提交按钮后刷新上一个表单表

时间：2016-09-10 03:21:33

标签： php sql refresh dropdown

单击“编辑现在按钮”时如何刷新表格？或者这更容易，我如何刷新刷新按钮？我对JSP不了解自动刷新。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！

enter image description here 

    <?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 个答案:

答案 0 :(得分：0)

我会使用2页view.php和edit.php。视图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");

重新加载view.php并显示新数据

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。如果您想要更多代码，请询问

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

编辑按要求包含脚本 - 抱歉它是旧代码而mysql不是mysqli

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 ());
?>

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>

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>

答案 1 :(得分：0)

这很容易。如果在编辑或删除操作后获取表的值，sql将带来新数据。

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