模拟multipart / form-data Express.JS请求对象

时间:2016-09-04 18:30:43

标签: node.js unit-testing express jasmine formidable

我想对Express中间件功能进行单元测试,然后使用node-formidable来处理多部分文件上传。

这是一个人为的例子:

function doUpload(req, res, next) {
    const form = new formidable.IncomingForm();
    form.uploadDir = this.mediaDir;
    form.keepExtensions = true;
    form.type = 'multipart';
    form.multiples = true;

    form.parse(req, function(err, fields, files) {
        res.writeHead(200, {'content-type': 'text/plain'});
        res.write('received upload:\n\n');
        res.end(util.inspect({fields: fields, files: files}));
    });
}

在使用Chrome和正在运行的Express应用程序进行测试时,此代码适用于我。

我希望我的测试代码看起来像我在模拟请求对象的地方,但我无法弄清楚如何使用表单数据模拟请求对象。强大的回调没有触发:

it(‘handles uploads’, (done) => {
    const mockReq = http.request({
        host: 'example.org',
    });
    mockReq.url = ‘/upload’;

    const res = jasmine.createSpyObj('response', [ 'json', 'send', 'status', 'render', 'header', ‘redirect’, ‘end’, ‘write;]);
    const next = jasmine.createSpy('next');

    //*how to emulate a request with a multipart file upload)

    doUpload(req, res, next);
    res.write.and.callFake(done);
});

我尝试使用form-data库来创建FormData对象并将其传递给请求,但没有运气,我不确定我是否在正确的轨道上或离开。像这样:

var form = new FormData();

const buff = fs.readFileSync(path.join(__dirname, '../fixtures/logo.png'));

form.append('file', buff, {
    filename: 'logo.jpg',
    contentType: 'image/png',
    knownLength: buff.length
});
form.append('name', 'Logo');

req.headers = form.getHeaders();

form.pipe(req);
doUpload(req, res, next);

2 个答案:

答案 0 :(得分:0)

您可以使用诸如supertest之类的请求测试器来完成。这是一个示例,假设您的主文件名为app.js:

const request = require('supertest');
const app = require('app.js');

it('Uploads a file', function(){
    request(app)
      .post('/upload')
      .field('name', 'Logo') //adds a field 'name' and sets its value to 'Logo'
      .attach('file', '/path/to/file') // attaches the file to the form
      .then(function(response){
          // response from the server
          console.log(response.status);
          console.log(response.body);
      })

})

答案 1 :(得分:0)

使用form-datamock-express-request的组合。这对我有用:

const fs = require('fs');
const MockExpressRequest = require('mock-express-request');
const FormData = require('form-data');

const form = new FormData();
form.append('my_file',
  fs.createReadStream(path.join(__dirname, 'fixtures', 'file-upload-test.txt'))
);
const request = new MockExpressRequest({
  method: 'POST',
  host: 'localhost',
  url: '/upload',
  headers: form.getHeaders()
});

form.pipe(request);
doUpload(request, response, next);