在同一页面上将变量从javascript获取到PHP

Question

I am completely new to JS and PHP programming. Please help me with the following code..

<?php

echo'
<button id="checkin" onclick="myFunction()">CheckIn</button>        
<script>
function myFunction() {
document.getElementById("checkin").disabled = true;
var CheckinoutStatus=true;
        $(function(){
        $.ajax({
              type: "POST",
              url: "functions.php",
              data: "&CheckinoutStatus="+CheckinoutStatus,
              success: function(data)
              {
                 alert("Checked In");
              }
            });
        });
   }
 </script>
 ';
 if(isset($_POST['CheckinoutStatus']))
 {

     $CheckinoutStatus=isset($_POST['CheckinoutStatus']);
     $update = $db->query("UPDATE Details SET     CheckinoutStatus='$CheckinoutStatus' where date='$date' ");

}
?>

Clicking the checkin button should disable the button, send a CheckinoutStatus variable value of true to the database using an AJAX call, and then show a "Checked In" alert after the database has been updated.

I am defining and setting the variable CheckinoutStatus to true in js, and using AJAX, am trying to get that value in PHP and update that data in an existing table. I have not been able to pass the value from js to php.

Answer 1

Unless your table only has one record, you will also need to pass a key of some sort so you can update the correct record. Fir the line below, I'll presume you have a hidden field with an id of "key" in your form.

Change the line:

data: "&CheckinoutStatus="+CheckinoutStatus,

To:

data: "&CheckinoutStatus=1&key="+$('#key').val(),

Then create a PHP file named functions.php which checks $_GET['CheckinoutStatus'] for the value and save it to your table using update with a where clause similar to:

UPDATE `mystatustable`
SET `checkinoutstatus` = $_GET['key']
WHERE `id` = $_GET['key'];

Answer 2

UPDATE: I actually got it working when I pass the values through links..

window.location.href = "exp1.php?w1=" + CheckinoutStatus + "&w2="+date ;

but I cannot able to do that by using the same php file.. I have to create a new php file exp1.php inorder for it to work..