Bash: echo extract variables

Question

Suppose there's a script called 'test.sh':

#!/bin/bash
while read line; do
    APP=/apps echo "$line"
done < ./lines

And the 'lines':

cd $APP && pwd

If I bash test.sh , it prints out 'cd $APP && pwd'.

But when I type APP=/apps echo "cd $APP && pwd" in the terminal, it prints out 'cd /apps && pwd'.

Is it possible using echo to extract variables which are reading from a regular file?

Answer 1

Depending on the contents of the file, you may want to use eval:

#!/bin/bash
APP=/apps
while read line; do
    eval "echo \"$line\""  # WARNING: dangerous
done < ./lines

However, eval is extremely dangerous. Although the quoting here will work for simple cases, it is quite easy to execute arbitrary commands by manipulating the input.

Answer 2

您应该使用eval评估从文件读取的字符串行

Answer 3

如果知道要替换的变量，只需替换它们即可。

sed 's%\$APP\>%/apps%g' ./lines