避免shell脚本中的单独文件

Question

我正在尝试执行以下命令，该命令运行良好

我正在执行test.sh，它将调用script1.sh。

成功运行，输出如下

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

这两个文件的内容如下

test.sh

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

script1.sh如下所示

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，我将得到如下语法错误

applicationuser@servername:/application> sudo ./test.sh ./test.sh:第1行:如果ech“右回送”2>/dev/null，则then' ./test.sh: line 1:超时10附近的语法错误；然后回送‘命令成功’；否则回送‘该死，有错误’；fi && echo“超时未发生”

如何消除script1.sh，并将其内容放在test.sh中，并在没有语法错误的情况下执行？

Answer 1

一个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"'