How to update a specific cell in phpmyadmin with jQuery after appending a specific div to a group, code is below
JavaScript
x
<div id="item1">Item 1
<input type="button" value="put me in Group1" name="I1G1"> <!-- I1=Item1 / G1=Group1-->
<input type="button" value="put me in Group2" name="I2G2">
</div><br><br>
<div id="item2">Item 2
<input type="button" value="put me in 1" name="I2G1">
<input type="button" value="put me in 2" name="I2G2">
</div><br><br>
<div id="Group1">Group 1</div><br>
<div id="Group2">Group 2</div><br>
<script>
$('input[name$="I1G1"]').click(function(){
$("#item1").appendTo("#Group1");
$(this).hide();
$('input[name$="I1G2"]').hide();
<?php $UPDATE = mysqli_query($conn, "UPDATE results SET round = 'Group 1' WHERE team = 'Item 1'"); ?>
});
$('input[name$="I1G2"]').click(function(){
$("#item1").appendTo("#Group2");
$(this).hide();
$('input[name$="I1G1"]').hide();
<?php $UPDATE = mysqli_query($conn, "UPDATE results SET round = 'Group 2' WHERE team = 'Item 1'"); ?>
});
//same jQuery code with 2nd item
</script>
the problem is when the item getting updated in database, it became always Group 2 which is the second update. So how to update the database with the selected group on a side question is it better to use dropdown list or buttons in selecting groups
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Answer
You need to create a <form>
element that uses the POST method to a php script.
JavaScript
<form id='my_form' action='my_db_script.php' method='POST'>
// form elements
</form>
In your php script you would use $_POST['my_option']
to get a form option by its name… you can execute your sql in the php file.
Since you tagged jquery, you could also use the jquery onsubmit event with $.ajax
: (documentation)
JavaScript
$("#my_form").onsubmit(function() {
$.post({
// my options
});
});