Question

我有这样的代码：

TEXT_TO_FILTER='I would like to replace this $var to proper value
                in multiline text'
var=variable

我想要的只是：

TEXT_AFTER_FILTERED="I'd like to replace this variable to proper value"

所以我做了：

TEXT_AFTER_FILTERED=`eval echo $TEXT_TO_FILTER`
TEXT_AFTER_FILTERED=`eval echo $(eval echo $TEXT_TO_FILTER)`

甚至更奇怪的东西，但没有任何影响。我记得有一天我有类似的问题而且我做了类似的事情：

cat << EOF > tmp.sh
echo $TEXT_TO_FILTER
EOF
chmod +x tmp.sh
TEXT_AFTER_FILTERED=`. tmp.sh`

但这个解决方案似乎要复杂得多。你有没有听说过更简单的解决方案？

Answer 1

对于security reasons，最好避免eval。这样的事情会更好：

TEXT_TO_FILTER='I would like to replace this %s to proper value'
var=variable
printf -v TEXT_AFTER_FILTERED "$TEXT_TO_FILTER" "$var"
# or TEXT_AFTER_FILTERED=$(printf "$TEXT_TO_FILTER" "$var")
echo "$TEXT_AFTER_FILTERED"

Answer 2

TEXT_AFTER_FILTERED="${TEXT_TO_FILTER//\$var/$var}"

或者，使用perl：

export var
TEXT_AFTER_FILTERED="$(echo "$TEXT_TO_FILTER" | perl -p -i -e 's/\$(\S+)/$ENV{$1} || $&/e')"

这比eval更安全。

解析另一个变量中包含的bash变量

2 个答案: