[英]How to include partial html file into another html file using gulp?
I have two folders, dist
and partials
, the 'dist' folder contains the index.html file and the 'partials' folder contains header.html, navbar.html, and footer.html files. 我想將這些部分文件包含到 index.html 中。 我嘗試了 gulp-file-include 插件,它工作正常,但我希望每當我對任何部分文件執行任何更改時,都應該更新 index.html 文件。 我無法使用 gulp-file-include 插件執行此操作,請提供任何其他解決方案...?
gulpfile.js
'use strict'
const fileinclude = require('gulp-file-include');
const gulp = require('gulp');
gulp.task('fileinclude', function() {
return gulp.src(['dist/index.html'])
.pipe(fileinclude({
prefix: '@@',
basepath: '@file'
}))
.pipe(gulp.dest('dist'));
});
索引.html
@@include('../partials/header.html')
@@include('../partials/navbar.html')
@@include('../partials/footer.html')
使用 gulp.watch(...) 跟蹤所有更改,它不僅適用於 html。 使用此方法跟蹤 gulp 中的任何文件。
const gulp = require('gulp');
const include_file = require('gulp-file-include');
gulp.task('include', () => {
return gulp.src('./res/*.html')
.pipe(include({
prefix: "@@",
basepath: "@file"
}))
.pipe(gulp.dest('./public'));
});
gulp.task('watch', () => {
gulp.watch('./res/*.html', gulp.series('include'));
})
也是我的變種:
const { src, dest, watch } = require('gulp'),
include_file = require('gulp-file-include');
function include() {
return src('./res/*.html')
.pipe(file_include({
prefix: '@',
basepath: '@file'
}))
.pipe(dest('./public/'));
}
function watching() {
watch('./res/*.html', include);
}
exports.watch = watching;
然后:
gulp watch
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