I need to grab the most recent file with src
in every subdirectory.
This is the task I have :
gulp.task('compile', function() {
return gulp.src('src/jade/**/*.js')
.pipe(jade({pretty: true}))
.pipe(gulp.dest('dist'));
});
Is there a way to get the freshest JS with gulp src ?
Well, without really having used gulp so far, what you should do would be something like:
src
method (which takes a string, or an array of files or a stream)One of my problems was how to handle the glob style path, so for that I installed the glob
npm package in my local package.json
What I then did was, get all the files using glob( path, callback )
, filter those according to the last modified using fs.stat
and the mtime
or mtimeMs
and only keeping the last one
const { src, dest } = require('gulp');
const glob = require('glob');
const path = require('path');
const fs = require('fs');
async function getFileLastModifiedBy( file ) {
return new Promise( (resolve, reject) => {
fs.stat( file, (err, stats) => {
if (err) {
reject( err );
return;
}
resolve( stats.mtimeMs );
});
} );
}
async function filterByNewest( input ) {
const filesByModifiedBy = {};
for (let file of input) {
let directory = path.dirname( file );
let lastModifiedBy = await getFileLastModifiedBy( file );
if (!filesByModifiedBy[directory] || filesByModifiedBy[directory].lastModifiedBy < lastModifiedBy) {
filesByModifiedBy[directory] = { file, lastModifiedBy };
}
}
return Object.values( filesByModifiedBy ).map( v => v.file );
}
and then you could integrate it with your current workflow like in the following way
gulp.task('compile', function() {
return new Promise( (resolve, reject) => {
glob('src/jade/**/*.js', async function( err, files ) {
if ( files ) {
const newest = await filterByNewest( files );
resolve(
src( newest )
.pipe(jade({pretty: true}))
.pipe(dest('dist'))
);
return;
}
reject( err );
})
});
});