Notice: Undefined index: image - unable to find the error - php

Question

here is my error.i dont know why it is not working.i checked all the parts but i was unable to find the error.
Notice: Undefined index: image in C:\xampp\htdocs\Final\places\Ressave.php on line 27
here is my html code that pass the name of input:
<form class="form-horizontal" action="Ressave.php" method="POST" autocomplete="on">
<div class="well">
<legend>Photos</legend>
<div class="control-group">
<label class="control-label">Upload Photo: </label>
<div class="controls">
<input name="image" type="file" />
</div>
</div>
</div>
<div class="form-actions">
<button type="submit" class="btn btn-primary">Submit</button>
<button type="button" class="btn">Cancel</button>
</div>
</form>
Ressave.php is here and can not recieve the name here,so the error occure.....
<?php
{ // Secure Connection Script
include('../Secure/dbConfig.php');
$dbSuccess = false;
$dbConnected = mysql_connect($db['hostname'],$db['username'],$db['password']);
if ($dbConnected) {
$dbSelected = mysql_select_db($db['database'],$dbConnected);
if ($dbSelected) {
$dbSuccess = true;
} else {
echo "DB Selection FAILed";
}
} else {
echo "MySQL Connection FAILed";
}
// END Secure Connection Script
}
if(! $dbConnected )
{
die('Could not connect: ' . mysql_error());
}
{ // File Properties
$file = $_FILES['image']['tmp_name']; //Error comes from here(here is the prob!)
if(!isset($file))
echo "Please Choose an Image.";
else {
$image = addslashes(file_get_contents($_FILES['image']['tmp_name']));
$image_size = getimagesize($_FILES['image']['tmp_name']);
if($image_size == FALSE)
echo "That is not an image.";
else
{
$lastid = mysql_insert_id();
echo "Image Uploaded.";
}
}
}
{ //join the post values into comma separated
$features = mysql_real_escape_string(implode(',', $_POST['features']));
$parking = mysql_real_escape_string(implode(',', $_POST['parking']));
$noise = mysql_real_escape_string(implode(',', $_POST['noise']));
$good_for = mysql_real_escape_string(implode(',', $_POST['good_for']));
$ambience = mysql_real_escape_string(implode(',', $_POST['ambience']));
$alcohol = mysql_real_escape_string(implode(',', $_POST['alcohol']));
}
$sql = "INSERT INTO prestaurant ( ResName, Rating, Food_serve, Features, Parking, noise, Good_For, Ambience, Alcohol, Addition_info, Name, Address1, Zipcode1, Address2, Zipcode2, Address3, Zipcode3, City, Mobile, phone1, phone2, phone3, phone4, Email1, Email2, Fax, Website, image)".
"VALUES ('$_POST[restaurant_name]','$_POST[star_rating]','$_POST[food_served]','$features','$parking','$noise','$good_for','$ambience','$alcohol','$_POST[description]','$_POST[name]','$_POST[address1]','$_POST[zipcode1]','$_POST[address2]','$_POST[zipcode2]','$_POST[address3]','$_POST[zipcode3]','$_POST[city]','$_POST[mobile]','$_POST[phone1]','$_POST[phone2]','$_POST[phone3]','$_POST[phone4]','$_POST[email1]','$_POST[email2]','$_POST[fax]','$_POST[url]','$image')";
mysql_select_db('place');
$retval = mysql_query( $sql );
if(! $retval )
{
die('Could not enter data: ' . mysql_error());
}
echo "Entered data successfully\n";
mysql_close($dbConnected);
?>

Answer 1

If a file was not uploaded the $_FILES array will be empty. Specifically, if the file image was not uploaded, $_FILES['image'] will not be set.
So
$file = $_FILES['image']['tmp_name']; //Error comes from here(here is the prob!)
should be:
if(empty($_FILES) || !isset($_FILES['image']))
update
You will also have issues because you're missing the enctype attribute on your form:
<form class="form-horizontal" action="Ressave.php" method="POST" autocomplete="on" enctype="multipart/form-data">

Answer 2

In order to be able to process files in your form you need to add the enctype attribute.
<form method='POST' enctype='multipart/form-data' >

Answer 3

Hey i guess you have forgotten one important setting in form
enctype="multipart/form-data" this option is used when used with files eg image file etc
<form name="image" method="post" enctype="multipart/form-data">
Apart from that to extract the contents of the file you can use following options
$tmp_img_path = $_FILES['image']['tmp_name'];
$img_name = $_FILES['image']['name'];
to print all the content of a file use:
print_r($_POST['image']);

Answer 4

this is the code i have which successfully insert the data into mysql and retrieve from the DB.
"Index.PHP"
<html>
<head>
<title>PHP & MySQL: Upload an image</title>
</head>
<body>
<form action="index.php" method="POST" enctype="multipart/form-data">
File: <input type="file" name="image" /><input type="submit" value="Upload" />
</form>
<?php
mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("registrations") or die(mysql_error());
if(!isset($_FILES['image']))
{
echo 'Please select an image.';
}
else {
$image = addslashes(file_get_contents($_FILES['image']['tmp_name']));
echo $_FILES['image']['tmp_name'];
$image_name = addslashes($_FILES['image']['name']);
$image_size = getimagesize($_FILES['image']['tmp_name']);
if($image_size==FALSE){
echo "That's not an image.";
} else {
if(!$insert = mysql_query("INSERT INTO test_image VALUES ('','$image_name','$image')"))
{
echo "Problem uploading image.";
} else {
$lastid = mysql_insert_id();
echo "Image uploaded.<p />Your image:<p /><img src=get.php?id=$lastid>";
}
}
}
?>
</body>
</html>
this is get.PHP
<?php
mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("registrations") or die(mysql_error());
$id = addslashes($_REQUEST['id']);
$image = mysql_query("SELECT * FROM test_image WHERE id=$id");
$image = mysql_fetch_assoc($image);
$image = $image['image'];
header("Content-type: image/jpeg");
echo $image;
?>
NOTE:- please changes your DB and table name accordingly..
Thanks,
Santanu