$_POST data from form being changed before entered into MySQL - php

Question

So new here to stack exchange but here goes nothing. So when I send a form to my apache server my data is only showing up as ones and zeros. Using var_dump[_$POST]; shows all of my data is correct before passing to MySQL.
My html form:
<form method="POST" action="submit.php" class="subForm">
<input type="text" name = "item1" value="0">
<input type="text" name="item2" value="0">
<input type="text" name="item3" value="no">
<input type="text" name="item4" value="no">
<input type="text" name="item5" value="no">
<input type="text" name="item6" value="no">
<input type="submit" id = "form2">
</form>
my php:
$connect = mysqli_connect('*****','*****','*****','*****');
if(!$connect){
die('Could not Connect: ' . mysqli_error($connect));
}
$NoR = isset($_POST["item1"]);
$CC = isset($_POST["item2"]);
$SD = isset($_POST["item3"]);
$HD = isset($_POST["item4"]);
$pack1 = isset($_POST["item5"]);
$pack2 = isset($_POST["item6"]);
$sql = "INSERT INTO form_test (item_1,item_2,item_3,item_4,item_5,item_6) VALUES (".$NoR.",".$CC.",".$SD.",".$HD.",".$pack1.",".$pack2.")";
mysqli_query($connect, $sql);
mysqli_close($connect);
var_dump($_POST)
var_dumb shows all data input is correct but in the table itself everything shows as 1s and 0s. Any advice?

Answer 1

In php, isset will return a boolean value (true if it's set, false if not), and when you try to print a boolean in php, it will display as 1 or 0. I suggest using the ternary comparison in your code, it's shorter and more readable than having a ton of if statements:
$NoR = isset($_POST["item1"]) ? $_POST["item1"] : '';
$CC = isset($_POST["item2"]) ? $_POST["item2"] : '';
$SD = isset($_POST["item3"]) ? $_POST["item3"] : '';
$HD = isset($_POST["item4"]) ? $_POST["item4"] : '';
$pack1 = isset($_POST["item5"]) ? $_POST["item5"] : '';
$pack2 = isset($_POST["item6"]) ? $_POST["item6"] : '';

Answer 2

Add this in your HTML form
<input type="text" name="action" value="submit">
And PHP code will be
if (isset($_POST['action']) && ($_POST['action']) == 'submit') {
$NoR = mysqli_real_escape_string($_POST["item1"]);
$CC = mysqli_real_escape_string($_POST["item2"]);
$SD = mysqli_real_escape_string($_POST["item3"]);
$HD = mysqli_real_escape_string($_POST["item4"]);
$pack1 = mysqli_real_escape_string($_POST["item5"]);
$pack2 = mysqli_real_escape_string($_POST["item6"]);
$sql = "INSERT INTO form_test (item_1,item_2,item_3,item_4,item_5,item_6) VALUES (".$NoR.",".$CC.",".$SD.",".$HD.",".$pack1.",".$pack2.")";
mysqli_query($connect, $sql);
mysqli_close($connect);
//var_dump($_POST)
}

Answer 3

isset will only give you if value is set or not. 1 if set else 0.
Use
if(isset($_POST["item1"])){
$NoR = $_POST["item1"];`
}
Updated: I made code for you to make sude ISSET values only goto INSERT query!
$arrColumns = $arrValues = array();
foreach($_POST as $key=>$value){
$arrColumns[] = key($key);
$arrValues[] = $value;
}
if(is_array($arrValues)){
$sql = "INSERT INTO form_test (implode(',',$arrColumns))
VALUES(implode(',',$arrValues)";
mysqli_query($connect, $sql);
}
mysqli_close($connect);