Sumon Bappi
Sumon Bappi

Reputation: 2019

save image to a folder under web-app in grails

I need to save image to a folder in my application. So far I have learned to save image into a database but I need to save it to a folder. How can I do this? Can anyone please help me on this please? here is my code below to save into database >>>

def upload={
  def user = User.findById(1)
    CommonsMultipartFile file = params.list("photo")?.getAt(0)
    user.avatar = file?.bytes
    user.save()
}

Upvotes: 1

Views: 4879

Answers (3)

Sumon Bappi
Sumon Bappi

Reputation: 2019

I have solve this so easily as follows. You will have to import followings :

import org.apache.commons.io.FileUtils

import org.springframework.web.multipart.commons.CommonsMultipartFile

import org.springframework.web.multipart.*

Good luck who needs this >>>

def saveImageToFolder = {
    String message = ""
    MultipartHttpServletRequest mpr = (MultipartHttpServletRequest)request;
    CommonsMultipartFile f = (CommonsMultipartFile) mpr.getFile("userPhoto")

    if(!f.empty) {
        def usr = User.findByUsername(1)
        if(!usr){
        User user = new User()
        user.username = params.username
        user.avatarType = f.getContentType()

        if(user.save()){
            def userId = user.id
            String username = user.username
            String fileName = username + "." + f.getContentType().substring(6)          // here my file type is image/jpeg
            byte[] userImage = f.getBytes()
            FileUtils.writeByteArrayToFile(new File( grailsApplication.config.images.location.toString() + File.separatorChar + fileName ), userImage )
            message = "User Created Successfully."
        }else{
            message = "Can not Create User !!!"
        }
        }else{
            message = "Username already exists. Please try another one !!!"
        }
    }
    else {
        message = 'file cannot be empty'
    }
    render(view: 'addUser', model:[message: message])
}

and in your config file paste this >>>

images.location = "web-app/images/userImages/"      // after web-app/folder name/folder name and go on if you want to add other folder

Upvotes: 0

Houcem Berrayana
Houcem Berrayana

Reputation: 3080

You have only to copy the MutipartFile into web-app folder. This is how :

MultipartHttpServletRequest mpr = (MultipartHttpServletRequest)request;
CommonsMultipartFile f = (CommonsMultipartFile) mpr.getFile("myfile");
String fileName = System.currentTimeMillis() + f.name
String destinationFileName = configService.getAbsoluteDocumentsPath() + fileName // We will put it on web-app/documents/xxxxx
f.renameTo(new File(destinationFileName))
//Save filename to database in 
user.avatar = fileName
user.save()

And in configService I have that (used to calculate paths)

class ConfigService {
def grailsApplication

     /**
     * @return absolute path of documents
     */
    def getAbsoluteDocumentsPath(){
        def asolutePath = grailsApplication.mainContext.servletContext.getRealPath('documents')
        return asolutePath.endsWith("/") ? asolutePath : asolutePath + "/"
    }
}

EDIT To make sure that your request is an instance of MutipartHttServletRequest Add the following test

if(request instanceof MultipartHttpServletRequest) {
//Do stuff here
}

Don't forget to check the the encoding of the form in which you put the file input.

Upvotes: 2

Saurabh Dixit
Saurabh Dixit

Reputation: 633

Find below for the step wise implementation, I have added a GSP page with the uploadForm(it will have multipart form submission by default), and then a controller function to handle file save request, and a service method to save file in a specified directory:

Step1: Create a form for file upload:

<g:uploadForm name="picUploadForm" class="well form-horizontal" controller="<your-controller-name>" action="savePicture">

    Select Picture: <input type="file" name="productPic"/>
    <button type="submit" class="btn btn-success"><g:message code="shopItem.btn.saveProductImage" default="Save Image" /></button>



</g:uploadForm>

Step2: Then in your controller's savePicture action:

String baseImageName = java.util.UUID.randomUUID().toString();
// Saving image in a folder assets/channelImage/, in the web-app, with the name: baseImageName
        def downloadedFile = request.getFile( "product.baseImage" )
        String fileUploaded = fileUploadService.uploadFile( downloadedFile, "${baseImageName}.jpg", "assets/channelImage/" )
        if( fileUploaded ){
            // DO further actions, for example make a db entry for the file name
        }

Step3: and in the file uploader service(User defined service with the name FileUploadService in this case):

def String uploadFile( MultipartFile file, String name, String destinationDirectory ) {

        def serveletContext = ServletContextHolder.servletContext
        def storagePath = serveletContext.getRealPath( destinationDirectory )

        def storagePathDirectory = new File( storagePath )

        if( !storagePathDirectory.exists() ){
            println("creating directory ${storagePath}")
            if(storagePathDirectory.mkdirs()){
                println "SUCCESS"
            }else{
                println "FAILED"
            }
        }

        // Store file

        if(!file.isEmpty()){
            file.transferTo( new File("${storagePath}/${name}") )
            println("Saved File: ${storagePath}/${name}")
            return "${storagePath}/${name}"
        }else{
            println "File: ${file.inspect()} was empty"
            return null
        }
    }

Upvotes: 4

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