bash :替换""内的变量值 [英] bash : replace variable value inside ' '

问题描述

对不起，如果问题很简单，但我是 shell 脚本的新手.我正在尝试写这样的东西:

for i in {1..20}做curl "某物 $i ........ -d '某物 "$i" 某物' "完毕

问题是单引号部分 '' 中的第二个 $i 没有被替换.应该怎么做才能让它工作?

解决方案

如上所说，参数在单引号内不展开，必须使用双引号.唯一的一点是，因为它出现在一个已经被双引号括起来的字符串中，你必须用反斜杠 (\) 将它们转义，就像这样:

<前>$ foo=bar$ eval "echo \"something \\\"$foo\\\"\""一些酒吧"

注意在最里面的 " 之前有三个 \，因为这将被扩展两次(在评估 eval 的参数时一次，并且评估 echo) 的参数时一次

Sorry if the question is very straight forward but am a newbie to shell scripting. I am trying to write something like this :

for i in {1..20}
do
   curl "something $i ........ -d  'something "$i" something' "
done

The problem is that the second $i inside the single quotes part '' is not being replaced. What should be done to get it working ?

解决方案

As said above, parameters are not expanded inside single quotes, you have to use double quotes. The only point is that since it occurs in a already double-quoted string, you have to escape them with a backslash (\), like this:

$ foo=bar
$ eval "echo \"something \\\"$foo\\\"\""
something "bar"

Note that there are three \ before the innermost ", as this will be expanded twice (once when evaluating the argument of eval and once when evaluating the argument of echo)

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