[英]How read data from file gulp.src('json/username.json') in node_modules?
如何读取数据文件gulp.src('json/username.json')
在node_modules(node_modules /大口向上/ index.js)?
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 /吞噬/ 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: 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)
我不太了解您的问题...但是,如果您想阅读json/username.json
文件,则可以只需要它,就像这样:
var json = require('json/username.json');
json变量将包含username.json文件中的数据。
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