Bash 等待作业并限制作业数量

Bash 等待作业并限制作业数量

可能的重复:
并行执行四个任务……我该怎么做?

假设一个循环调用一个命令

 grep -v '#' < files.m3u | sed 's/\\\\/\/\//g' | sed 's/\\/\//g' | while read line
 do
    filename=$(basename "$line")
    avconv -i "$line" "${filename%.*}.wav"
 done

在 avconv 之后放置 & 将为每个文件持续生成 avconv。现在我想做两件事:

  • 我想将生成的进程数限制为 4
  • 循环完成后,我想等待最后一个准备好

答案1

您可以记住每个新子进程的 PID($!启动后检查)。定期检查仍有多少子级存在(例如通过kill -0),如果数量下降,则生成一个新的,等等。最后,只需wait

这是我出于同样原因编写的脚本:

#! /bin/bash

## Tries to run commands in parallel. Commands are read from STDIN one
## per line, or from a given file specified by -f.
## Author: E. Choroba

file='-'
proc_num=$(grep -c ^processor'\b' /proc/cpuinfo)
prefix=$HOSTNAME-$USER-$$
sleep=10

children=()
names=()

if [[ $1 =~ ^--?h(elp)?$ ]] ; then
    cat <<-HELP
    Usage: ${0##*/} [-f file] [-n max-processes] [-p tmp-prefix] -s [sleep]
      Defaults:
        STDIN for file
        $proc_num for max-processes (number of processors)
        $prefix for tmp-prefix
        $sleep for sleep interval
    HELP
    exit
fi

function debug () {
    if ((DEBUG)) ; then
        echo "$@" >&2
    fi
}

function child_count () {
    debug Entering child_count "${children[@]}"
    child_count=0
    new_children=()
    for child in "${children[@]}" ; do
        debug Trying $child
        if kill -0 $child 2>/dev/null ; then
            debug ... exists
            let child_count++
            new_children+=($child)
        fi
    done

    children=("${new_children[@]}")
    echo $child_count
    debug Leaving child_count "${children[@]}"
}

while getopts 'f:n:p:s:' arg ; do
    case $arg in
        f ) file=$OPTARG ;;
        n ) proc_num=$((OPTARG)) ;;
        p ) prefix=$OPTARG;;
        s ) sleep=$OPTARG;;
        * ) echo "Warning: unknown option $arg" >&2 ;;
    esac
done

i=0
while read -r line ; do
    debug Reading $line
    name=$prefix.$i
    let i++
    names+=($name)

    while ((`child_count`>=proc_num)) ; do
        sleep $sleep
        debug Sleeping
    done

    eval $line 2>$name.e >$name.o &
    children+=($!)
    debug Running "${children[@]}"
done < <(cat $file)

debug Loop ended
wait
cat "${names[@]/%/.o}"
cat "${names[@]/%/.e}" >&2
rm "${names[@]/%/.o}" "${names[@]/%/.e}"

答案2

来自链接问题,根据您的变化量身定制:

sed -n -e '/#/!s,\\,/,g' files.m3u | xargs -d '\n' -I {} -P 4 \
    sh -c 'line=$1; file=${line##*/}; avconv -i "$line" "${file%.*}.wav"' avconv_sh {}

同样,GNUxargs或某些版本支持-d-P是必需的。还要注意输入文件中行首和行尾的多余空格 - 如果它们存在,此代码片段将保留它们,这可能会导致问题。

答案3

我是这样解决的。谢谢你的美元!尖端

#!/bin/bash
children[0]=0
children[1]=0
children[2]=0
children[3]=0

i=0
grep -v '#' < files.m3u | sed 's/\\\\/\/\//g' | sed 's/\\/\//g' | while read line
do
    filename=$(basename "$line")
    let k="$i%4"
    wait ${children[k]}
    avconv -i "$line" "${filename%.*}.wav" &
    children[k]=$!
    let i++
done

wait ${children[0]}
wait ${children[1]}
wait ${children[2]}
wait ${children[3]}

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