I'm looking for a way to pipe text continuously into a process like write. I do not want to buffer and pipe it all at once. This is my bash script so far:
#!/bin/bash
for i in `seq 1 10`; do
echo $i | write user
done
The problem is that write gets opened and closed i times. Does anyone know how I can keep it alive while looping?
Sure, just move the pipe outside the loop:
#!/bin/bash
for i in `seq 1 10`; do
echo "$i"
done | write user
As you've tagged with bash, I would suggest using a brace expansion for i in {1..10} rather than calling seq. If the numbers are variable, you can use something like for (( i = a; i < b; ++i )).
Related
Bash allows to use: cat <(echo "$FILECONTENT")
Bash also allow to use: while read i; do echo $i; done </etc/passwd
to combine previous two this can be used: echo $FILECONTENT | while read i; do echo $i; done
The problem with last one is that it creates sub-shell and after the while loop ends variable i cannot be accessed any more.
My question is:
How to achieve something like this: while read i; do echo $i; done <(echo "$FILECONTENT") or in other words: How can I be sure that i survives while loop?
Please note that I am aware of enclosing while statement into {} but this does not solves the problem (imagine that you want use the while loop in function and return i variable)
The correct notation for Process Substitution is:
while read i; do echo $i; done < <(echo "$FILECONTENT")
The last value of i assigned in the loop is then available when the loop terminates.
An alternative is:
echo $FILECONTENT |
{
while read i; do echo $i; done
...do other things using $i here...
}
The braces are an I/O grouping operation and do not themselves create a subshell. In this context, they are part of a pipeline and are therefore run as a subshell, but it is because of the |, not the { ... }. You mention this in the question. AFAIK, you can do a return from within these inside a function.
Bash also provides the shopt builtin and one of its many options is:
lastpipe
If set, and job control is not active, the shell runs the last command of a pipeline not executed in the background in the current shell environment.
Thus, using something like this in a script makes the modfied sum available after the loop:
FILECONTENT="12 Name
13 Number
14 Information"
shopt -s lastpipe # Comment this out to see the alternative behaviour
sum=0
echo "$FILECONTENT" |
while read number name; do ((sum+=$number)); done
echo $sum
Doing this at the command line usually runs foul of 'job control is not active' (that is, at the command line, job control is active). Testing this without using a script failed.
Also, as noted by Gareth Rees in his answer, you can sometimes use a here string:
while read i; do echo $i; done <<< "$FILECONTENT"
This doesn't require shopt; you may be able to save a process using it.
Jonathan Leffler explains how to do what you want using process substitution, but another possibility is to use a here string:
while read i; do echo "$i"; done <<<"$FILECONTENT"
This saves a process.
This function makes duplicates $NUM times of jpg files (bash)
function makeDups() {
NUM=$1
echo "Making $1 duplicates for $(ls -1 *.jpg|wc -l) files"
ls -1 *.jpg|sort|while read f
do
COUNT=0
while [ "$COUNT" -le "$NUM" ]
do
cp $f ${f//sm/${COUNT}sm}
((COUNT++))
done
done
}
Bash allows to use: cat <(echo "$FILECONTENT")
Bash also allow to use: while read i; do echo $i; done </etc/passwd
to combine previous two this can be used: echo $FILECONTENT | while read i; do echo $i; done
The problem with last one is that it creates sub-shell and after the while loop ends variable i cannot be accessed any more.
My question is:
How to achieve something like this: while read i; do echo $i; done <(echo "$FILECONTENT") or in other words: How can I be sure that i survives while loop?
Please note that I am aware of enclosing while statement into {} but this does not solves the problem (imagine that you want use the while loop in function and return i variable)
The correct notation for Process Substitution is:
while read i; do echo $i; done < <(echo "$FILECONTENT")
The last value of i assigned in the loop is then available when the loop terminates.
An alternative is:
echo $FILECONTENT |
{
while read i; do echo $i; done
...do other things using $i here...
