Post a form array multiple times

Question

I have an HTML form that contains a number of checkboxes. When the user clicks the submit button the checked array is posted and a query runs based on that result. I have a button allowing the user to save their result, when they insert their saved result into another table. But this means I need to post the form result again which it doesn't allow me to do. How would I fix this?

Can you separate the parts which you want to call in first submit and other one for better understanding — Aman Rawat
– Aman Rawat, Commented May 3, 2017 at 14:51
b.SportID and a.Tag = '$value'" Where is $value coming from? — Aman Rawat
– Aman Rawat, Commented May 3, 2017 at 14:59

Aman Rawat · Accepted Answer · 2017-05-03 15:47:34Z

1

PHP Code

<?php

if (isset($_POST['submit']) && isset($_POST['sport'])) {
    $class = $_POST["sport"];
    foreach ($class As $key => $value) {
        $query = "SELECT *
FROM sport b
join sport a
on a.Tag = b.Name
where a.SportID<> b.SportID and a.Tag = '$value'";

        $result = mysqli_query($con, $query) or die("Invalid Query");

        while ($row = mysqli_fetch_assoc($result)) {

            echo "* $row[Name]\n";
        }
    }
} else if (isset($_POST['saved'])) {


    $arr_class = $_POST["sport"];


    foreach ($arr_class As $key => $newvalue) {

        $query2 = "INSERT INTO save (Username, Name)
    SELECT '$username', b.Name
    FROM sport b 
    JOIN sport a 
    on a.Tag = b.Name
    where a.SportID <> b.SportID 
    and a.Name = '$newvalue'";

        $result2 = mysqli_query($con, $query2) or die('Result could not be saved!');


    }
    echo 'Result saved!';
}

JS Code

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script type="text/javascript">
    $(document).ready(function(){
        $('.submit-button').click(function (ev) {
            var $item = $(this);
            var formData  = $('form').serialize();
            $.ajax({
                data: formData+'&submit=submit',
                url:'test1.php',
                type: "POST",
                dataType: "html",
                success:function (data) {
                    console.log(data);
                    if(confirm('You want to save \n'+data+ ' as your sport')){
                        $.ajax({
                            data: formData+'&saved=saved',
                            url:'test1.php',
                            type: "POST",
                            dataType: "html",
                            success:function (data) {
                                console.log(data);
                            }
                        });
                    }
                }
            });
        });
    });
</script>

HTML Code

<form id="form" action="" method="post">
    <input type="checkbox" name="sport[]" value="Football">Football<br>
    <input type="checkbox" name="sport[]" value="Rugby">Rugby<br>
    <input type="checkbox" name="sport[]" value="Golf">Golf<br>
    <input type="checkbox" name="sport[]" value="Basketball">Basketball<br>

    <br> <input type="button" class="submit-button btn btn-info" name="submit" value="submit">
    <input type="submit" style="display:none;">
</form>

answered May 3, 2017 at 15:47

Aman Rawat

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9 Comments

Aman Rawat Over a year ago

Have you changed the url from url:'test1.php', to your one

Beck Over a year ago

Im being told that Notice: Undefined index: sport

Aman Rawat Over a year ago

on what line number?

Beck Over a year ago

$arr_class = $_POST["sport"];

Aman Rawat Over a year ago

Have you selected any sport?

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