I have multiple files for test case in a folder and I want to run only one file, how can I do this?
Now I am giving like this:
'test/**/*.spec.js'
in karma.conf.js
and the following code is in gulpfile.js
-
gulp.task('test', function (done) {
new Server({
configFile: __dirname + '/karma.conf.js',
singleRun: true
}, function () {
done();
}).start();
});
Here, test is my folder where all test case js files are present.
How can I specify one js file?
Along with configFile
you specify the options that will overwrite the ones from config file (e.g. singleRun
).
So it has to be
gulp.task('test', function (done) {
new Server({
configFile: __dirname + '/karma.conf.js',
files: ['test/some.spec.js'],
singleRun: true
}, function () {
done();
}).start();
});