Bash 脚本 - 将变量内容作为要运行的命令
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Bash script - variable content as a command to run
提问by Barata
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.
我有一个 Perl 脚本，它为我提供了一个对应于文件行的已定义随机数列表。接下来我想使用sed.
#!/bin/bash
count=$(cat last_queries.txt | wc -l)
var=$(perl test.pl test2 $count)
The variable varreturns 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 $var, but the output is not correct (if I run manually the command it works fine, so no problem there). What is the correct way to do this?
可变var返回一个输出，如：cat last_queries.txt | sed -n '12p;500p;700p'。问题是我无法运行最后一个命令。我试过$var，但输出不正确（如果我手动运行命令它工作正常，所以没问题）。这样做的正确方法是什么？
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.
PS：当然我可以用 Perl 做所有的工作，但我正在尝试以这种方式学习，因为它可以在其他情况下帮助我。
回答by hmontoliu
You just need to do:
你只需要做：
#!/bin/bash
count=$(cat last_queries.txt | wc -l)
$(perl test.pl test2 $count)
However, if you want to call your Perl command later, and that's why you want to assign it to a variable, then:
但是，如果你想稍后调用你的 Perl 命令，这就是你想将它分配给一个变量的原因，那么：
#!/bin/bash
count=$(cat last_queries.txt | wc -l)
var="perl test.pl test2 $count" # You need double quotes to get your $count value substituted.
...stuff...
eval $var
As per Bash's help:
根据 Bash 的帮助：
~$ 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.
回答by August Karlstrom
Your are probably looking for eval $var.
您可能正在寻找eval $var.
回答by nhed
line=$((${RANDOM} % $(wc -l < /etc/passwd)))
sed -n "${line}p" /etc/passwd
just with your file instead.
只是用你的文件代替。
In this example I used the file /etc/password, using the special variable ${RANDOM}(about which I learned here), and the sedexpression you had, only difference is that I am using double quotes instead of single to allow the variable expansion.
在这个例子中，我使用了文件 /etc/password，使用了特殊变量${RANDOM}（我在这里学到的）和sed你所拥有的表达式，唯一的区别是我使用双引号而不是单引号来允许变量扩展。
回答by Ajedi32
In the case where you have multiple variables containing the arguments for a command you're running, and not just a single string, you should notuse eval directly, as it will fail in the following case:
在这种情况下，你必须包含参数的命令，你正在运行，而不仅仅是一个单一的字符串，你应该多变量不能EVAL直接使用，因为它会在以下情况下会失败：
function echo_arguments() {
echo "Argument 1: "
echo "Argument 2: "
echo "Argument 3: "
echo "Argument 4: "
}
# Note we are passing 3 arguments to `echo_arguments`, not 4
eval echo_arguments arg1 arg2 "Some arg"
Result:
结果：
Argument 1: arg1
Argument 2: arg2
Argument 3: Some
Argument 4: arg
Note that even though "Some arg" was passed as a single argument, evalread it as two.
请注意，即使“Some arg”作为单个参数传递，也eval应将其读为两个。
Instead, you can just use the string as the command itself:
相反，您可以将字符串用作命令本身：
# The regular bash eval works by jamming all its arguments into a string then
# evaluating the string. This function treats its arguments as individual
# arguments to be passed to the command being run.
function eval_command() {
"$@";
}
Note the difference between the output of evaland the new eval_commandfunction:
请注意eval和新eval_command函数的输出之间的区别：
eval_command echo_arguments arg1 arg2 "Some arg"
Result:
结果：
Argument 1: arg1
Argument 2: arg2
Argument 3: Some arg
Argument 4:
