避免在 shell 脚本中使用单独的文件

Question

我正在尝试执行以下命令 效果很好

我正在执行 test.sh，然后调用 script1.sh。

成功运行，输出如下

applicationuser@servername:/application> sudo ./test.sh
damn, there was an error
timeout happened

两个文件的内容如下

test.sh

timeout 10 script1.sh  && echo "timeout not happened" || echo "timeout happened"

script1.sh如下

if ech "right echo" 2>/dev/null ; then echo 'command was successful'; else echo 'damn, there was an error'; fi

但是当我将两个脚本组合成一个文件 (test.sh) 时，如下所示，

test.sh

timeout 10 if ech "right echo" 2>/dev/null ; then echo 'command was successful'; else echo 'damn, there was an error'; fi  && echo "timeout not happened" || echo "timeout happened"

当我执行脚本时会出现如下语法错误

applicationuser@servername:/application> sudo ./test.sh ./test.sh: line 1: syntax error near unexpected token then' ./test.sh: line 1: timeout 10 if ech "right echo" 2>/dev/null ; then echo 'command was successful'; else echo 'damn, there was an error'; fi && echo "timeout not happened" || echo "timeout happened"'

如何消除script1.sh并将其内容放入test.sh并执行无语法错误？

Answer 1

一个解决方案可以是运行直接使用命令 bash:

timeout 10 bash -c 'if ech "right echo" 2>/dev/null ; then echo "command was successful"; else echo "damn, there was an error"; fi  && echo "timeout not happened" || echo "timeout happened"'

避免在 shell 脚本中使用单独的文件

Avoid seperate file in shell script

unix

linux

bash

shell

scripting