gulp 4.0 提示 Error: watching index.html: watch task has to be a function 解决办法
提示：Error: watching index.html: watch task has to be a function (optionally generated by using gulp.parallel or gulp.series)
gulp 4.0之后不能按照之前的写法了 大有改变
4.0之前的写法
Gulp.watch(“监听的文件”，[任务名字符串])
-
// gulp.task("watch",function(){
-
// gulp.watch("index.html",["copyindex"]);
-
// gulp.watch("js/**/*",["copyjs"]);
-
// gulp.watch("css/**/*.scss",["copycss"]);
-
// gulp.watch("html/**/*",["copyhtml"]);
-
// gulp.watch("php/**/*",["copyphp"]);
-
// gulp.watch("img/**/*",["copyimg"]);
-
// })
4.0之后的写法
Gulp.watch(“监听的文件”，回调函数)。
gulp.task("watch-all",async ()=>{
gulp.watch("index.html",async ()=>{
gulp.src("index.html")
.pipe(gulp.dest("D:\myphp_www\PHPTutorial\WWW\baidu"));
});
gulp.watch("js/**/*",async ()=>{
gulp.src("js/**/*")
.pipe(gulp.dest("D:\myphp_www\PHPTutorial\WWW\baidu\js"));
});
gulp.watch("css/**/*.scss",async ()=>{
gulp.src("css/**/*.scss")
.pipe(sass())
.pipe(gulp.dest("D:\myphp_www\PHPTutorial\WWW\baidu\css"));
});
gulp.watch("html/**/*",async ()=>{
gulp.src("html/**/*")
.pipe(gulp.dest("D:\myphp_www\PHPTutorial\WWW\baidu\html"));
});
gulp.watch("php/**/*",async ()=>{
gulp.src("php/**/*")
.pipe(gulp.dest("D:\myphp_www\PHPTutorial\WWW\baidu\php"));
});
gulp.watch("img/**/*",async ()=>{
gulp.src("img/**/*")
.pipe(gulp.dest("D:\myphp_www\PHPTutorial\WWW\baidu\img"));
});
});