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php upload multiple images using -'php,html'


php upload multiple images  using -'php,html'

I need to upload multiple images via form. I thought that I will do it with no problem, but I have one.

When I try to do foreach and get image by image it is not acting like I hoped it will.

HTML

<form method="post" action="" enctype="multipart/form-data" id="frmImgUpload">
    <input name="fileImage[]" type="file" multiple="true" />
    <br />
    <input name="btnSubmit" type="submit" value="Upload" />
</form>


PHP

<?php
if ($_POST)
{
    echo "<pre>";
    foreach ($_FILES['fileImage'] as $file)
    {
        print_r($file);
        die(); // I want it to print first image content and then die to test this out...
        //imgUpload($file) - I already have working function that uploads one image
    }
}


What I expected from it to print out first image, instead it prints names of all the images.

Example

Array
(
    [0] => 002.jpg
    [1] => 003.jpg
    [2] => 004.jpg
    [3] => 005.jpg
)


What I want it to output

Array
(
    [name] => 002.jpg
    [type] => image/jpeg
    [tmp_name] => php68A5.tmp
    [error] => 0
    [size] => 359227
)


So how can I select image by image in the loop so I can upload them all?

Okey I found solution and this is how I did it, probably not the best way but it works.

foreach ($_FILES['fileImage']['name'] as $f)
{
    $file['name'] = $_FILES['fileImage']['name'][$i];
    $file['type'] = $_FILES['fileImage']['type'][$i];
    $file['tmp_name'] = $_FILES['fileImage']['tmp_name'][$i];
    $file['error'] = $_FILES['fileImage']['error'][$i];
    $file['size'] = $_FILES['fileImage']['size'][$i];
    imgUpload($file);
    $i++;
}

    

asked Sep 8, 2015 by rajesh
0 votes
7 views



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