Question

在向我的数据库提交信息后，我想刷新页面以显示这些更改，就像表单处理完毕一样。提交后页面“重新加载”但不反映更改，所以我假设我需要在按下提交时添加刷新命令，但它似乎太快了？

所以我添加了一个刷新时间，但是即使它达到50，我也得到了相同的结果。

如果我按两次按钮，则会刷新正确的信息。有更好的方法吗？

<?php
ini_set('display_errors',1);
error_reporting(E_ALL);

include_once '../includes/conn.php';

if(!$user->is_loggedin()){
    $user->redirect('../users/login.php');
}

$id = $_SESSION['session'];
$stmt = $conn->prepare("SELECT * FROM users WHERE id=:id");
$stmt->execute(array(":id"=>$id));

$userRow=$stmt->fetch(PDO::FETCH_ASSOC);
$location = isset($_POST['location']) ? $_POST['location'] : '';
$about = isset($_POST['about']) ? $_POST['about'] : '';
$title = isset($_POST['title']) ? $_POST['title'] : '';

if($title!=''){
    $sql = "UPDATE users SET title=:title WHERE id=:id";    
    $stmt = $conn->prepare($sql); 
    if($stmt == false){ 
        $error = "User Title update failed. Please try again.";
    } 

    $result = $stmt->execute(array(":title"=>$title, ":id"=>$id));

    if($result == false) { 
        $error = "User Title update failed. Please try again.";
    } 
    $count = $stmt->rowCount();
} 

if($location!=''){
    $sql = "UPDATE users SET location=:location WHERE id=:id";  
    $stmt = $conn->prepare($sql); 
    if($stmt == false){ 
        $error = "User Location update failed. Please try again.";
    } 

    $result = $stmt->execute(array(":location"=>$location, ":id"=>$id));

    if($result == false) { 
        $error = "User location update failed. Please try again.";
    } 
    $count = $stmt->rowCount();
} 

if($about!=''){
    $sql = "UPDATE users SET about=:about WHERE id=:id";    
    $stmt = $conn->prepare($sql); 
    if($stmt == false){ 
        $error = "about Me update failed. Please try again.";
    } 

    $result = $stmt->execute(array(":about"=>$about, ":id"=>$id));

    if($result == false) { 
        $error = "about Me location update failed. Please try again.";
    } 
    $count = $stmt->rowCount();
} 
?>
<!DOCTYPE html>
<html lang="en">    
<head>
    <title>EpicOwl UK | CMS Users Edit Profile</title>
    <meta charset="utf-8">
    <link rel="shortcut icon" href="../images/favicon.ico" type="image/x-icon" />
    <link rel="stylesheet" type="text/css" href="../css/main.css">
</head>
<body>
<div id="header">
    <a href="index.php"><img id="logo" src="../images/logo.png" /></a>
    <div id="navigation">
        <ul>
            <a href="../index.php"><li>Home</li></a>
            <a href="./profile.php"><li>My Profile</li></a>
            <a href="../admin/index.php"><li>Admin Panel</li></a>
        </ul>
    </div>
</div>
<div id="content">
<form method="post"><br />
    <h2>Edit Profile</h2>
    <label><strong>User Title:</strong></label><br />
    <input type="text" name="title" maxlength="50" placeholder="<?php echo ($userRow['title']); ?>" /><br /><br />
    <label><strong>My Location:</strong></label><br />
    <input type="text" name="location" maxlength="50" placeholder="<?php echo ($userRow['location']); ?>" /><br /><br />
    <label><strong>About Me:</strong><label><br />
    <textarea name="about" rows="13" cols="60" maxlength="255" placeholder="<?php echo ($userRow['about']); ?>"></textarea><br /><br />
    <button type="submit" name="update">Update</button><br /><br /><br />
    <?php
    if(isset($_POST['submit'])){
        header('refresh:20; Location: '.$_SERVER['REQUEST_URI']);
    }
    ?>
</form>
</div>
<div id="footer">
    <p class="copyright">&copy; EpicOwl UK. All Rights Reserved.</p>
</div>
</body>
</html>

Answer 1

你做错了，你必须在显示HTML之前处理表单提交。 PHP正在逐行执行，因此在您的情况下，您首先显示数据，然后检查表单是否已提交。只需将此代码移到其他PHP代码所在的位置（您甚至可以删除刷新内容命令）：

if(isset($_POST['submit'])){
    header('Location: '.$_SERVER['REQUEST_URI']);
}

修改

当你混合使用HTML和PHP时，人们发明了MVC，因为像你这样的案例，并且想知道为什么事情不起作用。将PHP代码保存在文件的顶部，尽量不要在HTML中的任何地方编写PHP代码，这样可以省去很多麻烦。此外，在调用exit之后使用header来进一步停止代码执行。这是您的代码的更新版本，简化和更多＆＃34;算法＆＃34; （我希望你能看到并理解代码流的方式）：

