提交按钮后刷新上一个表单表

How do I refresh the table when I click EDIT NOW Button? or this is easier, how do I make refresh button to refresh the table? I have no knowledge on JSP to auto refresh. However, I manage to make an onchange event so when I choose menu, it will refresh upon changing. When I edit a data and submit, it does not refresh. How do I re-execute the echo of table? Thanks!

单击“编辑现在按钮”时如何刷新表格？或者这更容易，如何使刷新按钮刷新表？我不知道JSP自动刷新。但是，我设法创建一个onchange事件，所以当我选择菜单时，它会在更改时刷新。当我编辑数据并提交时，它不会刷新。如何重新执行表的回显？谢谢！

    <?php



    $selected='';

    function get_options($select)
    {
        $conn = new mysqli('localhost', 'root', 'jared17', 'hbadb') 
        or die ('Cannot connect to db');

        $result = $conn->query("select * from students");

        $options='';
        while ($row = $result->fetch_assoc()) 
        {
            $LRN = $row['LRN'];
             $Last = $row['Last_Name']; 
            $First = $row['First_Name'];
            if($LRN == $_GET['Students'])
            {
                $options.='<option value="'.$LRN.'" selected>'.$Last.', '.$First.'</option>';

            }
            else
            {
                $options.='<option value="'.$LRN.'">'.$Last.', '.$First.'</option>';
            }
        }
        return $options;
    }
        if (isset($_GET['Students'])) {
            $conn = new mysqli('localhost', 'root', 'jared17', 'hbadb') 
        or die ('Cannot connect to db');

        $result = $conn->query("select * from students");
        $lrn = $_GET['Students'];
        $stmt = $conn->prepare("SELECT Last_Name, First_Name, Level, Q1, Q2, Q3, Q4, FINAL FROM english WHERE LRN = ?");
        $stmt->bind_param('i', $lrn);
        $stmt->execute();
        $stmt->bind_result($last, $first, $level, $q1, $q2, $q3, $q4, $final);
        $stmt->fetch();
        echo "<table><tr><th>LRN</th><th>Name</th><th>Level</th><th>Q1</th><th>Q2</th><th>Q3</th><th>Q4</th><th>Final</th></tr>";
        echo "<tr><td>$lrn</td><td>$last, $first</td><td>$level</td><td>$q1</td><td>$q2</td><td>$q3</td><td>$q4</td><td>$final</td></tr></table>";
    }




    echo "<html>";
    echo "<body>";
    echo "<form method=GET>";
    echo "<select name=Students onchange=this.form.submit();>";

        echo get_options();



    echo "</select>";
    echo "</form>";

    echo "<form method=POST>";
    ///////////EDIT DATA
    echo "Edit Data: ";
    echo "<select name = 'Edit'>";

    echo '<option value=Q1>Q1</option>';
    echo '<option value=Q2>Q2</option>';
    echo '<option value=Q3>Q3</option>';
    echo '<option value=Q4>Q4</option>';
    echo '<option value=FINAL>FINAL</option>';
    echo '<input type="number" max="100" name="editdata" required>';
    echo "</select>";
    echo "<input type='submit' name='submit2' value='Edit Now'>";


    if (isset($_POST['Edit'])) {
        $conn2 = new mysqli('localhost', 'root', 'jared17', 'hbadb') 
        or die ('Cannot connect to db');
        $upd = $_POST['Edit'];
        $txt = $_POST['editdata'];
        $now = "UPDATE english SET $upd='$txt' WHERE LRN='$lrn'";
        $res = $conn2->query($now);
        if (!$conn2->error) {
            echo "Errormessage: $conn->error";
    }
        echo $now;



    }



    echo "</form>";





    echo "</body>";
    echo "</html>";
    ?>

2 个解决方案

#1

I would use 2 pages view.php and edit.php. View would display the data with link to edit. When you click edit it would open edit.php to load the form to edit data and save it to db. Then issue the command

我会使用2页view.php和edit.php。 View将显示带有要编辑的链接的数据。单击编辑时，将打开edit.php以加载表单以编辑数据并将其保存到db。然后发出命令

header("Location: view.php");

to reload the view.php and display the new data

重新加载view.php并显示新数据

I have uploaded the scripts I use as a basis for databases so you can see if this is what you want - feel free to amend any data it is just a test database. If you want more code just ask

