Question

我有一些PHP和HTML代码，可从mysql数据库中获取ID，名称和状态。

使用按钮和$ _POST我试图在说用户按钮被单击时更新MYSQL数据库（这是一个简单的输入/输出面板）

这是我的代码

<?php 
include 'confile.php';
if(isset($_POST['update'])) {

echo $_POST['update']. " "; //test to show correct name
echo $_POST['staffid']; //test to show the correct staffid << **THIS IS WHERE THE ISSUE IS**
//$incid = $_POST['staffid'];

//$sql = "SELECT status FROM staff WHERE id=$incid";
//$result = $conn->query($sql);
//echo $result; //show the status

} else {
    //do nothing.
}
?>
<html>
<head>
<link rel="stylesheet" type="text/css" href="styles.css">
<title>Staff Board</title>
<body>
<div align="center" class="header">
    <div class="header text">
        <h1>Staff Board</h1>
    </div>
    <div class="header logo">
    <img src="/assets/img/logo.gif" width="64px" height="64px">
    </div>
</div>
<div id="conbox" align="center" class="content">
<hr>

<?php
//get all staff and their statuses
$sql = "SELECT id, firstname, surname, status FROM $staff ORDER BY surname ASC";
$result = $conn->query($sql);

    while($row = $result->fetch_assoc()) {

        // assign results to values
        $id = $row["id"]; 
        $firstname = $row["firstname"];
        $surname = $row["surname"];
        $status = $row["status"];
        $fullname = $firstname . " " . $surname . " " . $id;  //The $id variable will be dropped from here... it's just for testing. note, it works here, the correct ID is added to the button value

        if ($status == 1) { //pick the correct color for the status
            $color = "butGreen";
        } else {
            $color = "butRed";
        }
?>
    <form class="staffGrid" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="POST">
    <input type="hidden" id="staffid" name="staffid" value="<?php echo htmlspecialchars($id); ?>"/> <!-- hidden input to pass the id to $_POST -->
    <input type="submit" id="update" name="update" value="<?php echo htmlspecialchars($fullname); ?>"/> <!-- submit button to trigger POST -->
    </form> <!-- added as per devpro & billyonecan -->
    <?php
    };
    ?>

</div>
</div>
</body>
</html>

当我第一次加载页面时，按钮正确显示，并且在页面顶部没有期望的测试输出。

但是，当我单击一个按钮时，页面会正确刷新，并显示所按按钮的正确名称（来自第5行的回显），但是给出了错误的Staffid。它为while循环提供LAST id，而不是该按钮的正确值。

我假设对于每次迭代，将为该特定元素（按钮）设置值...显然我在这里是不正确的。

为什么会发生这种情况以及如何解决？

其他信息

Confile.php在代码中使用了以下变量：-

$conn = new mysqli($server, $username, $password);
$staff = [Location of db table]

一些输出：-

echo $sql;
SELECT id, firstname, surname, status FROM inout.staff ORDER BY surname ASC


echo print_r($_POST);
Array ( [staffid] => 17 [update] => First Second 8 )

为什么要给出最后的迭代而不是当前的迭代？

0 个答案: