Issue
I have an npm script though I think this would apply to any bash command:
node file.js
The content of file.js
for the purpose of the question might be:
console.log('hello')
I would like to wrap the content of file.js
in some code so that the final "content" is:
console.log('start')
console.log('hello')
console.log('end')
Right now I have the start/end stuff in a separate file which is the one I call and it in turns, imports the file.js
file
// wrapper.js
const foo = import('./file.js')
console.log('start')
foo() // or whatever...
console.log('end')
Then I call it:
node wrapper.js
This approach works but I am trying to eliminate the wrapper file and pass the content directly - Essentially concatenate 2 strings with content of a file and then have the node
utility accept that like a file. (note that this is not specific to node cli)
Solution
edit: Charles Duffy's comment recommends a simpler solution if you just want to wrap a file and don't need anything more complex than that.
{
echo "console.log('start')"
cat ./file.js
echo "console.log('end')"
} > ./wrapped.js
You can use cat
for this.
printf '%s\n' "console.log('hello')" > ./file.js
cat <( printf '%s\n' "console.log('start')" ) \
./file.js \
<( printf '%s\n' "console.log('end')" ) \
> ./wrapped.js
cat ./wrapped.js
# outputs:
# console.log('start')
# console.log('hello')
# console.log('end')
This utilizes two process substitutions (the <( command )
syntax) to create temporary files containing the content you want to wrap around your script, then concatenates the three files together into a single file.
If you don't want to write the concatenated data out to a file, you could wrap the whole thing in another process substitution and pass it directly to your command in whatever manner it takes files.
node <(
cat <( printf '%s\n' "console.log('start')" ) \
./file.js \
<( printf '%s\n' "console.log('end')" )
)
Answered By - tjm3772 Answer Checked By - Marilyn (WPSolving Volunteer)