<?php ini_set('display_errors',1); error_reporting(E_ALL); include_once '../includes/conn.php'; if(!$user->is_loggedin()){ $user->redirect('../users/login.php'); } $id = $_SESSION['session']; $stmt = $conn->prepare("SELECT * FROM users WHERE id=:id"); $stmt->execute(array(":id"=>$id)); $userRow=$stmt->fetch(PDO::FETCH_ASSOC); if (isset($_POST['submit'])) { $location = isset($_POST['location']) ? $_POST['location'] : null; $about = isset($_POST['about']) ? $_POST['about'] : null; $title = isset($_POST['title']) ? $_POST['title'] : null; $sql_part = array(); $prepare = array(); if ($location) { $sql_part[] = 'location = :location'; $prepare[':location'] = $location; } if ($about) { $sql_part[] = 'about = :about'; $prepare[':about'] = $about; } if ($title) { $sql_part[] = 'title = :title'; $prepare[':title'] = $title; } $prepare[':id'] = $id; if (count($sql_part)) { $sql = 'UPDATE users SET '; $sql .= implode(', ', $sql_part); $sql .= ' WHERE id = :id'; $stmt = $dbh->prepare($sql); if ($stmt) { // Find another way too pass these through the refresh // $result = $stmt->execute($prepare); // $count = $stmt->rowCount(); header('Location: '. $_SERVER['REQUEST_URI']); exit; } } } ?> <!DOCTYPE html> <html lang="en"> <head> <title>EpicOwl UK | CMS Users Edit Profile</title> <meta charset="utf-8"> <link rel="shortcut icon" href="../images/favicon.ico" type="image/x-icon" /> <link rel="stylesheet" type="text/css" href="../css/main.css"> </head> <body> <div id="header"> <a href="index.php"><img id="logo" src="../images/logo.png" /></a> <div id="navigation"> <ul> <a href="../index.php"><li>Home</li></a> <a href="./profile.php"><li>My Profile</li></a> <a href="../admin/index.php"><li>Admin Panel</li></a> </ul> </div> </div> <div id="content"> <form method="post"><br /> <h2>Edit Profile</h2> <label><strong>User Title:</strong></label><br /> <input type="text" name="title" maxlength="50" placeholder="<?php echo ($userRow['title']); ?>" /><br /><br /> <label><strong>My Location:</strong></label><br /> <input type="text" name="location" maxlength="50" placeholder="<?php echo ($userRow['location']); ?>" /><br /><br /> <label><strong>About Me:</strong><label><br /> <textarea name="about" rows="13" cols="60" maxlength="255" placeholder="<?php echo ($userRow['about']); ?>"></textarea><br /><br /> <button type="submit" name="update">Update</button><br /><br /><br /> </form> </div> <div id="footer"> <p class="copyright">© EpicOwl UK. All Rights Reserved.</p> </div> </body> </html>

Answer 2

只需使用JavaScript＆＃39; s -

window.location.reload().

在PHP中，您可以使用 -

$page = $_SERVER['PHP_SELF'];
$sec = "10";
header("Refresh: $sec; url=$page");

Answer 3

我通过在数据库更新后添加header('Location: ./editprofile.php');来获得所需的结果。见下文：

if($title!=''){
    $sql = "UPDATE users SET title=:title WHERE id=:id";    
    $stmt = $conn->prepare($sql); 
    if($stmt == false){ 
        $error = "User Title update failed. Please try again.";
    } 

    $result = $stmt->execute(array(":title"=>$title, ":id"=>$id));

    if($result == false) { 
        $error = "User Title update failed. Please try again.";
    } 
    $count = $stmt->rowCount();
}

后：

if($title!=''){
    $sql = "UPDATE users SET title=:title WHERE id=:id";    
    $stmt = $conn->prepare($sql); 
    if($stmt == false){ 
        $error = "User Title update failed. Please try again.";
    } 

    $result = $stmt->execute(array(":title"=>$title, ":id"=>$id));

    if($result == false) { 
        $error = "User Title update failed. Please try again.";
    } 
    $count = $stmt->rowCount();
    header('Location: ./editprofile.php');  
}

sql更新后刷新页面

3 个答案: