如果选中则如何更新数据库1的复选框值，如果未选中则更新为0？
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This is my database
<?php
if(isset($_POST["insert"]))
{
$conn = mysqli_connect("localhost", "root", "", "databaseappfeature");
if(isset($_POST["insert"]) == "1"){
$query = "UPDATE appfeature SET feature_switch = ('".$_POST["insert"]."')";
$result = mysqli_query($conn, $query);
echo "Data Inserted Successfully!";
}
}
?>This is my javascript code
<script>
$(document).ready(function(){
$('#submit').click(function(){
var insert = [];
$('.get_value').each(function(){
if($(this).is(":checked"))
{
insert.push($(this).val());
}
});
insert = insert.toString();
$.ajax({
url: "insert.php",
method: "POST",
data:{insert:insert},
success:function(data){
$('#result').html(data);
}
});
});
});
</script>This is my checkbox code
<form action="" method="POST">
<h4 id="result"></h4>
<div class="container">
<h2 align="center">Table App Feature</h2>
<table id="appFeature" class="table table-hover" align="center" style="width:500px;margin:auto;">
<thead>
<tr>
<th>Firstname</th>
<th>Please check to enable the features</th>
</tr>
</thead>
<tbody>
<tr>
<td>Smarthome</td>
<td>
<input type="checkbox" class="get_value" value="1" />
</td>
</tr>
<tr>
<td>Intercom</td>
<td>
<input type="checkbox" class="get_value" value="1" />
</td>
</tr>
<tr>
<td>Visitors</td>
<td>
<input type="checkbox" class="get_value" value="1" />
</td>
</tr>
<tr>
<td>Booking</td>
<td>
<input type="checkbox" class="get_value" value="1" />
</td>
</tr>
<tr>
<td>Access</td>
<td>
<input type="checkbox" class="get_value" value="1" />
</td>
</tr>
<tr>
<td>Parcel</td>
<td>
<input type="checkbox" class="get_value" value="1" />
</td>
</tr>
<tr>
<td>Bills</td>
<td>
<input type="checkbox" class="get_value" value="1" />
</td>
</tr>
<tr>
<td>Invoices</td>
<td>
<input type="checkbox" class="get_value" value="1" />
</td>
<tr>
<td>Receipts</td>
<td>
<input type="checkbox" class="get_value" value="1" />
</td>
</tr>
<tr>
<td>Transactions</td>
<td>
<input type="checkbox" class="get_value" value="1" />
</td>
</tr>
<tr>
<td>Meeting</td>
<td>
<input type="checkbox" class="get_value" value="1" />
</td>
</tr>
<tr>
<td>Vote</td>
<td>
<input type="checkbox" class="get_value" value="1"/>
</td>
</tr>
<tr>
<td>Feedback</td>
<td>
<input type="checkbox" class="get_value" value="1" />
</td>
</tr>
</tbody>
</table><br />
<div align="center">
<button type="button" name="submit" id="submit">Update</button>
</div>
</div>
</form>如何将多个复选框的值更新到数据库中，如果选中该复选框，则值应为1;如果未选中该复选框，则值应为0？ This is my checkboxThis is jquery that i use to pass the value of checkboxThis is my database code请帮助我..我对此很新......我这样做了一个星期..
答案
请查看文档并了解表单的工作方式。有很多例子我会推荐Head First书系列和here你可以找到一个很好的例子。
此处还有您问题的示例代码
创建一个名为example.html的文件并保存此内容
<html>
<head>
<!-- link to jquery lib -->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
</head>
<body>
<form action="" method="POST">
<h4 id="result"></h4>
<div class="container">
<h2 align="center">Table App Feature</h2>
<label> Name </label>
<input type='text' name='name' id='name' value=''/>
<table id="appFeature" class="table table-hover" align="center" style="width:500px;margin:auto;">
<thead>
<tr>
<th>Firstname</th>
<th>Please check to enable the features</th>
</tr>
</thead>
<tbody>
<tr>
<td>Smarthome</td>
<td>
<!-- checkbox for smarthome value -->
<input type="checkbox" class="get_value" id='smarthome'/>
</td>
</tr>
<tr>
<td>Intercom</td>
<td>
<input type="checkbox" class="get_value" id='intercom'/>
</td>
</tr>
</tbody>
</table><br />
<div align="center">
<label>check if you want to update, unckeck if you want to insert</label>
<input type="checkbox" class="get_value" id='update'/>
<br>
<!-- button name -->
<button type="button" name="submit" id="submit">Update or Insert</button>
</div>
</div>
</form>
</body>
<script type="text/javascript">
$(document).ready(function(){
$('#submit').click(function(){
// get the value is checked from the form
$.ajax({
url: "insert.php",
method: "POST",
data:{intercom: $('#intercom').is(':checked'),smarthome: $('#smarthome').is(':checked'), name: $('#name').val(), update: $('#update').is(':checked')},
success:function(data){
$('#result').html(data);
}
});
});
});
</script>
</html>
php文件如下命名为insert.php。确保两个文件都位于同一目录和Apache服务器内。（localhost public directory）
<?php
$servername = "localhost";
$username = "YOURDBUSER";
$password = "YOURDBPASSWORD";
$dbname = "databaseappfeature"; // db name
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql ;
if($_POST['update']){
$sql = "UPDATE appfeature SET smarthome=".$_POST['smarthome'].", intercom=".$_POST['intercom']." WHERE name='".$_POST['name']."'";
}else{
$sql = "INSERT INTO appfeature (name, smarthome, intercom) VALUES ('".$_POST['name']."',".$_POST['smarthome'].",".$_POST['intercom'].")";
}
if ($conn->query($sql) === TRUE && !$_POST['update']) {
echo "New record created successfully";
}else if($conn->query($sql) === TRUE && $_POST['update']){
echo "record updated";
}else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
here是与我使用的数据库相关的sql文件
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