如何在bash中组合超时和eval命令 [英] How to combine timeout and eval commands in bash

问题描述

对于执行存储在变量中的命令，使用 eval 命令：

For executing command that is stored in variable the eval command is used:

└──> a="echo -e 'a\nb' | wc -l"
└──> eval $a
2

但是如何将其与结合超时命令？我尝试了以下操作，这给了我错误的输出：

But how can it be combined with timeout command? I've tried following which gives me wrong output:

└──> timeout 10 $a
'a
b' | wc -l

以下内容使我出错：

└──> timeout 10 "$a"
timeout: failed to run command `echo -e \'a\\nb\' | wc -l': No such file or directory

└──> timeout 10 $(eval $a)
timeout: failed to run command `2': No such file or directory

└──> timeout 10 $(eval "$a")
timeout: failed to run command `2': No such file or directory

问题也可以成立：如何确定以下命令是否正确执行？

The question can also stand: How can I be sure that following command is executed properly?

timeout 10 "$PROGRAM" "$OPT1" "$OPT2" ...

推荐答案

简单：

a="echo -e 'a\nb' | wc -l"
eval timeout 10 $a

输出：

2

这篇关于如何在bash中组合超时和eval命令的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！