I am searching for a way to update my sql database without refreshing the page by selecting the checkbox. I am trying to achieve the id to be passed to the dolead.php when the user selects the checkbox and than check whether the id is higher than 0 it must pass the value 1 otherwise the value 0. Afther that check i need the database to be updated with the where clause the posted id. Below my code, what am i doing wrong? i get no errors in my logs and i have the error logging stated as followed: mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
PHP the user is seeing to select the switch box to trigger a select and the jquery to trigger the post.
<head>
<script src="jquery-3.2.1.min.js"></script>
<link rel="stylesheet" href="switch.css" type="text/css">
</head>
<body>
<table id="myTable" align='left' class="deftable">
<thead>
<tr>
<th style="width:10px !important;">!!</th>
<th>Name</th>
<th>Notes</th>
<th>Category</th>
</tr>
</thead>
<tbody>
<? //Selecteer accounts.
$stmt = $mysqli->prepare("SELECT * FROM account WHERE purpose = 'Crew4' AND level <= ? AND betaald = 'yes'
AND group_name = ? AND secret = 'no' ");
$stmt->bind_param("is", $status,$groupname);
$stmt->execute();
$result = $stmt->get_result(); //only works when nd_mysli is set on the server!
while ($row = $result->fetch_assoc()) {
?>
<tr>
<td class="footer">
<label class="switch">
<input type="checkbox" <?php if($row['do_lead']=='1') {
echo 'checked="checked"';
} ?> value="<?php echo filter_var($row['id'], FILTER_VALIDATE_INT); ?>" />
<div class="slider round">
<span class="on">ON</span>
<span class="off">OFF</span>
</div>
</label></td>
<td><?= htmlspecialchars($row['name']) ?></td>
<td><?= htmlspecialchars($row['notes']) ?></td>
<td><?= htmlspecialchars($row['purpose']) ?></td>
</tr>
<? } ?>
</table>
</body>
<script>
$(document).ready(function()
{
var x=$("input[type=checkbox]").length;
$('input[type=checkbox]').on('change', function(i)
{
if($(this).is(':checked'))
{
x-=1;
if(x==0)
{
var val = [];
$(':checkbox:checked').each(function(i)
{
val[i] = $(this).val();
});
var jsonString = JSON.stringify(val);
$.ajax(
{
type: "POST",
url: "dolead.php",
data: {checkbox_value:jsonString},
cache: false,
success: function(data)
{
/* alert if post is success */
}
});
}
}
else
{
x+=1;
}
});
});
</script>
</html>
The dolead.php to retrieve the post value and update the database;
$data = json_decode(stripslashes($_POST['checkbox_value']));
foreach($data as $d){
if($d > 0){$leadupdate = 1;}
if($d == ''){$leadupdate = 0;}
$stmt48 = $mysqli->prepare("UPDATE account SET do_lead = ? WHERE id = ? ");
$stmt48->bind_param("ii", $leadupdate,$d);
$stmt48->execute();
$stmt48->close();
}
?>
PS: the connection file is included but not shown here..
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Answer
You could add a class to the input and set the value dynamically:-
View:-
<input class="active" type="checkbox" value="$row['do_lead']" >
jQuery/AJAX Call:-
<script type="text/javascript" src="https://code.jquery.com/jquery-3.4.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function(e){
$("input.active").click(function() {
// store the values from the form checkbox, then send via ajax below
var check_active = $(this).is(':checked') ? 1 : 0;
var check_id = $(this).attr('value');
$.ajax({
type: "POST",
url: "dolead.php",
data: {id: check_id, active: check_active}
success: function(response){
alert('Data Updated Successfully');
}
});
return true;
});
});
</script>
PHP:-
// CLIENT INFORMATION $active = mysqli_real_escape_string($_POST['active']); $id = mysqli_real_escape_string($_POST['id']); //update query. //execute query.
Note:- for more info regarding click()