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()