Question

我正在开发这个脚本：

#!/bin/bash

# If no command line arguments, display a tiny usage message.
if [[ $# -eq 0 ]]; then
  echo "Usage: $0 files..." >&2
  exit 1
fi

# Destroy child processes, when exiting.
cleanup() {
  kill `jobs -p` 2>/dev/null
}

# Call cleanup function on SIGINT, ie. Ctrl-C.
trap "cleanup; exit 0" INT TERM EXIT

# Start child processes continously outputting from the files given as command
# line arguments. The filename is prepended on the line.
for f in $*; do
  tail -f "$f" | awk '{ print "'"$f"':"$0 }' &
done

# Wait for child processes to exit. Just to be sure, kill them all afterwards.
wait
cleanup

我这样用：

tailfiles *.log

它的作用是交错所有文件的尾部输出，前缀为文件名（有点像grep -H）。但是，我无法管道脚本的输出。它很简单，没有给我任何东西：

tailfiles *.log | grep error

是否可以在bash中将来自多个子进程的流收集到一个输出中？

Answer 1

尝试以下方法：

# ...
rm -f fifo
mkfifo fifo
for f in $*; do
  tail -f "$f" | awk '{ print "'"$f"':"$0 }'  > fifo  &
done

tail -f fifo
# ...

Answer 2

您可能只需要将循环更改为：

for f; do
  tail -f "$f" | awk -v f="$f" '{ printf "%s : %s\n", f, $0 }' &
done

使用$ *会使其受到分词的影响。

在bash脚本中重定向多个子进程的输出

2 个答案: