如何在提供文件之前等待 Python shell 执行

Question

I've an express server route which receives a xml file, then parses and return it as json.

When a user sends a file, it saves in a directory './upload', parses it with a Python script then output json in './json-output', which is served.

When I first upload a file, the response comes empty. But when I do the same upload steps (there is a json already created from last upload on './json-output' dir), it serves the json. It seems some asynchronous issue but I couldn't fix it.

app.post('/upload', function(req, res) {
    upload(req, res, async function(err) {
        if (err) {
            res.json({ error_code: 1, err_desc: err });
            return;
        }

        if (!req.file) {
            res.json({ error_code: 1, err_desc: 'No file passed' });
            return;
        }

        let fileName = req.file.originalname;

        const options = {
            args: [ fileName ]
        };

        const parserPath = path.join(__dirname + '/parser/parser.py');

        const readFile = promisify(PythonShell.run);

        await readFile(parserPath, options);

        fileName = fileName.split('.')[0];
        res.sendFile(path.join(__dirname + `/json-output/${fileName}.json`));
    });
});

I'm running it inside a docker images

Answer 1

This a quite a "Dirty fix" in my eyes but you could do a while loop EG:

fileName = fileName.split('.')[0];
while (!fs.existsSync(path.join(__dirname + `/json-output/${fileName}.json`)){
    console.log('File does not exist!')
}
//Becareful you should delete the file once the res.send is done
res.sendFile(path.join(__dirname + `/json-output/${fileName}.json`));

Decided to read the python-shell docs here an idea: https://www.npmjs.com/package/python-shell#exchanging-data-between-node-and-python

So, in theory, you can start a new

let pyshell = new PythonShell(path.join(__dirname + '/parser/parser.py'),options);
pyshell.end(function (err,code,signal) {
  if (err) throw err;
  fileName = fileName.split('.')[0];
  res.sendFile(path.join(__dirname + `/json-output/${fileName}.json`));
});