Question

用户注册到网站。管理员将登录＆amp;查看用户列表。

我正在尝试为管理员选择复选框并通过Dropdown submit更改用户状态。当我尝试下面的代码时，我可以选择“是/否/ A”，点击submit按钮后显示消息“已成功添加”，但其未更新值< / strong>在数据库中。

表名：tbl_users，column：userStatus [enum]，值：Y，N，A

形式

<form method="post" action="ajax1.php"> <select name="userStatus"> <option value="N">N</option> <option value="Y">Y</option> <option value="A">A</option> </select> <button type="submit" name="submit" >Submit</button> </form>

ajax1.php

if(isset($_POST["submit"])) { $userStatus=$_POST["userStatus"]; $conn = new Database(); $stmt = $conn->dbConnection()->prepare("INSERT INTO tbl_users (userStatus) VALUES ('$userStatus')"); echo " Added Successfully "; }

显示用户的代码复选框，ID，姓名，电子邮件：

$stmt = $user_home->runQuery("SELECT * FROM tbl_users"); $stmt->execute(array(":uid" => $_SESSION['userSession'])); $row = $stmt->fetch(PDO::FETCH_ASSOC); $stmt->execute(); $status = array('Y'=>'Verified','N'=>'Not verified' , 'A'=>'Approved'); echo "<table>"; echo "<tr> <td></td> <td>id</td> <td>name</td> <td>email</td> </tr>"; while($data = $stmt->fetch()) { echo "<tr> <td> <input type='checkbox' > </td> <td>" . $data['userID'] . "</td> <td>" . $data['name'] . "</td> <td>" . $data['email'] . "</td> </tr>"; } echo "</table>";

我研究过＆amp;在发布问题之前尝试了很多，在某些链接中，我看到我们需要使用Javascript，但在其他一些链接中，他们提到我们只能用php实现。我是新编码，请帮帮我

修改

按照以下答案后，我将代码更新为$stmt = $conn->dbConnection()->prepare('UPDATE tbl_users SET userStatus = ? WHERE userID = ?');

Answer 1

您正在做的是使用以下行向表中插入新行：

"INSERT INTO tbl_users (userStatus) VALUES ('$userStatus')"

你应该做UPDATE，而不是插入。您基本上想要做的是UPDATE用户WHERE用户userID（或任何id列的名称）是所选用户的ID。

请参阅MySQL UPDATE。

Answer 2

    <form method="post" action="ajax1.php">
        <select name="userStatus" id="userStatus" onchange="UpdateStatus();">
            <option value="N">N</option>
            <option value="Y">Y</option>
            <option value="A">A</option>
        </select> 
        <button type="submit" name="submit" >Submit</button> 
    </form>
    <script> 
        function UpdateStatus(){
            var staus=$('#userStatus :selected').val();

            allCheckedBoxes="";
            $("input[id^='checkBoxId']:visible:checked").each(function(){ // this is to get checked checkbox vaues to update which user to update
                allCheckedBoxes=allCheckedBoxes+$(this).val()+","; // comaseparated valuse
            }); 

            var dataString="allCheckedBoxes="+allCheckedBoxes+"&staus="+staus;

            $.post("aupdate_status.php",'allCheckedBoxes='+allCheckedBoxes'&staus='+staus,function(result,status,xhr)
            {
                if( status.toLowerCase()=="error".toLowerCase() )
                { alert("An Error Occurred.."); }
                else
                {
                    alert(result);
                }
            })
            .fail(function(){ alert("something went wrong. Please try again") });
        }
    </script>
    update_status.php
    <?php 
    $staus=$_POST['staus'];
    $allCheckedBoxes=$_POST['allCheckedBoxes'];
    $ArrallCheckedBoxes=explode(",",$allCheckedBoxes);
    foreach($ArrallCheckedBoxes as $tempBoxes){
        $sqlQueryToUpdate=" UPDATE tbl_users SET userStatus = '".$staus."'  WHERE userID = '".$tempBoxes."' ;";
        $conn = new Database();
        $stmt = $conn->dbConnection()->prepare($sqlQueryToUpdate); 
        echo " ok success";
    }
    ?>

Please try this . This is working in my case. This will work you too. Don't forget add jquery in your coding.

使用表单提交

2 个答案: