我有一個腳本，可以一次運(yùn)行4個進(jìn)程的批處理，我不在乎獲取每個proc的返回碼。我永遠(yuǎn)不想同時運(yùn)行4個以上的proc。以下方法的問題在于它一次最多不能填充4個進(jìn)程。例如，如果proc2和proc3提早完成，我希望proc 5和6啟動，而不是僅在1-4完成后才啟動。我如何在bash中實(shí)現(xiàn)這一目標(biāo)？run_func_1 &run_func_2 &run_func_3 &run_func_4 &waitrun_func_5 &run_func_6 &run_func_7 &run_func_8 &wait
2 回答
富國滬深
TA貢獻(xiàn)1790條經(jīng)驗(yàn) 獲得超9個贊
我嘗試使用工作人員池和工作隊(duì)列來執(zhí)行自定義實(shí)現(xiàn)。新工人將在上一個工人完成任務(wù)后立即將其從隊(duì)列中移出。
您可能可以根據(jù)需要修改此腳本，但我希望您會明白我的意圖。
這是腳本：
#!/bin/bash
f1() { echo Started f1; sleep 10; echo Finished f1; }
f2() { echo Started f2; sleep 8; echo Finished f2; }
f3() { echo Started f3; sleep 12; echo Finished f3; }
f4() { echo Started f4; sleep 14; echo Finished f4; }
f5() { echo Started f5; sleep 7; echo Finished f5; }
declare -r MAX_WORKERS=2
declare -a worker_pids
declare -a jobs=('f1' 'f2' 'f3' 'f4' 'f5')
available_worker_index() {
# If number of workers is less than MAX_WORKERS
# We still have workers that are idle
declare worker_count="${#worker_pids[@]}"
if [[ $worker_count -lt $MAX_WORKERS ]]; then
echo "$worker_count"
return 0
fi
# If we reached this code it means
# All workers are already created and executing a job
# We should check which of them finished and return it's index as available
declare -i index=0
for pid in "${worker_pids[@]}"; do
is_running=$(ps -p "$pid" > /dev/null; echo "$?")
if [[ $is_running != 0 ]]; then
echo "$index"
return 0
fi
index+=1
done
echo "None"
}
for job in "${jobs[@]}"; do
declare worker_index
worker_index=$(available_worker_index)
while [[ $worker_index == "None" ]]; do
# Wait for available worker
sleep 3
worker_index=$(available_worker_index)
done
# Run the job in background
"$job" &
# Save it's pid for later
pid="$!"
worker_pids["$worker_index"]="$pid"
done
# Wait all workers to finish
wait
您只需更改MAX_WORKERS變量即可輕松更改工作池的大小。
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