Bash脚本-可变内容作为要运行的命令 [英] Bash script - variable content as a command to run

问题描述

我有一个Perl脚本，该脚本为我提供了一个与文件行相对应的已定义列表随机数.接下来，我想使用sed从文件中提取这些行.

I have a Perl script that gives me a defined list random numbers that correspond to the lines of a file. Next I want to extract those lines from the file using sed.

#!/bin/bash
count=$(cat last_queries.txt | wc -l)
var=$(perl test.pl test2 $count)

变量var返回类似cat last_queries.txt | sed -n '12p;500p;700p'的输出.问题是我无法运行最后一条命令.我尝试使用$var，但输出不正确(如果我手动运行该命令，它将正常运行，因此那里没有问题).正确的方法是什么?

The variable var returns an output like: cat last_queries.txt | sed -n '12p;500p;700p'. The problem is that I can't run this last command. I tried with $var, but the output is not correct (if I run manually the command it works fine, so no problem there). What is the correct way to do this?

P.S:当然，我可以在Perl中完成所有工作，但是我正在尝试学习这种方式，因为它可以在其他情况下为我提供帮助.

P.S: Sure I could do all the work in Perl, but I'm trying to learn this way, because it could help me in other situations.

推荐答案

您只需要这样做:

#!/bin/bash
count=$(cat last_queries.txt | wc -l)
$(perl test.pl test2 $count)

但是，如果要稍后再调用Perl命令，这就是为什么要将其分配给变量的原因，那么:

However, if you want to call your Perl command later, and that's why you want to assign it to a variable, then:

#!/bin/bash
count=$(cat last_queries.txt | wc -l)
var="perl test.pl test2 $count" # You need double quotes to get your $count value substituted.

...stuff...

eval $var

根据Bash的帮助:

~$ help eval
eval: eval [arg ...]
    Execute arguments as a shell command.

    Combine ARGs into a single string, use the result as input to the shell,
    and execute the resulting commands.

    Exit Status:
    Returns exit status of command or success if command is null.

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