}
The braces are an I/O grouping operation and do not themselves create a subshell. In this context, they are part of a pipeline and are therefore run as a subshell, but it is because of the |, not the { ... }. You mention this in the question. AFAIK, you can do a return from within these inside a function.
Bash also provides the shopt builtin and one of its many options is:
lastpipe
If set, and job control is not active, the shell runs the last command of a pipeline not executed in the background in the current shell environment.
Thus, using something like this in a script makes the modfied sum available after the loop:
FILECONTENT="12 Name
13 Number
14 Information"
shopt -s lastpipe # Comment this out to see the alternative behaviour
sum=0
echo "$FILECONTENT" |
while read number name; do ((sum+=$number)); done
echo $sum
Doing this at the command line usually runs foul of 'job control is not active' (that is, at the command line, job control is active). Testing this without using a script failed.
Also, as noted by Gareth Rees in his answer, you can sometimes use a here string:
while read i; do echo $i; done <<< "$FILECONTENT"
This doesn't require shopt; you may be able to save a process using it.
Jonathan Leffler explains how to do what you want using process substitution, but another possibility is to use a here string:
while read i; do echo "$i"; done <<<"$FILECONTENT"
This saves a process.
This function makes duplicates $NUM times of jpg files (bash)
function makeDups() {
NUM=$1
echo "Making $1 duplicates for $(ls -1 *.jpg|wc -l) files"
ls -1 *.jpg|sort|while read f
do
COUNT=0
while [ "$COUNT" -le "$NUM" ]
do
cp $f ${f//sm/${COUNT}sm}
((COUNT++))
done
done
}
Bash allows to use: cat <(echo "$FILECONTENT")
Bash also allow to use: while read i; do echo $i; done </etc/passwd
to combine previous two this can be used: echo $FILECONTENT | while read i; do echo $i; done
The problem with last one is that it creates sub-shell and after the while loop ends variable i cannot be accessed any more.
My question is:
How to achieve something like this: while read i; do echo $i; done <(echo "$FILECONTENT") or in other words: How can I be sure that i survives while loop?
Please note that I am aware of enclosing while statement into {} but this does not solves the problem (imagine that you want use the while loop in function and return i variable)
The correct notation for Process Substitution is:
while read i; do echo $i; done < <(echo "$FILECONTENT")
The last value of i assigned in the loop is then available when the loop terminates.
An alternative is:
echo $FILECONTENT |
{
while read i; do echo $i; done
...do other things using $i here...
}
The braces are an I/O grouping operation and do not themselves create a subshell. In this context, they are part of a pipeline and are therefore run as a subshell, but it is because of the |, not the { ... }. You mention this in the question. AFAIK, you can do a return from within these inside a function.
Bash also provides the shopt builtin and one of its many options is:
lastpipe
If set, and job control is not active, the shell runs the last command of a pipeline not executed in the background in the current shell environment.
Thus, using something like this in a script makes the modfied sum available after the loop:
FILECONTENT="12 Name
13 Number
14 Information"
shopt -s lastpipe # Comment this out to see the alternative behaviour
sum=0
echo "$FILECONTENT" |
while read number name; do ((sum+=$number)); done
echo $sum
Doing this at the command line usually runs foul of 'job control is not active' (that is, at the command line, job control is active). Testing this without using a script failed.
Also, as noted by Gareth Rees in his answer, you can sometimes use a here string:
while read i; do echo $i; done <<< "$FILECONTENT"
This doesn't require shopt; you may be able to save a process using it.
Jonathan Leffler explains how to do what you want using process substitution, but another possibility is to use a here string:
while read i; do echo "$i"; done <<<"$FILECONTENT"
This saves a process.
This function makes duplicates $NUM times of jpg files (bash)
function makeDups() {
NUM=$1
echo "Making $1 duplicates for $(ls -1 *.jpg|wc -l) files"
ls -1 *.jpg|sort|while read f
do
COUNT=0
while [ "$COUNT" -le "$NUM" ]
do
cp $f ${f//sm/${COUNT}sm}
((COUNT++))
done
done
}
Bash allows to use: cat <(echo "$FILECONTENT")
Bash also allow to use: while read i; do echo $i; done </etc/passwd
to combine previous two this can be used: echo $FILECONTENT | while read i; do echo $i; done
The problem with last one is that it creates sub-shell and after the while loop ends variable i cannot be accessed any more.
