在带有所有参数的脚本上运行 nohup 命令
我正在尝试使用 nohup 运行脚本，但该命令恰好需要一个 带有变量 $* 的整行参数。我尝试运行这样的命令：
nohup time ./build_all all &
但这在 nohup.out 中给了我以下错误：
./build_all: DISPLAY=ted:0.0: is not an identifier
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特德
=================================================== ===================================
我意识到彼得·约翰·阿克拉姆是对的。该错误不是因为 nohup，而是因为脚本，我不确定我做错了什么，因为语法对我来说似乎是正确的。还有一点奇怪的是，当我单独运行脚本时，我没有看到错误，但是当我尝试使用 nohup 运行时，我看到了奇怪的错误。
不管怎样，脚本的开头是这样的：
#!/bin/bash
export DISPLAY=ted:0.0 # sets the display
export RELEASE=v1.0
node=`uname -n`
I am trying to run a script with nohup, but the command happens to take an
entire line of parameters with the variable $*. I try running the command like this:
nohup time ./build_all all &
But this is giving me the following error in nohup.out:
./build_all: DISPLAY=ted:0.0: is not an identifier
Any help appreciated.
Ted
==================================================================================
I realize that Peter John Acklam was right. The error is not because of nohup, but because of the script, I am not sure what I am doing wrong because the syntax seems correct to me. It is also kind of strange that when I run the script on its own, I don't see the error, but when I try to run with nohup, I see the strange error.
Anyhow, the beginning of the script looks like this:
#!/bin/bash
export DISPLAY=ted:0.0 # sets the display
export RELEASE=v1.0
node=`uname -n`
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只需将“build_all”的参数放在命令行上，就像任何其他命令一样：
参数将传递给“build_all”，而不是“time”或“nohup”。 shell 将正确解释 & 符号，并且不会将其作为参数传递给任何命令。
Simply place the arguments to “build_all” on the command line, as for any other command:
and the arguments will be passed to “build_all”, not to “time” or “nohup”. The ampersand will be interpreted correctly by the shell, and will not be passed as an argument to any of the commands.
脚本参数（参数）只需跟在脚本名称后面并在 & 符号前面。
The script arguments (parameters) simply follow the script's name and preceed the ampersand.