如何在提交后检查复选框(切换)的状态

时间:2017-09-03 01:59:07

标签: php mysql checkbox

在一个表格中我放入一个复选框(开关),我如何在提交检查MySQL更新的开关状态后发布?谢谢

因此MySQL表不会改变。

来自复选框并提交

<form method="POST" action="process.php">
<label class="mdl-switch mdl-js-switch mdl-js-ripple-effect" for="switch-1">
<input type="checkbox" id="switch-1" class="mdl-switch__input" checked>
<span class="mdl-switch__label">Autenticazione a due fattori</span>
</label>
....
<input onclick="conferma();" class="mdl-button mdl-js-button 
mdl-button--raised mdl-js-ripple-effect mdl-button--accent" type="submit" 
value="Salva" name="submitBtn">
</form>

切换控制和MySQL更新(process.php)

<?php 
session_start();
mysql_connect(localhost) or die(mysql_error()); 
mysql_select_db("*******") or die(mysql_error()); 
$user = $_SESSION['users'];
if(isset($_POST['submitBtn'])) { //form submission occured
    if(!isset($_POST['switch-1'])){            
        $sql = "UPDATE `*******`.`login_users` SET `auth` = \'checked\' WHERE username = '$user'";
        header("location: https://*******.php");
    } else {
        $sql = "UPDATE `*******`.`login_users` SET `auth` = \'unchecked\' WHERE username = '$user'";
        header("location: https://*******.php");
    }
}

?>

1 个答案:

答案 0 :(得分:0)

1)修改表格

<form method="POST" action="process.php">

    <label class="mdl-switch mdl-js-switch mdl-js-ripple-effect" for="switch-1">
    <input type="checkbox" name="switch-1" id="switch-1" class="mdl-switch__input" checked>
    <span class="mdl-switch__label">Autenticazione a due fattori</span>
    </label>
    ....
    <input type="submit" value="Salva" name="submitBtn" class="mdl-button mdl-js-button 
    mdl-button--raised mdl-js-ripple-effect mdl-button--accent" >

</form>

2)在同一工作目录中创建一个新的process.php页面并添加它们。

修改

<?php

session_start();
$user = $_SESSION['users'];

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "YourDBName";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

if(isset($_POST['submitBtn'])) { //form submission occured

    if(isset($_POST['switch-1'])){                       
        $sql = "UPDATE login_users SET auth = 'checked' WHERE username = '$user'";
    } else {
        $sql = "UPDATE login_users SET auth = 'no' WHERE username = '$user'";
    }

    if ($conn->query($sql)) {
        echo "Updated successfully";
    } else {
        echo "Error: " . $sql . "<br>" . $conn->error;
    }

} else {
    echo "Form Submission Error";
}

$conn->close();
?>

希望它有所帮助。