bash - Shell 脚本打开多个终端并执行不同的命令
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bash - Shell script opening multiple terminals and executing distinct commands
提问by itachi42
I've tried to write my own shell script. So far I've managed to open 4 xterminals that can only execute ONE command because of the 'hold' option.
我尝试编写自己的 shell 脚本。到目前为止，由于 'hold' 选项，我已经设法打开了 4 个只能执行一个命令的 xterminal。
If i don't use this option, the terminals just disappear.
如果我不使用这个选项，终端就会消失。
Here is my code :
这是我的代码：
#!/bin/sh
xterm -title "App 1" -hold -e mycommand | mysecondcommand &
xterm -title "App 2" -hold -e mycommand | mysecondcommand &
xterm -title "App 3" -hold -e mycommand | mysecondcommand &
xterm -title "App 4" -hold -e mycommand | mysecondcommand
Not so sure if I'm supposed to execute the second command in the same terminal that way.
不太确定我是否应该以这种方式在同一个终端中执行第二个命令。
Any ideas ?
有任何想法吗 ？
Thank you
谢谢
回答by Josh Jolly
Without -hold, the xterm will close as soon as the command is completed. You can execute multiple commands by using double quotes and command separators (eg ;, &):
如果没有-hold，xterm 将在命令完成后立即关闭。您可以使用双引号和命令分隔符（例如;, &）来执行多个命令：
xterm -title "App 1" -e "mycommand; mysecondcommand"
