8

我正在使用强大的来处理我在 NodeJs 中的文件上传。我有点卡在解析字段值上。

如何将 project_id 的值传递给表单处理程序，以便可以在文件名中写入参数？

<input type="text" id="project_id" value="{{projects._id}}" readonly>

编辑

更具体地说，这是我的表单上传处理的详细视图：

app.post('/uploads/', function (req, res){
    var form = new formidable.IncomingForm();
    form.parse(req, function (err, fields, files) {
        res.writeHead(200, {'content-type': 'image/jpeg'});
        res.write('received upload: \n\n');
        var project = fields.project_id;
        res.end(util.inspect(project, {fields: fields, files: files}));
    });

    form.on('end', function(project, fields, files){ 
        console.log(project); 
        /*Temporary location of our uploaded file */
        var temp_path = this.openedFiles[0].path;
        /*The file name of the uploaded file */
        var file_name =  project + '.' + this.openedFiles[0].name;

我可以登录var project该form.parse部分。但我没有得到这form.on('end'...部分的变量。

HTML 表单 

<form   id="uploadForm"
    enctype="multipart/form-data"
    action="/uploads/"
    method="post">
    <input type="text" name="project_id" id="project_id" value="{{projects._id}}" readonly>
    <input multiple="multiple" type="file" name="upload" />
    <button type="submit">Upload</button>
</form>
4

2 回答 2

9

Formidable 的end回调不带任何参数，但如果您使用回调，我不确定您是否需要调用它parse。我认为你正在寻找的是这样的：

var fs = require('fs');
app.post('/uploads', function(req, res, next) {
    var form = new formidable.IncomingForm();
    form.parse(req, function(err, fields, files) {
        if (err) next(err);

        // TODO: make sure my_file and project_id exist    
        fs.rename(files.my_file.path, fields.project_id, function(err) {
            if (err) next(err);
            res.end();
        });
    });
});

end()如果您选择不使用parse回调，则需要监听事件，如下所示：

new formidable.IncomingForm().parse(req)
    .on('file', function(name, file) {
        console.log('Got file:', name);
    })
    .on('field', function(name, field) {
        console.log('Got a field:', name);
    })
    .on('error', function(err) {
        next(err);
    })
    .on('end', function() {
        res.end();
    });
于 2015-05-08T18:23:02.487 回答
3

客户端脚本：

    //Upload the file
    var fd = new FormData();
    //Take the first selected file
    fd.append("dbDocPath", 'invoices/' + file.name);
    fd.append("file", file);
    $http({
            method: 'POST',
            url: $rootScope.apiUrl + 'uploadDocToServer',
            data: fd,
            headers: {
                'Content-Type': undefined
            },
            //prevents serializing payload.  don't do it.
            transformRequest: angular.identity,
        }).success(function (response) {
           if (response.success) {
           }
   })

服务器端脚本：

var fileDir = path.join(__dirname, '/../uploads');

// create an incoming form object
var form = new formidable.IncomingForm();
var dbDocPath = '';
form.parse(req)
        .on('field', function (name, field) {
            //console.log('Got a field:', field);
            //console.log('Got a field name:', name);
            dbDocPath = field;
        })
        .on('file', function (name, file) {
            //console.log('Got file:', name);

            // specify that we want to allow the user to upload multiple files in a single request
            //form.multiples = true;

            // store all uploads in the /uploads directory
            form.uploadDir = fileDir;

            fs.rename(file.path, path.join(form.uploadDir, file.name));

            // every time a file has been uploaded successfully,
            // rename it to it's orignal name

            var bucket = new AWS.S3();
            //console.log(dbDocPath);

            var params = {
                Bucket: DocsConfig.bucketName,
                Key: dbDocPath,
                Body: fs.createReadStream(path.join(form.uploadDir, file.name)),
                ACL: 'public-read'
            };

            bucket.putObject(params, function (perr, pres) {
                if (perr) {
                    //console.log("Error uploading data: ", perr);
                } else {
                    fs.unlinkSync(path.join(form.uploadDir, file.name));
                    //console.log("Successfully uploaded data", pres);
                }
            });
        })
        .on('error', function (err) {
            res.send({'success': false, error: err});
        })
        .on('end', function () {
            res.send({'success': true});
        });
// parse the incoming request containing the form data
//form.parse(req);

请记住一件事，将参数发送到 formData() 的顺序应该与上面代码中提到的相同，因为文件上传需要上传到命运的路径。

于 2017-05-23T09:18:54.270 回答