; do
if (( hours >= currentHours )) && (( minutes >= currentMinutes )); then
break
fi
done <file.txt
In this form only one external command is run, date +"%H:%M", outside the loop. If you were only targeting bash 4.2 and newer (with built-in time formatting support), even this would be unnecessary:
printf -v currentHours '%(%H)T' -1
printf -v currentMinutes '%(%M)T' -1
...will directly place the current hour and minute into the variables currentHours and currentMinutes using only functionality built into modern bash releases.
See:
BashFAQ #1 - How can I read a file (data stream, variable) line-by-line (and/or field-by-field)?
BashFAQ #100 - How can I do native string manipulations in bash? (Subsection: "Splitting a string into fields")
To be honest I'm not sure why it's taking an extremely long time but there are certainly some things which could be made more efficient.
currentHours=$(date +"%H")
currentMinutes=$(date +"%M")
for time in "${times[#]}"; do
IFS=: read -r hours minutes seconds <<<"$time"
if [[ hours -ge currentHours && minutes -ge currentMinutes ]]; then
break
fi
done
This uses read, a built-in command, to split the text into variables, rather than calling external commands and creating subshells.
I assume that you want the script to run so quickly that it's safe to reuse currentHours and currentMinutes within the loop.
Note that you can also just use awk to do the whole thing:
awk -F: -v currentHours="$(date +"%H") -v currentMinutes="$(date +"%M")" '
$1 >= currentHours && $2 >= currentMinutes { print; exit }' file.txt
Just to make the program produce some output, I added a print, so that the last line is printed.
awk to the rescue!
awk -v time="12:12:00" '
function pad(x) {split(x,ax,":"); return (ax[1]<10)?"0"x:x}
BEGIN {time=pad(time)}
time>pad($0) {next}
{print; exit}' times
12:41:00
with 0 padding the hour you can do string only comparison.

How can I monitor the average number of lines added to a file per second in a bash shell? [closed]

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Improve this question
I'd like to monitor the average rate at which lines are being added to a log file in a bash shell.
I can currently monitor how many lines are in the file each second via the command
watch -n 1 'wc -l log.txt'
However, this gives me the total count of lines when I would prefer a rate instead. In other words, I would like a command to every so often output the number of lines that have been added to the file since the command was executed divided by the number of seconds the command has been running.
For a rough count of lines per second, try:
tail -f log.txt | { count=0; old=$(date +%s); while read line; do ((count++)); s=$(date +%s); if [ "$s" -ne "$old" ]; then echo "$count lines per second"; count=0; old=$s; fi; done; }
(Bash required.)
Or, as spread out over multiple lines:
tail -f log.txt | {
count=0
old=$(date +%s)
while read line
do
((count++))
s=$(date +%s)
if [ "$s" -ne "$old" ]
then
echo "$count lines per second"
count=0
old=$s
fi
done
}
This uses date to record the time in seconds. Meanwhile, it counts the number of lines produced by tail -f log.txt. Every time another second passes, the count of lines seen during that second is printed.
Demonstration
One one terminal, run the command:
while sleep 0.1; do echo $((count++)); done >>log.txt
This command writes one line to the file log.txt every roughly tenth of a second.
In another terminal, run:
$ tail -f log.txt | { count=0; old=$(date +%s); while read line; do ((count++)); s=$(date +%s); if [ "$s" -ne "$old" ]; then echo "$count lines per second"; count=0; old=$s; fi; done; }
15 lines per second
10 lines per second
10 lines per second
10 lines per second
9 lines per second
10 lines per second
Due to buffering, the first count is off. Subsequent counts are fairly accurate.
Simple script you can deploy:
Filename="log.txt"
ln_1=`wc -l $Filename | awk '{print $1}'`
while true
do
ln_2=${ln_1}
sleep 1
ln_1=`wc -l $Filename | awk '{print $1}'`
echo $(( ln_1-ln_2 )) lines increased
done
The tail command supports watching for appended lines via --follow option, which accepts a file descriptor, or file name. With this option, tail periodically checks for file changes. The interval of the checks depends on whether kernel supports inotify. Inotify-based implementations detect the changes promptly (I would say, almost instantly). If, however, the kernel doesn't support inotify, tail resorts to periodic checks. In the latter case, tail sleeps for one second by default. The sleep interval can be changed with --sleep-interval option.
I wouldn't rely on the sleep interval in calculations, however:
When ‘tail’ uses inotify, this polling-related option is usually ignored.
Especially because Bash has a built-in seconds counter, the SECONDS variable (see info bash SECONDS):
This variable expands to the number of seconds since the shell was started. Assignment to this variable resets the count to the value assigned, and the expanded value becomes the value assigned plus the number of seconds since the assignment.
Thus, you can initialize SECONDS to 1, run a loop reading the output of tail, and calculate the speed as number_of_lines / $SECONDS. But this will produce average for the entire execution time. Average for the last N seconds is much more practical. It is also easy to implement, as Bash allows to reset the seconds counter.
Example
The following example implements the idea. The also features watch-like output in interactive mode.
# The number of seconds for which we calculate the average speed
timespan=8
# The number of lines
lines=0
# We'll assume that the shell is running in interactive mode,
# if the standard output descriptor (1) is attached to the terminal.
# See http://www.tldp.org/LDP/abs/html/intandnonint.html
if [ -t 1 ]; then
is_interactive=1
format='%d lines/sec'
else
is_interactive=
format='%d lines/sec\n'
fi
# Reset the built-in seconds counter.
# Also, prevent division by zero in the calculation below.
SECONDS=1
# Save cursor position, if interactive
test -n "$is_interactive" && tput sc
while read line; do
if [[ $(( $SECONDS % $timespan )) -eq 0 ]]; then
SECONDS=1
lines=0
fi
if test -n "$is_interactive"; then
# Restore cursor position, then delete line
tput rc; tput el1
fi
printf "$format" $(( ++lines / SECONDS ))
done < <(tail -n0 -F log.txt)
P.S.
There are many other ways to get an offset in seconds. For example, you can fetch the current Unix time using the built-in printf function:
# -1 represents the current time
# %s is strftime's format string for the number of seconds since the Epoch
timestamp=$(builtin printf '%(%s)T' -1)
Another way is to invoke the date command: date +%s.
But I believe that reading from the SECONDS variable is faster and cleaner.

