Question

如何从 node_modules （ node_modules / gulp-up / index.js ）中的gulp.src('json/username.json')读取数据文件？

username.json

{
  "name": "Rikardo",
  "city": "Mosscov",
  "age": "18"
}

gulpfile.js

upmodul = require("gulp-up");

gulp.task('gulp-ren', function () {
    return gulp.src('json/username.json')
       .pipe(myfunction());
});

gulp.task('ge', ['gulp-ren']);

node_modules / gulp-up / index.js

'use strict';
var Stream = require('stream');


function loger() {
    //The needed maybe not Transform and Read file
    var stream = new Stream.Transform({ objectMode: true });


    stream._transform = function (file, unused, callback) {
        console.log('test !!!  - ' + file.contents);

        return stream;
    }
}

module.exports =  loger;

我在控制台命令中写道：gulp ge，并给出错误：

    [00:19:39] TypeError: Cannot read property 'on' of undefined
        at DestroyableTransform.Readable.pipe (E:\Developers\WebDeveloper\OpenServer
    -WebProg\domains\progectapi2\node_modules\vinyl-fs\node_modules\readable-stream\
    lib\_stream_readable.js:516:7)
        at Gulp.<anonymous> (E:\Developers\WebDeveloper\OpenServer-WebProg\domains\p
    rogectapi2\gulpfile.js:159:9)
        at module.exports (E:\Developers\WebDeveloper\OpenServer-WebProg\domains\pro
    gectapi2\node_modules\orchestrator\lib\runTask.js:34:7)
        at Gulp.Orchestrator._runTask (E:\Developers\WebDeveloper\OpenServer-WebProg
    \domains\progectapi2\node_modules\orchestrator\index.js:273:3)
        at Gulp.Orchestrator._runStep (E:\Developers\WebDeveloper\OpenServer-WebProg
    \domains\progectapi2\node_modules\orchestrator\index.js:214:10)
        at Gulp.Orchestrator.start (E:\Developers\WebDeveloper\OpenServer-WebProg\do
    mains\progectapi2\node_modules\orchestrator\index.js:134:8)
        at C:\Users\Tiki
\AppData\Roaming\npm\node_modules\gulp\bin\gulp.js:129:20
        at nextTickCallbackWith0Args (node.js:433:9)
        at process._tickCallback (node.js:362:13)
        at Function.Module.runMain (module.js:432:11)

Answer 1

我真的不明白你的问题...但是如果你想阅读function display(){ $.get('https://URL', function(data){ var parser = new DOMParser() , doc = parser.parseFromString(data, "text/xml"); console.log(doc); var elements = doc.querySelectorAll( 'body *' ); console.log(elements[0].innerHTML); }) } display()文件，那么你可以这样做，就像这样：

json/username.json

json变量将包含username.json文件中的数据。

如何从node_modules中的文件gulp.src（'json / username.json'）读取数据？

1 个答案: