How to upload image to database in php? - php

here i am trying to upload image to directory and add the path in database.
here i am first adding some product details and trying to upload the image to uploads directory and add the uploaded image path in database.
here is what i have done:
<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "root", "", "wan_products_box");
// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
// Escape user inputs for security
$product_name = mysqli_real_escape_string($link, $_POST['product_name']);
$product_category = mysqli_real_escape_string($link, $_POST['product_category']);
$product_price = mysqli_real_escape_string($link, $_POST['product_price']);
$pro_url = mysqli_real_escape_string($link, $_POST['pro_url']);
$co_owners = mysqli_real_escape_string($link, $_POST['co_owners']);
// attempt insert query execution
$sql = "INSERT INTO product_list (product_name, product_category, product_price,product_referrence_URL,product_co_owners) VALUES ('$product_name', '$product_category', '$product_price', '$pro_url', '$co_owners')";
if(mysqli_query($link, $sql)){
echo "New product created.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
//image upload code
if($_POST)
{
if($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br />";
}
else
{
if(file_exists("uploaded_images/" . $_FILES["file"]["name"]))
{
echo $_FILES["file"]["name"] . " already exists. ";
}
else
{
if(move_uploaded_file($_FILES["file"]["tmp_name"],"uploaded_images/" . $_FILES["file"]["name"]))
{
$query_image = "INSERT INTO product_list (product_image_url) values ('".$_FILES['file']['name']."')";
if(mysql_query($query_image))
{
echo "Stored in: " . "uploaded_images/" . $_FILES["file"]["name"];
}
else
{
echo 'Unable to store';
}
}
}
}
}
mysqli_close($link);
?>
this is my html form:
<div id="menu2" class="tab-pane fade">
<h4>Add new product</h4>
<div class="col-xs-12 col-sm-12 col-md-12 col-lg-12">
<div class="panel panel-default">
<div class="panel-body">
<form class="form-horizontal" method="post" action="files/insert.php" role="form">
<div class="form-group">
<label class="control-label col-sm-2" for="product_name">Product Name</label>
<div class="col-sm-10">
<input type="text" class="form-control" name="product_name" id="product_name" placeholder="Iphone 5c" required>
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="pwd">Product Category</label>
<div class="col-sm-10">
<select class="btn-btn-primary form-control" name="product_category">
<option>Mobile</option>
<option>Television</option>
<option>Printer</option>
<option>Watch</option>
<option>Monitor</option>
</select>
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="product-pic">Upload your profile picture</label>
<div class="col-sm-10">
<input type="file" class="form-control" name="file" id="file" />
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="product_price">Product Price</label>
<div class="col-sm-10">
<input type="text" class="form-control" name="product_price" id="product_price" placeholder="Rs.36,000" required>
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="pro_url">Reference URL</label>
<div class="col-sm-10">
<input type="URL" class="form-control" name="pro_url" id="pro_url" placeholder="http://www.amazon.com" required>
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="co_owners">Co-owners</label>
<div class="col-sm-10">
<select class="btn-btn-primary form-control" name="co_owners">
<option>Select no. owners</option>
<option>1</option>
<option>2</option>
<option>3</option>
<option>4</option>
<option>5</option>
<option>6</option>
<option>7</option>
<option>8</option>
<option>9</option>
<option>10</option>
<option>11</option>
<option>12</option>
<option>13</option>
<option>14</option>
<option>15</option>
</select>
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-2 col-sm-10">
<button type="submit" class="btn btn-success">Add product</button>
</div>
</div>
</form>
</div>
</div>
</div>
</div>
the data is inserted to database but the image upload part is not happening...
how can i do this? how can i modify the code to upload the image to database and add path to database?
when i run this the output i get is
New product created.
Notice: Undefined index: file in D:\xampp\htdocs\wan\files\insert.php on line 31
Notice: Undefined index: file in D:\xampp\htdocs\wan\files\insert.php on line 37
Notice: Undefined index: file in D:\xampp\htdocs\wan\files\insert.php on line 43
Notice: Undefined index: file in D:\xampp\htdocs\wan\files\insert.php on line

