How to add values of generated textbox in database using mysqli - php

Question

This code somehow stores only the value of last textbox. When we generate 3 textboxes, only the value of 3rd is stored in database instead of all. I want to store all the generated textbox values in the database
<?php if(isset($_POST['submit'])){
$num_cat = $_POST['num_cat'];
$text = $_POST['category'];
foreach ($text as $key) {
// echo $key."\n";
}
$insertquery = mysqli_query($con, "INSERT INTO accounts (accountusername, accountemail) VALUES('".$num_cat."', '".$key."')");
if(!$insertquery){
echo "Error".mysqli_error($con);
}
} ?>
<script type="text/javascript">
//when the webpage has loaded do this
$(document).ready(function() {
//if the value within the dropdown box has changed then run this code
$('#num_cat').change(function(){
//get the number of fields required from the dropdown box
var num = $('#num_cat').val();
var i = 0; //integer variable for 'for' loop
var textboxes = ''; //string variable for html code for fields
//loop through to add the number of fields specified
for (i=1;i<=num;i++) {
//concatinate number of fields to a variable
textboxes += 'Category'+i+': <input type="text" name="category[]' + i + '" placeholder = "category_' + i + '" /><br/>';
}
//insert this html code into the div with id catList
$('#catList').html(textboxes);
});
});
</script>
<form method="post" action="">
Number of fields required:
<select id="num_cat" name="num_cat">
<option value="0">- SELECT -</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
<option value="6">6</option>
<option value="7">7</option>
<option value="8">8</option>
<option value="9">9</option>
<option value="10">10</option>
</select>
<div id="catList"></div>
<input type="submit" name="submit" value="Submit"/>
</form>

Answer 1

If you update your javascript to write this (remove ' + i + ' from the name):
textboxes += 'Category'+i+': <input type="text" name="category[]" placeholder = "category_' + i + '" /><br/>';
then $_POST['category'] will be an array. It then depends how you want to store the values, but assuming you'll want three rows, you'll need to do this:
foreach ($_POST['category'] as $key) {
$insertquery = mysqli_query($con, "INSERT INTO accounts (accountusername, accountemail) VALUES('".$num_cat."', '".$key."')");
if(!$insertquery){
echo "Error".mysqli_error($con);
}
}
This is explained in the PHP manual.
Please have a look at how can I prevent SQL-injection in PHP? as well, because your current code is vulnerable to simple exploits.