Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
467 views
in Technique[技术] by (71.8m points)

eval - Indirect variable assignment in bash

Seems that the recommended way of doing indirect variable setting in bash is to use eval:

var=x; val=foo
eval $var=$val
echo $x  # --> foo

The problem is the usual one with eval:

var=x; val=1$'
'pwd
eval $var=$val  # bad output here

(and since it is recommended in many places, I wonder just how many scripts are vulnerable because of this...)

In any case, the obvious solution of using (escaped) quotes doesn't really work:

var=x; val=1"$'
'pwd"
eval $var="$val"  # fail with the above

The thing is that bash has indirect variable reference baked in (with ${!foo}), but I don't see any such way to do indirect assignment -- is there any sane way to do this?

For the record, I did find a solution, but this is not something that I'd consider "sane"...:

eval "$var='"${val//'/'"'"'}"'"
Question&Answers:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

A slightly better way, avoiding the possible security implications of using eval, is

declare "$var=$val"

Note that declare is a synonym for typeset in bash. The typeset command is more widely supported (ksh and zsh also use it):

typeset "$var=$val"

In modern versions of bash, one should use a nameref.

declare -n var=x
x=$val

It's safer than eval, but still not perfect.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...