There are a few things wrong in your code.
Firstly, your form is missing a valid enctype enctype="multipart/form-data" it is required when dealing with files.
http://php.net/manual/en/features.file-upload.post-method.php
You're also mixing MySQL APIs in your second query.
$query_image = "INSERT INTO product_list (product_image_url) values ('".$_FILES['file']['name']."')";
if(mysql_query($query_image))
So it will fail.
Use the same method you used in your first query.
Those different APIs do not intermix.
So change that to if(mysqli_query($link, $query_image))
Also add or die(mysqli_error($link)) to mysqli_query() to check for errors.
Also make sure the folder has proper permissions to be written to.
Another thing, your <select>'s options have no values
<option>Select no. owners</option>
<option>1</option>
...
You need to add those
<option value="empty_value">Select no. owners</option>
<option value="1">1</option>
...
and do the same for the others.
Same thing for <option>Mobile</option>. There should be values for those too.
<option value="mobile">Mobile</option>
...
You will not get anything entered in your database for those.

Apart from the problems in HTML part like the lack of enctype="multipart/form-data". There is a problem in your second Query and this is the cause of these errors:
$query_image = "INSERT INTO product_list (product_image_url) values ('".$_FILES['file']['name']."')";
You need to change this to an UPDATE query and update the already created row in database.
Also it is a better approach to use the row's index number for the image name.

Related

how do I foreach through html input form and insert multiple rows or one based on a selected date field?

how do I foreach through html input form and insert multiple rows or one based on a selected date field? in other words when a user enters "name" "description" and "shift" and then selects either one date or more then one. PHP will then enter the same information for either one new row or multiples based on how many dates were selected.
<?php
if(isset($_REQUEST['submit']))
{
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "xxxx", "xxxx", "xxxx");
// Check connection
if($link === false){
die("| ERROR: Could not connect. " . mysqli_connect_error());
}
$name = mysqli_real_escape_string($link, $_REQUEST['name']);
$desc = mysqli_real_escape_string($link, $_REQUEST['description']);
$shift = mysqli_real_escape_string($link, $_REQUEST['shift']);
$date = mysqli_real_escape_string($link, $_REQUEST['daterange']);
$sql = "insert into db (name,description,shift,evdate) values ('$name', ' $desc','$shift','$date')";
$sql2 = "insert into db (name,description,shift,evdate) values ('$name', ' '$desc','$shift','$insert')";
if ($date=0) {
$result = mysqli_query($link, $sql);
}else{
$daterange = explode(',',$date);
foreach($daterange as $insert) {
$result = mysqli_query($link, $sql2);
}
}
if(mysqli_query($link, $sql)){
echo "";
} else{
echo "| ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
if ($link->multi_query($sql) === TRUE) {
echo "It Worked..... Maybe!!!!!!";
} else {
echo "Error: " . $sql . "<br>" . $link->error;
}
}
$link->close();
?>
<form action="test_insert.php" method="post">
<div class="col col-lg-2 col-lg-offset-0">
<div class="form-group col-lg-offset-0 col-lg-12">
<label for="Name">Employee Name:</label>
<input type="text" name="name" placeholder="First & Last Name" id="name" required>
<p class="help-block col-lg-12">First and Last Name Please.</p>
</div>
</div>
<div class="col col-lg-offset-0 col-lg-2">
<div class="form-group col-lg-12">
<label for="description">Description:</label>
<input type="text" name="description" id="description" placeholder="description..." required>
<p class="help-block">For Example: "Vacation Full Day" or "PTO 2 Hours." </p>
</div>
</div>
<div class="col col-lg-offset-0 col-lg-3">
<label for="shift">Shift:</label><br>
<input type="radio" name="shift" value="First Shift" id="shift" checked> First Shift |
<input type="radio" name="shift" value="Second Shift" id="shift"> Second Shift |
<input type="radio" name="shift" value="Third Shift" id="shift"> Third Shift
<p class="help-block">Select Correct Shift Worked.</p>
</div>
<div class="col col-lg-offset-0 col-lg-3">
<div class="form-group col-lg-10">
<label for="date2">Date/Dates:</label>
<input type="text" id="datepicker1" name="daterange" placeholder="Select Your Date" />
</div>
<div class="form-group col-lg-10">
<label for="date2">Date/Dates:</label>
<input type="text" id="datepicker2" name="daterange" placeholder="Select Your Date" />
</div>
<div class="form-group col-lg-10">
<label for="date2">Date/Dates:</label>
<input type="text" id="datepicker3" name="daterange" placeholder="Select Your Date" />
</div>
<div class="form-group col-lg-10">
<label for="date2">Date/Dates:</label>
<input type="text" id="datepicker4" name="daterange" placeholder="Select Your Date" />
</div>
<div class="form-group col-lg-10">
<label for="date2">Date/Dates:</label>
<input type="text" id="datepicker5" name="daterange" placeholder="Select Your Date" />
</div>
<div class="form-group col-lg-6">
<input type="submit" name="submit" class= "btn btn-primary">
</div>
</div>
</div>
</form>
Think the best way is to use AJAX,
Then with the response as a string you make a table or foreach in php as a string and then use the .html adapter to output the newly made data.
function submitForm(form){
var url = form.attr("action");
var formData = {};
$(form).find("input[name]").each(function (index, node) {
formData[node.name] = node.value;
});
$.post(url, formData).done(function (data) {
$('#showresults').html(result);
});
}