My question is:
How to achieve something like this: while read i; do echo $i; done <(echo "$FILECONTENT") or in other words: How can I be sure that i survives while loop?
Please note that I am aware of enclosing while statement into {} but this does not solves the problem (imagine that you want use the while loop in function and return i variable)
The correct notation for Process Substitution is:
while read i; do echo $i; done < <(echo "$FILECONTENT")
The last value of i assigned in the loop is then available when the loop terminates.
An alternative is:
echo $FILECONTENT |
{
while read i; do echo $i; done
...do other things using $i here...
}
The braces are an I/O grouping operation and do not themselves create a subshell. In this context, they are part of a pipeline and are therefore run as a subshell, but it is because of the |, not the { ... }. You mention this in the question. AFAIK, you can do a return from within these inside a function.
Bash also provides the shopt builtin and one of its many options is:
lastpipe
If set, and job control is not active, the shell runs the last command of a pipeline not executed in the background in the current shell environment.
Thus, using something like this in a script makes the modfied sum available after the loop:
FILECONTENT="12 Name
13 Number
14 Information"
shopt -s lastpipe # Comment this out to see the alternative behaviour
sum=0
echo "$FILECONTENT" |
while read number name; do ((sum+=$number)); done
echo $sum
Doing this at the command line usually runs foul of 'job control is not active' (that is, at the command line, job control is active). Testing this without using a script failed.
Also, as noted by Gareth Rees in his answer, you can sometimes use a here string:
while read i; do echo $i; done <<< "$FILECONTENT"
This doesn't require shopt; you may be able to save a process using it.
Jonathan Leffler explains how to do what you want using process substitution, but another possibility is to use a here string:
while read i; do echo "$i"; done <<<"$FILECONTENT"
This saves a process.
This function makes duplicates $NUM times of jpg files (bash)
function makeDups() {
NUM=$1
echo "Making $1 duplicates for $(ls -1 *.jpg|wc -l) files"
ls -1 *.jpg|sort|while read f
do
COUNT=0
while [ "$COUNT" -le "$NUM" ]
do
cp $f ${f//sm/${COUNT}sm}
((COUNT++))
done
done
}
Attempting to make a "simple" parallel function in bash. The problem is currently that when the line to capture the output is backgrounded, the output is lost. If that line is not backgrounded, the output is captured fine, but this of course defeats the purpose of the function.
#!/usr/bin/env bash
cluster="${1:-web100s}"
hosts=($(inventory.pl bash "$cluster" | sort -V))
cmds="${2:-uptime}"
parallel=10
cx=0
total=0
for host in "${hosts[#]}"; do
output[$total]=$(echo -en "$host: ")
echo "${output[$total]}"
output[$total]+=$(ssh -o ConnectTimeout=5 "$host" "$cmds") &
cx=$((cx + 1))
total=$((total + 1))
if [[ $cx -gt $parallel ]]; then
wait >&/dev/null
cx=0
fi
done
echo -en "***** DONE *****\n Results\n"
for ((i=0; i<= $total; i++)); do
echo "${output[$i]}"
done
That's because your command (the assignment) is run in a subshell, so this assignment can't influence the parent shell. This boils down to this:
a=something
a='hello senorsmile' &
echo "$a"
Can you guess what the output is? the output is, of course,
something
and not hello senorsmile. The only way for the subshell to communicate with the parent shell is to use an IPC (interprocess communication), in one form or another. I don't have any solution to propose, I only tried to explain why it fails.
If you think of it, it should make sense. What do you think of this?
a=$( echo a; sleep 1000000000; echo b ) &
The command immediately returns (after forking)... but the output is only going to be fully available in... over 31 years.
Assigning a shell variable in the background this way is effectively meaningless. Bash does have built in co-processing which should work for you:
http://www.gnu.org/software/bash/manual/bashref.html#Coprocesses