Get user input from a form

我正在尝试创建一个表单，允许用户在前端更新其信息。我刚刚开始学习PHP（我对C非常了解）。如果我使用此代码在WordPress页面上创建表单，当用户按下“更新”按钮，然后更新数据库时，我如何从用户那里获取信息？

function Insert_AccountTable_1()
{
    $current_user = wp_get_current_user();

    if ($current_user->ID != 0)
    {
        return <<<HTML
            <form class="AlumniForm" method="post">
            <h3>Name</h3>
            <br>
            <table>
                <tbody>
                    <tr>
                        <th><label for="Username">Username</label></th>
                        <td><span>$current_user->user_login</span></td>
                        <td><span class="description">Usernames cannot be changed.</span></td>
                    </tr>
                    <tr>
                        <th><label for="current_user_firstname">First Name</label></th>
                        <td><input id="current_user_firstname" type="text" name="current_user_firstname" value="$current_user->user_firstname"/></td>
                    </tr>
                    <tr>
                        <th><label for="current_user_lastname">Last Name</label></th>
                        <td><input id="current_user_lastname" name="current_user_lastname" type="text" value="$current_user->user_lastname"/></td>
                    </tr>
                    <tr>
                        <th><label for="current_user_displayname">Display Name<span class="description">(required)</span></label></th>
                        <td><input id="current_user_displayname" type="text" value="$current_user->display_name"/></td>
                    </tr>
                </tbody>
            </table>

            <h3>Contact Info</h3>
            <br>
            <table>
                <tbody>
                    <tr>
                        <th><label for="current_user_email">E-mail<span class="description">(Required)</span></label></th>
                        <td><input id="current_user_email" type="text" value="$current_user->user_email"/></td>
                    </tr>
                    <tr>
                        <th><label for="current_user_url">Website</label></th>
                        <td><input id="current_user_email" type="text" value="$current_user->user_url"/></td>
                    </tr>
                </tbody>
            </table>
            <input type="submit" name="current_user_submitupdates">Update</input>
            </form>
HTML;
    }
    else
    {
        return <<<HTML
        <h3>Error - User not logged in</h3>
HTML;
    }
}

add_shortcode(\'InsertAccountTableI\', \'Insert_AccountTable_1\');

add_shortcode( \'userform\', \'wpse_75723_userform\' ); add_action( \'admin_post_update_user_email\', \'wpse_75723_update\' ); /** * Create the form. */ function wpse_75723_userform() { $here = esc_url( home_url( $_SERVER[\'REQUEST_URI\'] ) ); if ( ! is_user_logged_in() ) return \'You have to <a href="\' . wp_login_url( $here ) . \'">log in</a> to use this page.\'; $action = admin_url( \'admin-post.php\'); $user_id = get_current_user_id(); return "<form method=\'post\' action=\'$action\'> <input type=\'hidden\' name=\'action\' value=\'update_user_email\'> <input type=\'hidden\' name=\'redirect\' value=\'$here\'> <input type=\'hidden\' name=\'user_id\' value=\'$user_id\'> <input type=\'email\' name=\'email\' size=\'15\'> <input type=\'submit\'> </form>"; } /** * Update user email */ function wpse_75723_update() { if ( ! isset ( $_POST[\'user_id\'] ) ) die( \'no id\' ); $user_id = absint( $_POST[\'user_id\'] ); if ( ! current_user_can( \'edit_user\', $user_id ) ) die( \'not allowed\' ); if ( ! isset ( $_POST[\'email\'] ) ) die( \'no email\' ); if ( ! is_email( $_POST[\'email\'] ) ) die( \'invalid email\' ); $user = get_userdata( $user_id ); if ( empty ( $user->user_login ) ) die( \'user denied\' ); global $wpdb; $wpdb->query( $wpdb->prepare( "UPDATE {$wpdb->users} SET user_email = %s WHERE user_login = %s", $_POST[\'email\'], $user->user_login ) ); $location = isset ( $_POST[\'redirect\'] ) ? urldecode( $_POST[\'redirect\'] ) : home_url( \'/\' ); wp_redirect( $location, 303 ); exit; }

Ajax for non-logged-in users

我正在开发一个插件，它可以从未登录用户（如Like按钮）捕获一些数据，并根据post\\u id将其输入数据库。我的插件对登录用户很好，但非登录用户在AJAX中会产生“-1”响应。我猜这与写入数据库的身份验证障碍有关，但我正在寻找一种解决方法。有什么想法吗？编辑：以下是代码：http://pastebin.com/ghtvJNiW我是根据Love-It AJAX plugin 皮平。他在教程中说，这只适用于登录用户，但我需要让它也适用于未登录用户。

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Ajax for non-logged-in users