Why do I keep getting Notice: Array to string conversion?

I am trying to update a mysql database but every time I get the Notice: Array to string conversion. I really can't figure out what I'm doing wrong here. Can someone kindly help?
My form
<div class="panel-body">
<form role="form" method="post" action="client_post.php" class="form-horizontal">
<fieldset>
<div class="form-group">
<label class="col-sm-3 control-label">Hospital No:
</label>
<div class="col-sm-5">
<input required class="form-control" name="Hospital_no" placeholder="Patients' Hospital Number">
</div>
</div>
<div class="form-group">
<label class="col-sm-3 control-label">Date of New Status</label>
<div class="form-group">
<label class="col-sm-3 control-label">New Status</label>
<div class="col-sm-5">
<select required name ="art_status" class="form-control">
<option></option>
<option value = "ART Restart" name="ART Restart">ART Restart</option>
<option value = "ART Transfer Out" name="ART Transfer Out" >ART Transfer Out</option>
<option value = "Pre ART Transfer Out" name="Pre ART Transfer Out">Pre ART Transfer Out</option>
<option value = "Lost to Followup" name="Lost to Followup">Lost to Followup</option>
<option value = "Stopped Treatment" name="Stopped Treatment">Stopped Treatment</option>
<option value = "Known Death" name="Known Death">Known Death</option>
</select>
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-3 col-sm-5">
<input class="btn btn-primary" type="submit" value="Update" name="submit" >
</div>
</div>
</fieldset>
</form>
I'm trying to make the update from this page (client.php)
if($_SERVER['REQUEST_METHOD'] === 'POST')
{
$Hosp_no = ['Hospital_no'];
$art_status = ['art_status'];
$art_date = ['art_start_date'];
$update_client = "UPDATE `pat_reg` SET `art_status` = '$art_status', `art_start_date` = '$art_date' WHERE `Hospital_no` = '$Hosp_no'";
if (mysqli_query($dbcon,$update_client))
{
echo"<script>alert('Record Updated')</script>";
//echo"<script>window.open('client_update.php','_self')</script>";
}
$dbcon->close();
}
Please help.
Using the $_POST superglobal allows you to access the values submitted in the form.
Change the following:
$Hosp_no = ['Hospital_no'];
$art_status = ['art_status'];
$art_date = ['art_start_date'];
to this:
$Hosp_no = $_POST['Hospital_no'];
$art_status = $_POST['art_status'];
$art_date = $_POST['art_start_date'];
PS. You should be escaping the user's input before querying the database (or, ideally, use prepared statements) or you may find yourself being hacked very quickly if released to the public.

