How do I force gulp calls to run synchronously?
Solution 1:
You can use async as a control flow for your calls to get them in only one task, also avoiding you to get a "pyramid effect". So something like this should be good for your use-case:
var async = require('async');
gulp.task('yeah', function (cb) {
async.series([
function (next) {
gulp.src('...')
.pipe(gulp.dest('...')
.on('end', next);
},
function (next) {
gulp.src('...')
.pipe(gulp.dest('...')
.on('end', next);
},
function (next) {
gulp.src('...')
.pipe(gulp.dest('...')
.on('end', next);
}
], cb);
});
That will also allow you to have some error handling and target better where a problem occured.
Solution 2:
Well, it's just streams so you could listen for the end event (Watch out for the pyramid of doom!)
gulp.task("dostuff", function (callback) {
gulp
.src("...")
.pipe(gulp.dest("..."))
.on('end', function () {
gulp
.src("...")
.pipe(gulp.dest("..."))
.on('end', function () {
gulp
.src("...")
.pipe(gulp.dest("..."))
.on('end', callback);
});
});
});
But it's probably a better pattern to split it up in multiple tasks each one with a dependency on the previous one.
Solution 3:
run-sequence:
Runs a sequence of gulp tasks in the specified order. This function is designed to solve the situation where you have defined run-order, but choose not to or cannot use dependencies.
npm install --save-dev run-sequence
// runSequence will ensure this task will run the following tasks in the listed order
gulp.task('things-to-do', callback => runSequence(
'clean-up-workspace', // clean up before copying new files
'copy-new-files', // wait till copy-new-files done before linting
'lint', // wait till lint is done before running minify
'minify', // wait till minify is done before starting laundry and dinner
['do-laundry', // do laundry and cook dinner at the same time
'cook-dinner'],
'bath-cat', // don't bath the cat till both laundry and dinner are done
callback
));
https://www.npmjs.com/package/run-sequence