For loop with an argument based range

I want to run certain actions on a group of lexicographically named files (01-09 before 10). I have to use a rather old version of FreeBSD (7.3), so I can't use yummies like echo {01..30} or seq -w 1 30.
The only working solution I found is printf "%02d " {1..30}. However, I can't figure out why can't I use $1 and $2 instead of 1 and 30. When I run my script (bash ~/myscript.sh 1 30) printf says {1..30}: invalid number
AFAIK, variables in bash are typeless, so how can't printf accept an integer argument as an integer?
Bash supports C-style for loops:
s=1
e=30
for i in ((i=s; i<e; i++)); do printf "%02d " "$i"; done
The syntax you attempted doesn't work because brace expansion happens before parameter expansion, so when the shell tries to expand {$1..$2}, it's still literally {$1..$2}, not {1..30}.
The answer given by #Kent works because eval goes back to the beginning of the parsing process. I tend to suggest avoiding making habitual use of it, as eval can introduce hard-to-recognize bugs -- if your command were whitelisted to be run by sudo and $1 were, say, '$(rm -rf /; echo 1)', the C-style-for-loop example would safely fail, and the eval example... not so much.
Granted, 95% of the scripts you write may not be accessible to folks executing privilege escalation attacks, but the remaining 5% can really ruin one's day; following good practices 100% of the time avoids being in sloppy habits.
Thus, if one really wants to pass a range of numbers to a single command, the safe thing is to collect them in an array:
a=( )
for i in ((i=s; i<e; i++)); do a+=( "$i" ); done
printf "%02d " "${a[#]}"
I guess you are looking for this trick:
#!/bin/bash
s=1
e=30
printf "%02d " $(eval echo {$s..$e})
Ok, I finally got it!
#!/bin/bash
#BSD-only iteration method
#for day in `jot $1 $2`
for ((day=$1; day<$2; day++))
do
echo $(printf %02d $day)
done
I initially wanted to use the cycle iterator as a "day" in file names, but now I see that in my exact case it's easier to iterate through normal numbers (1,2,3 etc.) and process them into lexicographical ones inside the loop. While using jot, remember that $1 is the numbers amount, and the $2 is the starting point.

binary execution in a while loop.for loop through shell script

Hello i am new to shell script. want to execute a binary through loop in a shell script.
wrote a pgm which looked like:
i="1"
while [ $i -lt 100 ]
do
/home/rajni/BUFFER_SEND_STUB/build/buffer_send.exe
i=`expr $i +1`
done
doubt it is not working fine. Can anyone suggest????
Thanks.
expr won't like the fact you've used +1 rather than the space-separated + 1.
I also tend to use [[ and ]] rather than the single ones since they're definitely bash-internal and more powerful than the external [/test.
In any case, there's a more efficient way if you're using a relatively recent bash:
for i in {1..100} ; do
echo $i
done
which will do something with each value 1 through 100 inclusive (your current loop does 1 through 99 so you may have to adjust for that).
Changing that 100 to a 5 shows how it works, generating:
1
2
3
4
5
you can use the for loop
for i in {1..100}
do
/home/rajni/BUFFER_SEND_STUB/build/buffer_send.exe
done

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