Picture not being stored inside mysql database properly

A value is being stored in the database but it is not the one i am expecting. I tried many methods before but this one kind of seems to work but the file name is not being stored and when I try to download the file directly from the database, it downloads a .bin file format which looks something like table_Name-column_Name.bin. The file name being stored is BLOB - ## B.
My Form
<form class="form-horizontal" method="post" action="productsValidate.php" name="myForm" enctype="multipart/form-data">
<fieldset>
<legend>Add Product</legend>
<div class="form-group">
<label for="Product_Name" class="col-lg-2 control-label">Product Name</label>
<div class="col-lg-10">
<input type="text" class="form-control" id="Product_Name" placeholder="Name" required="required" name="Product_Name">
</div>
</div>
<div class="form-group">
<label for="Size" class="col-lg-2 control-label">Size</label>
<div class="col-lg-10">
<input type="text" class="form-control" id="Size" placeholder="Size" required="required" name="Size">
</div>
</div>
<div class="form-group">
<label for="Color" class="col-lg-2 control-label">Color</label>
<div class="col-lg-10">
<input type="text" class="form-control" id="Color" placeholder="Size" required="required" name="Color">
</div>
</div>
<div class="form-group">
<label for="price" class="col-lg-2 control-label">Price</label>
<div class="col-lg-10">
<input type="number" class="form-control" id="price" placeholder="price" required="required" name="price">
</div>
</div>
<div class="form-group">
<label for="image" class="col-lg-2 control-label">Select Image</label>
<div class="col-lg-10">
<input type="file" name="image" id="image">
</div>
</div>
<div class="form-group">
<label for="categoryId" class="col-lg-2 control-label">Category Id</label>
<div class="col-lg-10">
<?php
//your connection to the db and query would go here
include "../include/settings.php";
$conn = new mysqli($host, $user, $pwd, $sql_db);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT distinct Category_Id FROM products";
$result = mysqli_query($conn, $sql);
?>
<select id="categoryId" name="categoryId">
<option value = ""></option>
<?php
while($row = mysqli_fetch_array($result)) {
echo '<option value='.$row['Category_Id'].'>'.$row['Category_Id'].'</option>';
}
?>
</select>
</div>
</div>
<div class="form-group">
<label for="description" class="col-lg-2 control-label">Description</label>
<div class="col-lg-10">
<textarea type="text" class="form-control" id="description" placeholder="Description" required="required" name="description" pattern="[\sA-Za-z]+"></textarea>
</div>
</div>
<div class="form-group">
<div class="col-lg-6 col-lg-offset-2">
<button type="submit" class="btn btn-primary">Add Product</button>
</div>
</div>
</fieldset>
</form>
My Form validation
<?php
$name = $_POST["Product_Name"];
$size = $_POST["Size"];
$color = $_POST["Color"];
$price = $_POST["price"];
$image = addslashes($_FILES['image']['tmp_name']);
$image = file_get_contents($image);
$image = base64_encode($image);
$image=basename( $_FILES["image"]["tmp_name"],".jpg");
$category = $_POST['categoryId'];
$description = $_POST['description'];
insertProduct($name, $size, $color, $price, $image, $category, $description);
function insertProduct($name, $size, $color, $price, $image, $category, $description){
require_once ("../include/settings.php"); // Load MySQL log in credentials
$conn = #mysqli_connect ($host,$user,$pwd,$sql_db); // Log in and use database
if ($conn) { // check is database is avialable for use
$query = "INSERT INTO products
(Product_Id, Product_Name, Size, Color, Price, Picture, Category_Id, Description)
VALUES ('', '$name', '$size', '$color', '$price', '$image', '$category', '$description')";
$result = mysqli_query ($conn, $query);
if ($result) { // check if query was successfully executed
echo 'Successfully Added';
} else {
echo 'Product could not be added';
}
mysqli_close ($conn); // Close the database connect
} else {
echo "<p>Unable to connect to our database for adding the product.</p>";
}
}
?>
I guess you're trying to store the actual encoded image in the database, not a pointer to it. It looks to me like your eleven-byte BLOB has the pointer in it instead.
Your code contains this sequence of lines.
$image = addslashes($_FILES['image']['tmp_name']);
$image = file_get_contents($image);
$image = base64_encode($image);
$image=basename( $_FILES["image"]["tmp_name"],".jpg");
The third line puts an encoded, not binary, version of the image into a text string. That's close to what you want, but you probably should not base64-encode it if you're putting in a BLOB.
The fourth line discards the image itself and overwrites it with an image name. I think that's wrong.
If you're going to use BLOB data this way, you also need to use mysqli's facilities to prepare your SQL statements, and then bind your parameters. The bind_param() function gives you a way to declare a parameter to be a blob. That's better than trying to trick php's string processing into accepting it.
All that being said, most people use a file system or content server rather than BLOBs to store and serve images to web clients. BLOB programming is a pain in the neck. Also, using a DBMS to store and retrieve images quickly becomes a performance bottleneck in an application that scales up.

