Checkbox insert in mysql and check if checked - php

Question

After i searched for this solution on this site, nothing i found. Here is basic php code, just for testing.
index.php
<input type="checkbox" name="chk" id="chk" value="1" />
<button type="button" name="" id="submit">TEST</button>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
$('#submit').click(function(){
var check = $('#chk').val();
$.ajax({
url:"qry.php",
method:"POST",
data: {check:check},
success:function(data)
{
//alert(data);
window.location.reload();
}
});
});
});
</script>
qry.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "check";
// Create connection
$connect = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($connect->connect_error) {
die("Connection failed: " . $connect->connect_error);
}
if($_REQUEST['chk'] == true){
$stok = '1';
}
elseif ($_REQUEST['chk'] == false) {
$stok = '0';
}
$query = "INSERT INTO test(checked) VALUES('$stok')";
$result = mysqli_query($connect, $query);
if ($result === TRUE) {
echo "Zahtev je uspešno poslat!";
} else {
echo "Error: " . $query . "<br>" . $connect->error;
}
?>
How to set checkbox checked true or false into mysql and then echo if in html? It's always set to 0 in mysql boolen.
<input type="hidden" name="chk" id="chk" value="1" <?php if ($checked == '1') {echo 'checked';} else {} ?>/>
I tried everything from this site and nothing works.

Answer 1

Try this Jquery. This will get rid of the always value 1 problem you're having. What this code does is when you click on the "submit" button it check the status of your check box. If the check box is checked then the code will take the checked check box value and send that in the ajax function if it's not checked then the value 0 get assigned and that will be sent using the ajax.
Doing this will reduce the work has to be done by the back end PHP. I also made some changes to your PHP code as well.
$(document).ready(function() {
$(document).on('click', '#submit', function() {
if ($("#chk").is(":checked")) {
var chk = $('#chk').val();
}else{
chk = 0;
}
$.ajax({
url: "qry.php",
method: "POST",
data: {
check: chk
},
success: function(data) {
//alert(data);
window.location.reload();
}
});
});
})
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="checkbox" name="chk" id="chk" value="1" />
<button type="button" name="" id="submit">TEST</button>
You will see that the way I'm inserting the data is a bit different from the way you have done. I'm using mysqli_ prepared statements which makes SQL injection a hard to do.
$query = $connect -> prepare("INSERT INTO test(checked) VALUES (?)";
$query -> bind_param("i", $_REQUEST['check']);
if ($query -> execute()) {
echo "Zahtev je uspešno poslat!";
} else {
echo "Error: " . $query . "<br>" . $connect->error;
}
jsFiddle if you want to test it.

Answer 2

You write the value into check, but read it back from $_REQUEST['chk']. That won't work. Change that to $_REQUEST['check'].
You are using val() to get the state of the checkbox, you should use checked.
Also, you are possibly open to SQL injection, start using prepared statements.

Answer 3

Add another line while sending the AJAX request.
While sending the value of checkbox, send 1 or 0.
It will reduce our work at PHP end.
So, the code should be:
var check = $('#chk').is(":checked");
check = (check) ? 1 : 0;
Final code should be:
<input type="checkbox" name="chk" id="chk" value="1" />
<button type="button" name="" id="submit">TEST</button>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
$('#submit').click(function(){
var check = $('#chk').val();
check = (check) ? 1 : 0;
$.ajax({
url:"qry.php",
method:"POST",
data: {check:check},
success:function(data) {
//alert(data);
window.location.reload();
}
});
});
});
</script>
And PHP:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "check";
// Create connection
$connect = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($connect->connect_error) {
die("Connection failed: " . $connect->connect_error);
}
$stok = $_REQUEST['chk'];
$query = "INSERT INTO test(checked) VALUES('$stok')";
$result = mysqli_query($connect, $query);
if ($result === TRUE) {
echo "Zahtev je uspešno poslat!";
}
else {
echo "Error: " . $query . "<br>" . $connect->error;
}
?>