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，它又调用 script1.sh。

运行成功，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"

执行脚本时会出现如下语法错误

applicationuser@servername:/application> sudo ./test.sh ./test.sh: 行 1: 意外标记then' ./test.sh: line 1: timeout 10 附近出现语法错误 if ech "right echo" 2>/dev/null ;然后 echo ' 命令是 成功的'; else echo '该死，出现错误';菲 && 回声 “没有发生超时” || echo "发生超时"'

如何去掉script1.sh，把它的内容放到test.sh中执行，不会出现语法错误？

Answer 1

解决方案可以直接使用 bash 运行命令：

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"'