An empty row getting inserted in database

Hey I am trying to get this code running for the past few days now. I do not know what is the problem. Whenever I run the code I can see it running but an empty row gets inserted. Basically I ave tried to hard code the data and the data gets inserted. Here is the HTML form:
<form action="register.php" id="contactForm" type="post">
<div class="row">
<div class="form-group">
<div class="col-md-6">
<label>First name *</label>
<input type="text" class="form-control" name="fname" >
</div>
<div class="col-md-6">
<label>Last name *</label>
<input type="text" class="form-control" name="lname" >
</div>
</div>
</div>
<div class="row">
<div class="form-group">
<div class="col-md-6">
<label>Gender *</label><br>
<select name="gender">
<option> Male </option>
<option> Female </option>
</select>
</div>
<div class="col-md-6">
<label>Stream *</label><br>
<select name="stream">
<option> B-Tech </option>
<option> M-Tech </option>
</select>
</div>
</div>
</div>
<div class="row">
<div class="form-group">
<div class="col-md-6">
<label>Email *</label>
<input type="text" class="form-control" name="email" >
</div>
<div class="col-md-6">
<label>Mobile *</label>
<input type="text" class="form-control" name="mobile">
</div>
</div>
</div>
<div class="row">
<div class="form-group">
<div class="col-md-6">
<label>College *</label>
<input type="text" class="form-control" name="college" >
</div>
<div class="col-md-6">
<label>Job Kind *</label>
<input type="text" class="form-control" name="job" >
</div>
</div>
</div>
<div class="row">
<div class="col-md-12">
&nbsp&nbsp&nbsp&nbsp
<input type="submit" value="Register" class="btn btn-primary btn-lg"
data-loading-text="Loading..." name="submit">
</div>
</div>
</form>
Here is the registration.php
<?php
$connection = mysql_connect("EDITED by billy, was an I.P and port number", "user", "password"); // Establishing Connection with Server
$db = mysql_select_db("Registrations_connect", $connection); // Selecting Database from Server
$first_name = $_POST["fname"];
$last_name = $_POST["lname"];
$sex = $_POST["gender"];
$field = $_POST["stream"];
$contact = $_POST["mobile"];
$eaddress = $_POST["email"];
$institute = $_POST["college"];
$naukri = $_POST["job"];
$query = mysql_query("insert into students(fname, lname, gender, stream, mobile, email, college, job)
values ('$name', '$last_name', '$sex', '$field','$contact', '$eaddress', '$intitute', '$naukri')");
echo "<br/><br/><span>Data Inserted successfully...!!</span>";
mysql_close($connection); // Closing Connection with Server
?>
After running; In the inspect element I checked the response:- It shows Data Inserted successfully but actually an empty row is getting inserted. Basically what i think I am not able to correctly grab the data properly from form. Can somebody please check what is the problem. It will be a great help.
The attribute is method, not type. This typo is causing your form to process a GET rather than a POST. So all your variable assignments are wrong.
$first_name = $_POST["fname"];
would be
$first_name = $_GET["fname"];
or you could use the $_REQUEST; or you can just correct the attribute,
<form action="register.php" id="contactForm" method="post">
Your code also is wide open to SQL injections and is using the deprecated mysql_ functions. You should update to mysqli or pdo and be using prepared statements with parameterized queries.
More on SQL injections:
http://php.net/manual/en/security.database.sql-injection.phpHow can I prevent SQL injection in PHP?https://www.owasp.org/index.php/SQL_Injection_Prevention_Cheat_Sheet#Defense_Option_1:_Prepared_Statements_.28Parameterized_Queries.29

