我目前正在尝试为每个页面使用一个svg精灵,以便它只为每个单独的页面加载必要的svg。当前正在工作,但我想创建一个数组,以便为我想要的N个页面调用任务get。
我目前在gulpfile.js (版本4)中有这个结构:
var autoprefixer = require('autoprefixer'),
browsersync = require('browser-sync').create(),
cssnano = require('cssnano'),
sourcemaps = require('gulp-sourcemaps'),
del = require('del'),
gulp = require('gulp'),
imagemin = require('gulp-imagemin'),
newer = require('gulp-newer'),
plumber = require('gulp-plumber'),
postcss = require('gulp-postcss'),
rename = require('gulp-rename'),
sass = require('gulp-sass'),
svgSprite = require('gulp-svg-sprite');
//Create SVG's into one single image file
function svgs(){
return gulp
.src('app/images/src/pageName/*')
.pipe(svgSprite({
shape: {
spacing: {
padding: 5
}
},
mode: {
css: {
dest: './',
layout: 'diagonal',
sprite: 'images/dist/pageName/sprite.svg',
bust: false,
render: {
scss: {
dest: 'sass/modules/sprites/_pageName.scss',
template: 'app/sass/modules/sprites/_pageNameTemplate.scss'
}
}
}
},
variables: {
mapname: 'icons'
}
}))
.pipe(gulp.dest('app/'));
}
exports.svgs = svgs;我们的想法是创建一个包含页面名称的数组,并执行for循环:
var pages = ['page1', 'page2',...];
function svgs(){
for (var i = 0; i < pages.length; i++) {
return gulp
.src('app/images/src/' + pages[i] + '/*').........
}然后在控制台中只调用任务,但return只在一次迭代中执行,有人以前这样做过吗?
谢谢!
发布于 2019-10-22 01:13:40
尝试glob.sync它返回一个数组:
const glob = require('glob');
const pageArray = glob.sync('path/to/your/*.pages');
function svgs(done) {
pageArray.forEach( page => {
return gulp.src(page)
...
};
done();
}https://stackoverflow.com/questions/58490977
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