could not inserting data to mysql table using php - php

Question

I have following form in html
<form method="post" enctype="multipart/form-data" />
<fieldset>
<legend>Activate Scheme</legend>
<p>Date Of Draw</p>
<p><input type="text" name="dateOfDraw" class="textBox" style="width:150px" /></p>
<p>Time Of Draw</p>
<p><input type="text" class="textBox" name="timeOfDraw" style="width:150px" /></p>
<p>Enter scheme name</p>
<p><input type="text" class="textBox" name="schemeName" style="width:150px" />
</p>
<p>Upload Image</p>
<p><input type="file" name="image" id="image" /></p>
<p><input name="scheme_button" type="submit" class="button1" value="Submit"></p>
</fieldset>
</form>
Problem is that their is no error when I execute the following query
<?php
if(isset($_POST['submit_button']) && count($_POST)>0) {
print_r($_POST);
$dateOfDraw = $_POST['dateOfDraw'];
$timeOfDraw = $_POST['timeOfDraw'];
$schemeName = $_POST['schemeName'];
$imageName = $_FILES['image']['name'];
$destination = '../images/'.$imageName;
$source = $_FILES['image']['tmp_name'];
if(move_uploaded_file($source, $destination)) {
echo 'file uploaded';
} else {
echo ' file not uploaded';
}
$sSQL = "INSERT INTO landing_page(dateOfDraw, timeOfDraw, schemeName, image) VALUES('$dateOfDraw','$timeOfDraw','$schemeName','$imageName')";
echo $sSQL;
if(!$sSQL){
die(mysqli_query($con));
}
mysqli_query($con,$sSQL);
//header("location: list_products.php");
}
?>
but when I check my data in a mysql database the rows are still empty. Please check it am I missing something or the php code is wrong.
Note: I am using php version 5.5.16

Answer 1

The name of your submit button is scheme_button and that's exactly the thing that you need to $_POST.
if(isset($_POST['scheme_button']) && count($_POST) > 0)
You form doesn't have an action and it doesn't know where to go. So add one.
<form method="post" action="index.php" enctype="multipart/form-data" />

Answer 2

firstly you have to name of submit button
if(isset($_POST['scheme_button']) && count($_POST) > 0)
secondly if insert code in same page then
if insert code in another page then
action = "filename"