CODEWITHSUNDEEP
extjs4.2
CARTIC
CARTIC

Reputation: 573

File upload in extjs 4.2 without form.submit()

I'm trying to upload a file (as of now of any extension) in extjs. I have a model and store. the file upload happens from a window and I dont have a form in the window. All example I tried in the net are with form.submit(). I instead use and Ajax call as below to send the data to the server. 

Ext.Ajax.request({

            url : 'qaf/saveSetupDetails.action',

            params : {
                'data' : recordsToSend
            },
            failure : function(response){
                //console.log('error connecting controller');
            },
            success : function(response){
                //console.log('successfully submitted');
            }
        });

The records to send in the data is got as below.

var store = Ext.getStore('SomeStore');
        var modifiedRecords = store.getModifiedRecords();
        var recordsToSend = [];
        if(modifiedRecords.length > 0){
            Ext.each(modifiedRecords, function(record){
                recordsToSend.push(record.data);//I'm sure that this is so dump but this is how I do it for other records which are string and not sure how to do it for a file...
            });
        }
        Ext.USE_NATIVE_JSON = true;
        recordsToSend = Ext.encode(recordsToSend);

While setting the record in the model, I use the below code..

var rec = Ext.create('QAF.model.MyModel');
rec.set('modelField',Ext.getCmp('fileUploadCompID').value);

I received a 500 status error with the response "Cannot convert value of type [java.lang.String] to required type [org.springframework.web.multipart.commons.CommonsMultipartFile]"

I'm sure that this is because of the way I set the value to the model as Ext.getCmp('fileUploadCompID').value gives the file name. Please let me know how to set the file to the model and what data type I have to specify for the field in the model.

Below is how I try to retrieve the file in the spring controller.

@RequestMapping (value = "/qaf/saveSetupDetails.action")
    public @ResponseBody
    void saveSetupDetails(@RequestParam CommonsMultipartFile data)throws Exception{
        log.info("Enter into saveSetupDetails method..." + data.getOriginalFilename());
    }

Please let me know what I'm doing wrong and what has to be done to fix this...

Upvotes: 7

Views: 16948

Answers (4)

khmurach
khmurach

Reputation: 484

Yes, you can use Ajax and FormData API:

var file = s.fileInputEl.dom.files[0],
     data = new FormData();
data.append('file', file);
Ext.Ajax.request({
   url: '/upload/files',
   rawData: data,
   headers: {'Content-Type':null}, //to use content type of FormData
   success: function(response){
   }
});

See my demo here

Upvotes: 2

dev
dev

Reputation: 1387

ExtJs version 6.0.1 - Using iframe

Ext.define('xxx.yyy.UploadData', {
    extend : 'Ext.form.Panel',
    alias  : 'widget.uploaddata',

    initComponent : function(){        
        var me = this;        

        me.items = [{
          xtype      : 'filefield',
          margin     : '20 0 0 20',
          name       : 'excelfile',
          fieldLabel : 'Choose file',
          msgTarget  : 'side',
          allowBlank : false,
          anchor     : '30%',
          buttonText : 'Select',
          defaultButtonTarget : 'fileframe'
        },{
          xtype : 'panel',
          html  : '<iframe width="340" height="340" style="display: none" name="fileframe"></iframe>'
        },{
          xtype : 'button',
          text  : 'Import',
          handler : function(){
            var form = this.up('form').getForm();
            if(form.isValid()){
                form.submit({
                    url     : './upload.php',
                    waitMsg : 'uploading...',
                    success : function(fp, o) {
                        alert("OK");                            
                    }
                });
            }
        }
    }];

    me.callParent();        
    }    
   });

Upvotes: 0

Francis Ducharme
Francis Ducharme

Reputation: 4987

If you want to still use ExtJS's fileuploadfield and upload through an AJAX call using HTML5 FileReader, you can do it like such:

launchUpload: function () {
    //get a handle of the "file" input in the widget itself...
    var fileInput = document.getElementById(yourUploadField.button.fileInputEl.id);
    var fileReader = New FileReader();
    var fileToUpload = fileInput.files[0]; //assuming your only uploading one file...
    var me = this

    fileReader.onload = function (e) {
         me.onLoadFile(e, me, fileToUpload.name);
    }

    fileReader.readAsDataURL(fileToUpload);

}, 
onLoadFile: function (e, scope, filename) {

     //I carry the scope around for functionality...

     Ext.Ajax.request({
        method: 'POST',
        url: 'url',
        scope: scope,
        jsonData: { fileNameParameter: filename, fileDatainBase64: e.target.result},
        success: function (response, operation) {
            //success..
        },
        failure: function (response, operation) {
            //failure...
        }
    });       

}

Upvotes: 2

user2292083
user2292083

Reputation:

You can't do it with filefield of Extjs

filefield of Extjs return string url from select file.

I think you need the file selected as occurs in change event but filefield haven't this event

you can use this solution, maybe you get one idea from the solution

Example: http://jsfiddle.net/e3M3e/e8V7g/

var itemFile = null;
Ext.create('Ext.panel.Panel', {
    title: 'Hello',
    width: 400,
    html: "<input id='inputFile' type='file' name='uploaded'/>",
    renderTo: Ext.getBody(),
    listeners: {
        afterrender: function() {
            itemFile = document.getElementById("inputFile");            
            itemFile.addEventListener('change', EventChange, false);
        }
   }
});

function EventChange(e){    
    var files = itemFile.files;
    console.log(files);
}

Upvotes: 0

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