php MySQL查詢更新列到數據庫..但它無法顯示更新的列值到頁面

Question

你好我是PHP和MYSQL的新手。 想要了解更多。 今天我面臨一些問題，比如將值更新到數據庫中。 但我解決了它。 我的代碼將每行的特定列名“status”更新到數據庫中。 但它仍然顯示我在頁面中的相同價值

<div class="widget-body" style="height: 290px;">
    <table class="table ">
        <thead>

            <tr>
                <th>User Name</th>
                <th>Starts</th>
                <th>Length</th>
                <th>Status</th>
                <th>Update</th>
            </tr>
        </thead>
        <tbody>   
            <?php
            include 'sql.php';

            $result = mysql_query("SELECT vacation.id, vacation.start_date, vacation.length, vacation.status, userinfo.* FROM vacation INNER JOIN userinfo ON vacation.user_id = userinfo.user_id ");
            if ($result === FALSE) {
                die(mysql_error()); // TODO: better error handling
            }
            while ($db_field = mysql_fetch_assoc($result)) {
                $id = $db_field['id'];
                $uid = $db_field['user_id'];
                $uname = $db_field['username'];
                $sdate = $db_field['start_date'];
                $len = $db_field['length'];
                $stat = $db_field['status'];
                echo("<tr>");
                echo("<td>$uname</td>");
                echo("<td>$sdate</td>");
                echo("<td>$len</td>");
                echo("<td>$stat</td>");
                ?> <td>

                <form method='post' action = '<?php echo $_SERVER['PHP_SELF'];?>' >    
                    <input  type='hidden' name="vacation_id" value='<?php echo $id; ?>'/>
                    <button type='submit' class='btn btn-success btn-mini' name='btn-accept' >Accept</button>
                </form>
                <form method='post' action = '<?php echo $_SERVER['PHP_SELF'];?>' >    
                    <input  type='hidden' name="vacation_id" value='<?php echo $id; ?>'/>
                    <button type='submit' class='btn btn-danger btn-mini'  name='btn-deny' >Deny</button>
                </form>

            </td><?php
                echo("</tr>");

            } //end while-loop
                ?>

            <?php
                if (isset($_POST['btn-accept'])) {
                    $v_id = $_POST['vacation_id'];
                    $SQL = "UPDATE vacation SET status = 1 WHERE id = $v_id ";
                    $result = mysql_query($SQL);
                    if ($result === FALSE) {
                        die(mysql_error()); // TODO: better error handling
                    }
                }

                if (isset($_POST['btn-deny'])) {
                    $v_id = $_POST['vacation_id'];
                    $SQL = "UPDATE vacation SET status = 0 WHERE id = $v_id ";
                    $result = mysql_query($SQL);
                    if ($result === FALSE) {
                        die(mysql_error()); // TODO: better error handling
                    }
                }
             ?>

        </tbody>

    </table>
</div>

我知道它很老了，但我正在學習。 提前致謝。

Answer 1

我的意思是這樣的：

        if (isset($_POST['btn-accept'])) {
            $v_id = $_POST['vacation_id'];
            $SQL = "UPDATE vacation SET status = 1 WHERE id = $v_id ";
            $result = mysql_query($SQL);
            if ($result === FALSE) {
                die(mysql_error()); // TODO: better error handling
            }
       }

       if (isset($_POST['btn-deny'])) {
            $v_id = $_POST['vacation_id'];
            $SQL = "UPDATE vacation SET status = 0 WHERE id = $v_id ";                   
            $result = mysql_query($SQL)
            if ($result === FALSE) {
                die(mysql_error()); // TODO: better error handling
            }
        }

        $result = mysql_query("SELECT vacation.id, vacation.start_date, vacation.length, vacation.status, userinfo.* FROM vacation INNER JOIN userinfo ON vacation.user_id = userinfo.user_id ");
        if ($result === FALSE) {
            die(mysql_error()); // TODO: better error handling
        }

        while ($db_field = mysql_fetch_assoc($result)) {