Onclick link to save multiple data without refresh

Question

With this code I can see empty data saving into my db once I click on the link button. But i need to the GET's data in my database. Any solution to this.

<a href="?id=1&pid=238874" id="day" onclick="this.form.submit();><button type="button" class="label label-danger" >Select</button></a>

<script>
$('#day').click(function(e) 
       {
         e.preventDefault();
          $.ajax({
               type: 'post',
               url: "index.php",
               data: $("select.day").serialize(),
           });
            return false;
       });
</script>

For PHP Code to save data

<?php
$a = $_GET['id'];
$b = $_GET['pid'];
// query
$sql = "INSERT INTO hos_patient(reg_id,pid) VALUES ('$a','$b')";
mysqli_query($db, $sql); 
?>

Thanks

Have you found solution? Please vote for the answer(s) that helps you. — Meloman
– Meloman, Commented Feb 6, 2018 at 6:55
No solution yet, I have tried all solution provided and its not working — owebindex
– owebindex, Commented Feb 6, 2018 at 11:13

Zammy · Accepted Answer · 2018-02-02 12:00:25Z

1

You specified using POST in your ajax function. But then you try to get the data by GET in your PHP-Script. I'd suggest you just use POST for this.

Alter your PHP-Script to use $_POST instead of $_GET

answered Feb 2, 2018 at 12:00

Zammy

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4 Comments

Zammy Over a year ago

By the way.. your code is very confusing. You try to make a GET request on your link whilst trying to submit some kind of form and then you got an ajax request where I have no idea how you submit that one.

owebindex Over a year ago

<a href="?id=1&pid=238874" id="day" onclick="this.form.submit();><button type="button" class="label label-danger" >Select</button></a>

id="day" miss out. I have change it to POST not working yet

Zammy Over a year ago

Have you checked the output of $("select.day").serialize(). Will this return any data at all?

owebindex Over a year ago

No... I have use some values to serialize it. My aim is to get those id=1&pid=238874 on a click

Meloman · Accepted Answer · 2018-02-02 12:50:49Z

1

Try with this :

...
$.ajax({
    type: 'post',
    url: "index.php?id=1&pid=238874",
    data: $("select.day").serialize(),
});
...

answered Feb 2, 2018 at 12:50

Meloman

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Comments

Ms.KV · Accepted Answer · 2018-02-02 12:57:57Z

0

Use this in script file and in PHP file use below code

<script>
$('#day').click(function(e) 
       {
         e.preventDefault();
          $.ajax({
               type: 'post',
               url: "index.php?id=1&pid=238874",
               data: $("select.day").serialize(),
           });
            return false;
       });
</script>


    <?php
    $a = $_POST['id'];
    $b = $_POST['pid'];
    // query
    $sql = "INSERT INTO hos_patient(reg_id,pid) VALUES ('$a','$b')";
    mysqli_query($db, $sql); 
    ?>

answered Feb 2, 2018 at 12:57

Ms.KV

2913 silver badges13 bronze badges

Comments

owebindex · Accepted Answer · 2018-02-06 11:37:25Z

0

<a href="?id=1&pid=238874" class="login_form_1"><button type="button" class="label label-danger" >Select</button></a>

<script>
 $(function () {
    $('.login_form_1').on('click', function (e) {
      e.preventDefault();
          $.ajax({
            type: 'GET',
            url: 'show.php?id=22&pid=33',
            data: $('.login_form_1').serialize(),
            success: function (data) {
              $('div.logerrors').html(data);
            }
          });
    });
  });
</script>

For PHP output

<?php
$a = $_GET['id'];
$b = $_GET['pid'];
// query
$sql = "INSERT INTO hos_patient(reg_id,pid) VALUES ('$a','$b')";
mysqli_query($db, $sql); 
?>

answered Feb 6, 2018 at 11:37

owebindex

174 bronze badges

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