Have this:
style.scss:
.example {
background: linear-gradient(to top, #45678a, #ab4563);
}
gulpfile.js:
var gulp = require('gulp');
var sass = require('gulp-sass');
var autoprefixer = require('gulp-autoprefixer');
gulp.task('sass', function(){
return gulp.src('scss/*.scss')
.pipe(sass({
outputStyle: 'expanded',
}))
.pipe(autoprefixer({
browsers: ['> 0%'], cascade: false
}))
.pipe(gulp.dest('css'))
});
Get:
.example {
background: -webkit-gradient(linear, left bottom, left top, from(#45678a), to(#ab4563));
background: -webkit-linear-gradient(bottom, #45678a, #ab4563);
background: -moz-linear-gradient(bottom, #45678a, #ab4563);
background: -o-linear-gradient(bottom, #45678a, #ab4563);
background: linear-gradient(to top, #45678a, #ab4563);
}
Want:
.example {
background: #45678a;
background: -webkit-gradient(linear, left bottom, left top, from(#45678a), to(#ab4563));
background: -webkit-linear-gradient(bottom, #45678a, #ab4563);
background: -moz-linear-gradient(bottom, #45678a, #ab4563);
background: -o-linear-gradient(bottom, #45678a, #ab4563);
background: linear-gradient(to top, #45678a, #ab4563);
}
So i want to have background: #45678a; in case when linear-gradient doesn't support (i.g. ie8).
Does it possible with autoprefixer or some other 'auto...' not to write extra background property?
Thanks.