Question

So I am trying to do, what I thought, was a simple file read and import. So the way my code works now is that I have a file stored in a folder on the server, select it, and import it. I select the file through a multipartfile:

    <g:form method="post"  enctype="multipart/form-data" controller="fileManager">
        <label>Company Name:</label><input id="cname" type="text" name="cname" /><br />
        <label>Select a CSV file to upload:</label> <input type="file" id="newFile" name="newFile" /> 
        <g:actionSubmit action="importContactsFromFile" value="Import" />
    </g:form>

So what I want to do is to:

• Have user select a file
• Download the file to a directory on the server (input/output stream?)
• Read the file contents (CSVReader) and import into DB (this works)
• Possibly clean up file after

So I have a massive headache with that second step. I pull the file into a def variable:

def newFile = request.getFile('newFile')

How do I bridge this gap? Slamming my head over this! Thanks!

Answer 1

Do you need to save the file as an intermediate step before parsing it? Most CSV parsers (OpenCSV for example) will read from a Reader or InputStream directly. So you can do something like:

def newFile = request.getFile('newFile')
def csv = new CSVReader(newFile.inputStream.newReader())
...

If you want to save the file, you can use the transferTo method:

def newFile = request.getFile('newFile')
newFile.transferTo(new File('newFile.csv'))
def csv = new CSVReader(new File('newFile.csv').newReader())

In that case, you’ll have to ensure two requests don’t overwrite each other by ensuring unique a file name.

You can find more information in the grails manual: http://grails.org/doc/latest/guide/theWebLayer.html#6.1.9%20Uploading%20Files