我已经上传了我用作数据库基础的脚本，因此您可以看到这是否是您想要的 - 随意修改任何数据它只是一个测试数据库。如果你想要更多的代码，请问

<a href="http://www.cambodia.me.uk/php/view.php">www.cambodia.me.uk/php/view.php</a>

Edited to include scripts as requested - sorry it is old code and mysql not mysqli

编辑包括按要求包含脚本 - 对不起它是旧代码和mysql而不是mysqli

Connect script

连接脚本

<?php
/*
CONNECT-DB.PHP
Allows PHP to connect to your database
*/
// Database Variables (edit with your own server information)
$server = 'server';
$user = 'user';
$pass = 'pass';
$db = 'database';
// Connect to Database
$connection = mysql_connect($server, $user, $pass)
or die ("Could not connect to server ... \n" . mysql_error ());
mysql_select_db($db)
or die ("Could not connect to database ... \n" . mysql_error ());
?>

View script

查看脚本

<?php    
include('remote-connect.php');
$result = mysql_query("SELECT * FROM stats")
or die(mysql_error());
echo "<p><b>View All</b> | <a href='view-paginated.php?page=1'>View Paginated</a></p>";
echo "<table border='1' cellpadding='10'>";
echo "<tr> <th>date</th> <th>Home Team</th> <th></th><th></th><th>Away Team</th> <th></th>  </tr>";
while($row = mysql_fetch_array( $result )) {
echo "<tr>";
echo '<td>' . $row['date'] . '</td>';
echo '<td>' . $row['hometeam'] . '</td>';
echo '<td>' . $row['fthg'] . '</td>';
echo '<td>' . $row['ftag'] . '</td>';
echo '<td>' . $row['awayteam'] . '</td>';
echo '<td><a href="edit.php?id=' . $row['id'] . '">Edit</a></td>';
//echo '<td><a href="delete.php?id=' . $row['id'] . '">Delete</a></td>';
echo "</tr>";
}
echo "</table>";
?>
<p><a href="new.php">Add a new record</a></p>
</body>
</html>

Edit script

编辑脚本

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<?php
function renderForm($id, $hometeam, $awayteam, $error)
{
if ($error != '')
{
echo '<div style="padding:4px; border:1px solid red; color:red;">'.$error.'</div>';
}
?>
<form action="" method="post">
<input type="hidden" name="id" value="<?php echo $id; ?>"/>
<div>
<p><strong>ID:</strong> <?php echo $id; ?></p>
<strong>First Name: *</strong> <input type="text" name="hometeam" value="<?php echo $hometeam; ?>"/><br/>
<strong>Last Name: *</strong> <input type="text" name="awayteam" value="<?php echo $awayteam; ?>"/><br/>
<p>* Required</p>
<input type="submit" name="submit" value="Submit">
</div>
</form>
</body>
</html>
<?php
}
include('remote-connect.php');
if (isset($_POST['submit']))
{
if (is_numeric($_POST['id']))
{
$id = $_POST['id'];
$hometeam = mysql_real_escape_string(htmlspecialchars($_POST['hometeam']));
$awayteam = mysql_real_escape_string(htmlspecialchars($_POST['awayteam']));
if ($hometeam == '' || $awayteam == '')
{
$error = 'ERROR: Please complete all mandatory fields!';
renderForm($id, $hometeam, $awayteam, $error);
}
else
{
mysql_query("UPDATE stats SET hometeam='$hometeam', awayteam='$awayteam' WHERE id='$id'")
or die(mysql_error());
// Go back to view page and redisplay the edited data
header("Location: view.php");
}
}
else
{
echo 'Error!';
}
}
else
{
if (isset($_GET['id']) && is_numeric($_GET['id']) && $_GET['id'] > 0)
{
$id = $_GET['id'];
$result = mysql_query("SELECT * FROM stats WHERE id=$id")
or die(mysql_error());
$row = mysql_fetch_array($result);
if($row)
{
$hometeam = $row['hometeam'];
$awayteam = $row['awayteam'];
renderForm($id, $hometeam, $awayteam, '');
}
else
{
echo "No results!";
}
}
else
{
echo 'Error!';
}
}
?>
</body>
</html>

#2

It's so easy. If you fetch your table's values after edit or delete operation, sql will bring fresh data.

它是如此容易。如果在编辑或删除操作后获取表的值，sql将带来新的数据。

#1