show error next to input field - php

Question

I would like to show the errors next to input field. I tried echoing the error but it just displays the word Array. How can I show the error next to input field? Also I would like to let the user upload 5 photos and only the first photo is required. I mean the user can upload 5 photos but all the 5 photos aren't required only one. is that possible? thanks
<?php
$out['error'][]=''; //this is what I added
function uploadFile ($file_field = null, $check_image = false, $random_name = false) {
//Config Section
//Set file upload path
$path = 'productpic/'; //with trailing slash
//Set max file size in bytes
$max_size = 2097152;
//Set default file extension whitelist
$whitelist_ext = array('jpg','png','gif');
//Set default file type whitelist
$whitelist_type = array('image/jpeg', 'image/png','image/gif');
//The Validation
// Create an array to hold any output
$out = array('error'=>null);
if (!$file_field) {
$out['error'][] = "Please specify a valid form field name";
}
if (!$path) {
$out['error'][] = "Please specify a valid upload path";
}
if (count($out['error'])>0) {
return $out;
}
//Make sure that there is a file
if((!empty($_FILES[$file_field])) && ($_FILES[$file_field]['error'] == 0)) {
// Get filename
$file_info = pathinfo($_FILES[$file_field]['name']);
$name = $file_info['filename'];
$ext = $file_info['extension'];
//Check file has the right extension
if (!in_array($ext, $whitelist_ext)) {
$out['error'][] = "Invalid file Extension";
}
//Check that the file is of the right type
if (!in_array($_FILES[$file_field]["type"], $whitelist_type)) {
$out['error'][] = "Invalid file Type";
}
//Check that the file is not too big
if ($_FILES[$file_field]["size"] > $max_size) {
$out['error'][] = "We are sorry, the image must be less than 2MB";
}
//If $check image is set as true
if ($check_image) {
if (!getimagesize($_FILES[$file_field]['tmp_name'])) {
$out['error'][] = "The file you trying to upload is not an Image, we only accept images";
}
}
//Create full filename including path
if ($random_name) {
// Generate random filename
$tmp = str_replace(array('.',' '), array('',''), microtime());
if (!$tmp || $tmp == '') {
$out['error'][] = "File must have a name";
}
$newname = $tmp.'.'.$ext;
} else {
$newname = $name.'.'.$ext;
}
//Check if file already exists on server
if (file_exists($path.$newname)) {
$out['error'][] = "The image you trying to upload already exists, please upload only once";
}
if (count($out['error'])>0) {
//The file has not correctly validated
return $out;
}
if (move_uploaded_file($_FILES[$file_field]['tmp_name'], $path.$newname)) {
//Success
$out['filepath'] = $path;
$out['filename'] = $newname;
return $out;
} else {
$out['error'][] = "Server Error!";
}
} else {
$out['error'][] = "Please select a photo";
return $out;
}
}
?>
<?php
if (isset($_POST['submit'])) {
$file = uploadFile('file', true, false);
if (is_array($file['error'])) {
$message = '';
foreach ($file['error'] as $msg) {
$message .= '<p>'.$msg.'</p>';
}
} else {
$message = "File uploaded successfully";
$sub=1;
}
echo $message;
}
?>
<form action="" method="post" enctype="multipart/form-data" name="form1" id="form1">
<?php
ini_set( "display_errors", 0);
if($sub==0)
{
?>
<input name="file" type="file" size="20" /><span><?php echo $out['error'] ;?></span> //It displays here just the word Array
<input name="submit" type="submit" value="Upload" />
<?php
}
?>
</form>

Answer 1

Because $out['error'] is an array, echoing it will output Array, as you noticed. To output it as a string you'll need to convert it first; one option to do so is using implode. I'd suggest using <br> as the 'glue' so that each error will show on a different line.
So,
<input name="file" type="file" size="20" /><span><?php echo implode('<br>', $out['error']) ;?></span>
<input name="submit" type="submit" value="Upload" />
However, perhaps a better solution would be to store the error as a string ($out['error'] = 'There was an error' rather than $out['error'][] = 'There was an error') and using a proper control structure to ensure that once an error is found the validation check ends and the form with the error message is output.
For the control structure you could do:
if ($first_check)
{
$out['error'] = 'First error message';
}
elseif ($second_check)
{
$out['error'] = 'Second error message';
}
else
{
$out['success'] = 'Success message';
}