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I've been working on this all day now, and I just can't get it working.

I have basically got a simple ajax request using the jQuery library and I want to send the data which I post through a mutlipart/form-data file input, however, I have tried everything I can think of.

My File upload script is in place awaiting the file name as a parameter (tried without also), but it just doesn't want to get the data from the file input box itself.

Could someone please enlighten me on how to do this without another plugin (multiple upload, etc).

Here is my jQuery Code for this bit:

function uploadTimesheets(){

$('#waiting').show();

var error = '';

var msg = '';

//Performs the Ajax Request
 var data = $.ajax({
    type        :   'POST',
    url         :   '/ajax/timesheet/uploadNewTimesheets.php',
    dataType    :   'json',
    contentType :   'multipart/form-data',
    data        :   data,
    error       :   error,
    msg         :   msg,
    success     :   function(data){

        if(!data){
            $('#notification').removeClass(this).addClass('notification-success').html(data).show().delay(1200).fadeOut(800);
            getActiveTimesheets(getSelectedPage());
        }else{
            $('#notification').removeClass().addClass('notification-error').html(data.msg + data.errorList).show();
            alert('PHHAIL');
        }

        $('#waiting').hide();
        function(xhr, status, errorThrown){
            $('#waiting').hide();
        }
    }
});

}

And here is my PHP upload script:

    /**
 * Creates a directory in the active directory with the given folder name
 *
 * @author  RichardC
 * @param   string    $dirName
 * @return  boolean
 */
public function createDir( $dirName ) {

    $docRoot = getenv('DOCUMENT_ROOT');

    if (!is_dir(sprintf('%s/%s', $docRoot, $dirName))) {
        $makeDir = mkdir(sprintf('%s/%s', $docRoot, $dirName));
        echo sprintf('Creating a folder called \'/%s/\' ...', $dirName);
        if ($makeDir) {
            echo '<br />Successfully created the folder.<br />';
            return true;
        } else {
            echo sprintf('<br /> Sorry, please create the folder manually at: %s/%s', $docRoot, $dirName);
            return false;
        }
    }
}

/**
 * Uploads either a CSV or an EXCEL file to a temporary directory
 *
 * @author  RichardC
 * @param   Resource    $file
 * @return  Boolean     true/false
 */
public function upload( $filename ) {

    $filename = (!isset($filename)) ? $this->file : $filename;

    //Get the document root
    $docRoot = getenv('DOCUMENT_ROOT');

    $this->createDir('uploads');

    if (($_FILES['file']['type'] == 'application/vnd.ms-excel') || ($_FILES['file']['type'] == 'application/csv') || ($_FILES['file']['type'] == 'text/csv') || ($_FILES['file']['type'] == 'text/ma-separated-values') || ($_FILES['file']['type'] == 'application/excel') &&
            ($_FILES["file"]["size"] < 1000000)) {
        if ($_FILES["file"]["error"] > 0) {
            echo "Return Code: " . $_FILES["file"]["error"] . "<br />";
        } else {
            if (file_exists($docRoot . "upload/" . $_FILES["file"]["name"])) {
                echo $_FILES["file"]["name"] . " already exists. ";
                $this->file = $docRoot . '/upload/' . $_FILES["file"]["name"];
            } else {
                move_uploaded_file($_FILES["file"]["tmp_name"], $docRoot . "/upload/" . $_FILES["file"]["name"]);
                $this->file = $docRoot . '/upload/' . $_FILES["file"]["name"];
            }
        }
    } else {
        echo "Invalid file";

        return false;
    }

    //Remove the unwanted file now
    $this->fileContents = file_get_contents($this->file);
    @unlink($this->file);
    unset($this->file);

    return true;
}

If anyone can help on this, it'd be much appreciated!

I've been working on this all day now, and I just can't get it working.

I have basically got a simple ajax request using the jQuery library and I want to send the data which I post through a mutlipart/form-data file input, however, I have tried everything I can think of.

My File upload script is in place awaiting the file name as a parameter (tried without also), but it just doesn't want to get the data from the file input box itself.

Could someone please enlighten me on how to do this without another plugin (multiple upload, etc).

