如何使用gulp watch中的参数运行gulp任务?
我有一个像这样的怀表:如何使用gulp watch中的参数运行gulp任务?,gulp,gulp-watch,Gulp,Gulp Watch,我有一个像这样的怀表: gulp.watch(somepath, {interval: 500}, ['buildScripts']); gulp.task('buildScripts', function (path) { //compiles the file from watch }); 构建脚本如下所示: gulp.watch(somepath, {interval: 500}, ['buildScripts']); gulp.task('buildScripts', fun
gulp.watch(somepath, {interval: 500}, ['buildScripts']);
gulp.task('buildScripts', function (path) {
//compiles the file from watch
});
构建脚本如下所示:
gulp.watch(somepath, {interval: 500}, ['buildScripts']);
gulp.task('buildScripts', function (path) {
//compiles the file from watch
});
如何传递已更改文件的值,以便
构建脚本
可以编译这些文件?您必须将构建脚本
代码分解为自己的函数,并从手表和任务中调用该函数。如果你不需要这项任务,你当然可以放弃它
var gulp = require('gulp');
function buildScripts(changedFile) {
if (changedFile) {
// called from watch
// compile the changed file
} else {
// called from task
// compile all files
}
}
gulp.task('buildScripts', function() {
return buildScripts();
});
gulp.task('watch', function() {
gulp.watch('*.js', {interval:500}, buildScripts);
});
您必须将
buildScripts
代码分解成它自己的函数,并从手表和任务中调用该函数。如果你不需要这项任务,你当然可以放弃它
var gulp = require('gulp');
function buildScripts(changedFile) {
if (changedFile) {
// called from watch
// compile the changed file
} else {
// called from task
// compile all files
}
}
gulp.task('buildScripts', function() {
return buildScripts();
});
gulp.task('watch', function() {
gulp.watch('*.js', {interval:500}, buildScripts);
});