xargs-我在查找中替换了str -exec sed

Question

I'm on GNU/Linux system with Bash 4.4.23 and I have a file with some Bash variables:

ex.

export VARONE="var1"
export VARTWO="var2"
...

I have to replace the ${VARIABLE} with {{VARIABLE}} in other files with ".template" extension under "templates" folder.

I try to execute this command:

cat varfile \
  | grep export \
  | awk -F '=' '{print $1}' \
  | awk '{print $2}' \
  | xargs -I var -- \
      find templates -type f -iname "*.template" \
        -exec sed 's/${var}/{{var}}/g' {} \;

but I had no luck :( it seems that sed does not match with this LHS, infact it prints the original file as output.

Someone can explain what is wrong with the command above?

Answer 1

The only issue seems like the lack of -i flag and the unescaped $ for sed.

Here's a simpler, tested, initial pipeline with aforementioned fixes:

sed -nE '/^export /{s///;s/([^=]+).*/\1/;p;}' varfile |
    xargs -I var -- find templates -name '*.template' -exec sed -i '' 's/\${var}/{{var}}/g' {} +

For assurance, I added a g flag if the variable appears more than once per line.