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问题描述

限时删除!!

我一直在尝试使用gulp-typescript取得一定程度的成功,但我有一个小问题。我所有的代码都存储在'src'下,我希望这些代码被编译为'.tmp',但不包含'src'。

这里是我的代码,我认为问题在于传递值(glob)到tsProject.src不支持,所以我得到/.tmp/src/aTypescriptFile.js例如



这个代码我直接从github回购,我真的不明白为什么gulp.src被tsProject.src替换



任何想法?我确实需要合并我的tsconfig.json文件。

  let tsProject = plugins.typescript.createProject('./ tsconfig。 JSON'); 
return tsProject.src('/ src / ** / *。ts')
.pipe(plugins.typescript(tsProject))
.pipe(plugins.sourcemaps.init())
.pipe(plugins.sourcemaps.write('。'))
.pipe(gulp.dest('。tmp'));

**编辑**



更多info,我已经设法限制它使用glob通过替换

  return tsProject.src('/ src / ** / * .ts')

with

  return gulp.src('/ src / ** / *。ts')

问题现在我得到一个关于缺少类型的错误。

  src / testme.ts(4,10 ):错误TS2304:找不到名字'require'。 

我的TSCONFIG.JSON文件在这里,里面有类型。

  {
compilerOptions:{
target:ES6,
module:commonjs ,
moduleResolution:node,
sourceMap:true,
emitDecoratorMetadata:true,
experimentalDecorators:true,
removeComments :false,
noImplicitAny:false
},
files:[
typings / main.d.ts,
src / testme .ts

}


解决方案

所有路径都应该传递给gulp.src - 源代码和类型。



让我们有一些路径:

  var paths = {
lib:./wwwroot/,
ts:[./sources/**/*.ts],
样式:[./sources/**/*.scss],
模板:[./sources/**/*.html],
类型:。 /typings/**/*.d.ts,
// svg:./sources/**/*.svg,
};

我们可以将一组源文件路径传递给gulp-typescript:

  gulp.task(?typescript:demo:debug,function(){
var tsResult = gulp.src([
paths .typings,
//< ...其他路径...>
] .concat(paths.ts))
.pipe(sourcemaps.init())
.pipe(ts({
target:ES5,
experimentalDecorators:true,
noImplicitAny:false
}));

return tsResult.js
.pipe(concat(package.name +.js))
.pipe(sourcemaps.write({sourceRoot:}))
.pipe(gulp。 dest(paths.lib));
})

我正在传递

  gulp.src([paths.typings,< ... some other paths ...>]。concat(paths.ts ))

但当然,它也可以以更简单的方式完成:

  gulp.src([paths.typings,paths.ts])


I have been trying to use gulp-typescript with some degree of success but I have a small issue. All my code is stored under 'src' and I want these to be compiled to '.tmp' but without the 'src' being included.

here is my code, I think the issue is that passing a value (glob) to tsProject.src isn't supported so I get /.tmp/src/aTypescriptFile.js for example

This code I got directly from the github repo, what I really didn't understand is why gulp.src is replaced with tsProject.src

Any ideas ? I do really need to incorporate my tsconfig.json file.

    let tsProject = plugins.typescript.createProject('./tsconfig.json');
    return tsProject.src('/src/**/*.ts')
        .pipe(plugins.typescript(tsProject))
        .pipe(plugins.sourcemaps.init())
        .pipe(plugins.sourcemaps.write('.'))
        .pipe(gulp.dest('.tmp'));

** EDIT **

More info, I have managed to confine it using a glob by replacing the

             return tsProject.src('/src/**/*.ts')

with

             return gulp.src('/src/**/*.ts')

problem is now that I get an error about missing typings.

        src/testme.ts(4,10): error TS2304: Cannot find name 'require'.

my TSCONFIG.JSON file is here, which has the typings in there.

{
  "compilerOptions": {
    "target": "ES6",
    "module": "commonjs",
    "moduleResolution": "node",
    "sourceMap": true,
    "emitDecoratorMetadata": true,
    "experimentalDecorators": true,
    "removeComments": false,
    "noImplicitAny": false
  },
  "files": [
    "typings/main.d.ts",
    "src/testme.ts"
  ]
}
解决方案

All paths should be passed to gulp.src -- sources and typings.

Let us have some paths:

var paths = {
    lib: "./wwwroot/",
    ts: ["./sources/**/*.ts"],
    styles: ["./sources/**/*.scss"],
    templates: ["./sources/**/*.html"],
    typings: "./typings/**/*.d.ts",
    //svg: "./sources/**/*.svg",
};

We can pass an array of source paths to gulp-typescript:

gulp.task("?typescript:demo:debug", function () {
    var tsResult = gulp.src([
          paths.typings,
          // <...some other paths...>
        ].concat(paths.ts))
       .pipe(sourcemaps.init())
       .pipe(ts({
           target: "ES5",
           experimentalDecorators: true,
           noImplicitAny: false
       }));

    return tsResult.js
        .pipe(concat(package.name + ".js"))
        .pipe(sourcemaps.write({ sourceRoot: "" }))
        .pipe(gulp.dest(paths.lib));
})

I'm passing

gulp.src([paths.typings, <...some other paths...>].concat(paths.ts))

but of course, it can also be done in a simpler way:

gulp.src([paths.typings, paths.ts])

这篇关于gulp-typescript:使用createProject的问题的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

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09-06 12:31