display error messages next to input field

Question

Can anybody help me to display the error messages next to the input field? I tried by echoing as below but it doesn't work... All I get next to the input is the word Array. Any help is appreciated.

<?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>

MC Emperor · Accepted Answer · 2013-02-10 23:05:06Z

2

Use

foreach ($out['error'] as $e) {
    echo $e;
}

instead of echo $out['error'].

You might also want to either cast it as an array, or make sure that $out is not empty before doing a foreach on it.

edited Feb 10, 2013 at 23:05

MC Emperor

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answered Feb 10, 2013 at 22:54

Ethan

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1 Comment

alte Over a year ago

Tried as you said but it still doesn't display the errors next to the input.

Maykonn · Accepted Answer · 2013-02-10 23:14:37Z

0

$sub never be equals to 0:

$sub=0;
if (is_array($file['error'])) {
    $message = '';
    foreach ($file['error'] as $msg) {
        $message .= '<p>'.$msg.'</p>';    
    }
} else {
    $message = "File uploaded successfully";
    $sub=1;
}

answered Feb 10, 2013 at 23:14

Maykonn

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Comments

alte · Accepted Answer · 2013-02-10 23:19:43Z

0

I solved my problem by moving the second line of php code next to input field and it works fine.. I did as below..

<input name="file" type="file" size="40" /><span><?php

if (isset($_POST['submit'])) {

$file = uploadFile('file', true, false);
if (is_array($file['error'])) {
$message = '';
foreach ($file['error'] as $msg) {
  $message .= '<span>'.$msg.'</span>';    
}
} else {
$message = "File uploaded successfully";
$sub=1;
}
echo $message;
}
?></span>

<input name="submit" type="submit" value="Upload" />

answered Feb 10, 2013 at 23:19

alte

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