为多个进程创建管道

Question

我尝试管道几个脚本，但是真的不明白如何正确地完成它。

  mkfifo pipe1
  cat ./script1 > pipe1 &
  cat ./script2 > pipe1 &
  cat ./script3 > pipe1 &
 ./script1 < pipe1

所以问题是我可以这样做吗？我的意思是将所有脚本写在一个管道中，并在我需要时只读取其中一个

Answer 1

回答原始问题

cat可以处理多个文件，并依次读取每个文件：

cat ./script1 ./script2 ./script3 | ./videoplaylist

更新：在问题的修订版中，目标是script1本身：

cat ./script1 ./script2 ./script3 | ./script1

回答修订问题

以下代码将运行三个脚本，合并其输出并将其发送到masterscript：

mkfifo pipe1
bash ./script1 > pipe1 &
bash ./script2 > pipe1 &
bash ./script3 > pipe1 &
bash ./masterscript < pipe1

工作示例

让我们从四个脚本开始：

$ cat script1
#!/bin/sh
while sleep 1;do echo $0 $(date); done
$ cat script2
#!/bin/sh
$ cat masterscript 
#!/bin/sh
while read line; do echo "$0 received: $line"; done
while sleep 1;do echo $0 $(date); done
$ cat script3
#!/bin/sh
while sleep 1;do echo $0 $(date); done

现在让我们执行它们：

$ mkfifo pipe1
$ bash script1 >pipe1 & bash script2 >pipe1 & bash script3 >pipe1 &
[1] 29154
[2] 29155
[3] 29156
$ bash masterscript <pipe11
masterscript received: script2 Sat Apr 11 15:39:37 PDT 2015
masterscript received: script1 Sat Apr 11 15:39:37 PDT 2015
masterscript received: script3 Sat Apr 11 15:39:37 PDT 2015
masterscript received: script2 Sat Apr 11 15:39:38 PDT 2015
masterscript received: script1 Sat Apr 11 15:39:38 PDT 2015
masterscript received: script3 Sat Apr 11 15:39:38 PDT 2015
masterscript received: script1 Sat Apr 11 15:39:39 PDT 2015
masterscript received: script3 Sat Apr 11 15:39:39 PDT 2015
masterscript received: script2 Sat Apr 11 15:39:39 PDT 2015
masterscript received: script1 Sat Apr 11 15:39:40 PDT 2015
masterscript received: script2 Sat Apr 11 15:39:40 PDT 2015
masterscript received: script3 Sat Apr 11 15:39:40 PDT 2015
^C

如您所见，FIFO成功将每个脚本的输出发送到主脚本。