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php - PHP jQuery Tabledit 选择选项

问题描述

我正在尝试使用 jQueryTabledit插件制作一个动态表。所以我有一张包含用户的表和一张包含用户类型（管理员、员工、客户）的表。这两个表之间的关系是 via user_type_id。我需要的是 select with options from table user_type，所以当管理员想要编辑一些用户的信息时，他可以user_type在 select 中看到 table 中的值，当他从 select 中选择选项时，数据库中的数据将被更新。

这是html表格的样子 在此处输入图像描述

这是两个表之间的关系 employess.php在此处输入图像描述 

<?php 
$con = mysqli_connect("localhost", "root", "", "gym_project_db");
$query = "SELECT * FROM user WHERE user_type_id = 1";
$result = mysqli_query($con, $query);

$query2 = "SELECT * FROM user_type";
$result2 = mysqli_query($con, $query2);
?>
<!DOCTYPE html>
<html>
<head>
    <title></title>
    <link rel="stylesheet" type="text/css" href="css/main_page.css">
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.0/jquery.min.js"></script>
    <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" />  
    <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/js/bootstrap.min.js"></script>      
    <script src="js/jquery.tabledit.js"></script>     
    <script src="js/jquery.multi-select.js"></script>  
</head>
<body>
    <div class="wrapper">
        <div class="sidebar">
            <h2>Gym management system</h2>
            <ul>
                <li><a href="#"><i class="fas fa-home"></i>Dashboard</a></li>
                <li><a href="employess.php"><i class="fas fa-user"></i>Employees</a></li>
                <li><a href="#"><i class="fas fa-address-card"></i>Clients</a></li>
                <li><a href="#"><i class="fas fa-project-diagram"></i>Subscriptions</a></li>
                <li><a href="#"><i class="fas fa-blog"></i>Branches</a></li>
                <li><a href="#"><i class="fas fa-address-book"></i>Cards</a></li>
                <li><a href="#"><i class="fas fa-address-book"></i>Profile</a></li>
                <li><a href="#"><i class="fas fa-map-pin"></i>Logout</a></li>
            </ul> 
        </div>
        <div class="main_content">
            <div class="header">Welcome!! Have a nice day.</div>  
            <div class="info">

                <div class="employess">

                    <table id="editable_table" class="table table-bordered table-striped">
                        <thead>
                            <tr>
                                <th>User_id</th>
                                <th>Username</th>
                                <th>Password</th>
                                <th>Name</th>
                                <th>Surname</th>
                                <th>EGN</th>
                                <th>Email</th>
                                <th>Phone</th>
                                <th>Address</th>
                                <th>User_type_id</th>
                            </tr>
                        </thead>
                        <tbody>

                            
                            <?php while($row = mysqli_fetch_array($result)){ ?>
                                <tr>
                                    <td> <?= $row["user_id"] ?></td>
                                    <td> <?= $row["username"] ?></td>
                                    <td> <?= $row["password"] ?></td>
                                    <td> <?= $row["name"] ?></td>
                                    <td> <?= $row["surname"] ?></td>
                                    <td> <?= $row["egn"] ?></td>
                                    <td> <?= $row["email"] ?></td>
                                    <td> <?= $row["phone"] ?></td>
                                    <td> <?= $row["address"] ?></td>
                                    <?php 
                                    $user_type_query = mysqli_query($con, "SELECT user_type_id, user_type FROM user_type WHERE user_type_id = $row[9]");
                                    $typerow = mysqli_fetch_array($user_type_query);
                                    ?>
                                    <td><?= $typerow[1]; ?></td>

                                </tr>


                            <?php }?>

                        </tbody>
                    </table>
                </div>
            </div>
        </div>
    </body>
    </html>

    <script>  
        $(document).ready(function(){ 
            $('#editable_table').Tabledit({
                url:'employee_action.php',
                columns:{
                    identifier:[0, "user_id"],
                    editable:[[1, 'username'], 
                    [2, 'password'],
                    [3, 'name'],
                    [4, 'surname'],
                    [5, 'egn'],
                    [6, 'email'],
                    [7, 'phone'],
                    [8, 'address'],
                    [9, 'user_type_id']
                    ]
                },
                restoreButton:false,
                onSuccess:function(data, textStatus, jqXHR)
                {
                    if(data.action == 'delete')
                    {
                        $('#'+data.id).remove();
                    }
                }
            });

        });
</script>

employee_action.php

<?php  

$connect = mysqli_connect("localhost", "root", "", "gym_project_db");

$input = filter_input_array(INPUT_POST);


if($input["action"] === 'edit')
{
 $query = "
 UPDATE user
 SET 

 username = '". $input["username"] ."', 
 password = '". $input["password"] ."', 
 name = '". $input["name"] ."', 
 surname = '". $input["surname"] ."', 
 egn = '". $input["egn"] ."', 
 email = '". $input["email"] ."', 
 phone = '". $input["phone"] ."', 
 address = '". $input["address"] ."'
 
 WHERE user_id = '".$input["user_id"]."'
 ";

 mysqli_query($connect, $query);

}



if($input["action"] === 'delete')
{
 $query = "
 DELETE FROM user 
 WHERE user_id = '".$input["user_id"]."'
 ";
 mysqli_query($connect, $query);
}

echo json_encode($input);

?>

标签： phpjquerytabledit

解决方案

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