Uploaded Image file extension not getting stored in database

here i am trying to upload image to directory and add the path in database.
when i fill form and click submit the image gets uploaded to directory but the file extension doesn't getting saved in database.
for example if i upload image by name mobile-image-small-1.png the file name in database will be just mobile-image-small-1 and it is causing image not to display on web page.what is the problem here
i am using mysql database
here is my php code
// Escape user inputs for security
$product_name = mysqli_real_escape_string($link, $_POST['product_name']);
$product_category = mysqli_real_escape_string($link, $_POST['product_category']);
$product_price = mysqli_real_escape_string($link, $_POST['product_price']);
$pro_url = mysqli_real_escape_string($link, $_POST['pro_url']);
$co_owners = mysqli_real_escape_string($link, $_POST['co_owners']);
// attempt insert query execution
// close connection
// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
if($_POST)
{
if($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br />";
}
else
{
if(file_exists("uploaded_images/" . $_FILES["file"]["name"]))
{
echo $_FILES["file"]["name"] . " already exists. ";
}
else
{
if(move_uploaded_file($_FILES["file"]["tmp_name"],"uploaded_images/" . $_FILES["file"]["name"]))
{
$sql = "INSERT INTO product_list (product_name, product_category, product_price,product_referrence_URL,product_co_owners,product_image_url) VALUES ('$product_name', '$product_category', '$product_price', '$pro_url', '$co_owners', 'files/uploaded_images/".$_FILES['file']['name']."')";
if(mysqli_query($link, $sql)){
header("Location: ../index.php");
}else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
if(mysqli_query($sql))
{
echo "Stored in: " . "uploaded_images/" . $_FILES["file"]["name"];
}
else
{
echo 'Unable to store';
}
}
}
}
}
mysqli_close($link);
?>
here is my form
<form class="form-horizontal" method="post" action="files/insert.php" role="form" enctype="multipart/form-data">
<div class="form-group">
<label class="control-label col-sm-2" for="product_name">Product Name</label>
<div class="col-sm-10">
<input type="text" class="form-control" name="product_name" id="product_name" placeholder="Iphone 5c" required>
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="pwd">Product Category</label>
<div class="col-sm-10">
<select class="btn-btn-primary form-control" name="product_category">
<option>Mobile</option>
<option>Television</option>
<option>Printer</option>
<option>Watch</option>
<option>Monitor</option>
</select>
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="product-pic">Upload your profile picture</label>
<div class="col-sm-10">
<input type="file" class="form-control" name="file" id="file" />
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="product_price">Product Price</label>
<div class="col-sm-10">
<input type="text" class="form-control" name="product_price" id="product_price" placeholder="Rs.36,000" required>
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="pro_url">Reference URL</label>
<div class="col-sm-10">
<input type="URL" class="form-control" name="pro_url" id="pro_url" placeholder="http://www.amazon.com" required>
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="co_owners">Co-owners</label>
<div class="col-sm-10">
<select class="btn-btn-primary form-control" name="co_owners">
<option>Select no. owners</option>
<option>1</option>
<option>2</option>
<option>3</option>
<option>4</option>
<option>5</option>
<option>6</option>
<option>7</option>
<option>8</option>
<option>9</option>
<option>10</option>
<option>11</option>
<option>12</option>
<option>13</option>
<option>14</option>
<option>15</option>
</select>
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-2 col-sm-10">
<button type="submit" class="btn btn-success">Add product</button>
</div>
</div>
</form>
As per discussion your database length is 30 and type is varchar and your image length is greater the 30. Thats the reason you half data updated to your database.
So increase the length of your column up to 100 it will take length uo to 100 length.
I think you can use the type field to get this value
$_FILE['file']['type']
Check out the documentation here

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