Here is my jQuery Code for this bit:

function uploadTimesheets(){

$('#waiting').show();

var error = '';

var msg = '';

//Performs the Ajax Request
 var data = $.ajax({
    type        :   'POST',
    url         :   '/ajax/timesheet/uploadNewTimesheets.php',
    dataType    :   'json',
    contentType :   'multipart/form-data',
    data        :   data,
    error       :   error,
    msg         :   msg,
    success     :   function(data){

        if(!data){
            $('#notification').removeClass(this).addClass('notification-success').html(data).show().delay(1200).fadeOut(800);
            getActiveTimesheets(getSelectedPage());
        }else{
            $('#notification').removeClass().addClass('notification-error').html(data.msg + data.errorList).show();
            alert('PHHAIL');
        }

        $('#waiting').hide();
        function(xhr, status, errorThrown){
            $('#waiting').hide();
        }
    }
});

}

And here is my PHP upload script:

    /**
 * Creates a directory in the active directory with the given folder name
 *
 * @author  RichardC
 * @param   string    $dirName
 * @return  boolean
 */
public function createDir( $dirName ) {

    $docRoot = getenv('DOCUMENT_ROOT');

    if (!is_dir(sprintf('%s/%s', $docRoot, $dirName))) {
        $makeDir = mkdir(sprintf('%s/%s', $docRoot, $dirName));
        echo sprintf('Creating a folder called \'/%s/\' ...', $dirName);
        if ($makeDir) {
            echo '<br />Successfully created the folder.<br />';
            return true;
        } else {
            echo sprintf('<br /> Sorry, please create the folder manually at: %s/%s', $docRoot, $dirName);
            return false;
        }
    }
}

/**
 * Uploads either a CSV or an EXCEL file to a temporary directory
 *
 * @author  RichardC
 * @param   Resource    $file
 * @return  Boolean     true/false
 */
public function upload( $filename ) {

    $filename = (!isset($filename)) ? $this->file : $filename;

    //Get the document root
    $docRoot = getenv('DOCUMENT_ROOT');

    $this->createDir('uploads');

    if (($_FILES['file']['type'] == 'application/vnd.ms-excel') || ($_FILES['file']['type'] == 'application/csv') || ($_FILES['file']['type'] == 'text/csv') || ($_FILES['file']['type'] == 'text/ma-separated-values') || ($_FILES['file']['type'] == 'application/excel') &&
            ($_FILES["file"]["size"] < 1000000)) {
        if ($_FILES["file"]["error"] > 0) {
            echo "Return Code: " . $_FILES["file"]["error"] . "<br />";
        } else {
            if (file_exists($docRoot . "upload/" . $_FILES["file"]["name"])) {
                echo $_FILES["file"]["name"] . " already exists. ";
                $this->file = $docRoot . '/upload/' . $_FILES["file"]["name"];
            } else {
                move_uploaded_file($_FILES["file"]["tmp_name"], $docRoot . "/upload/" . $_FILES["file"]["name"]);
                $this->file = $docRoot . '/upload/' . $_FILES["file"]["name"];
            }
        }
    } else {
        echo "Invalid file";

        return false;
    }

    //Remove the unwanted file now
    $this->fileContents = file_get_contents($this->file);
    @unlink($this->file);
    unset($this->file);

    return true;
}

If anyone can help on this, it'd be much appreciated!

• php
• javascript
• jquery
• ajax
• multipartform-data

Share Improve this question Follow asked Sep 22, 2011 at 14:24 DarkMantisDarkMantis 1,51655 gold badges1919 silver badges4040 bronze badges 1

• 1 I am not sure that will allow for file fields... however, there's a JQuery plugin that handles submitting forms with files: jquery.malsup./form. Another option is using a file upload plugin like uploadify.. – Michael C. Gates Commented Sep 22, 2011 at 14:29

Add a ment  | 

1 Answer 1

Sorted by: Reset to default 6

In order to make your multipart/formdata work, you must add some extra stuff in your ajax-request:

cache: false,
contentType: false,
processData: false,

You can easily create your data-field by doing this:

var uploadData = $("#uploadFile").prop("files")[0];
var newData = new FormData();

$.each($('#uploadFile').prop("files"), function(i, file) {
    newData.append('file-'+i, file);
});

in your ajax-request you'll have to set this:

data: newData

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