php - 我有一个带有 JQuery 的表单，一旦提交两次就会刷新页面(当用户出现错误时)

Question

刚开始使用 JQuery 进行 AJAX 调用，我以为是一个简单的脚本，但现在却让我困惑了两天。基本上我希望将表单数据发布到服务器，如果用户输入的数据正常，则返回一条“谢谢”消息，或者返回一条消息说有错误。

在这两个事件中，我想要返回的是 HTML。然而，这种情况正在发生，当表单数据中有错误，并且再次提交表单时，会出现页面刷新(我知道这是因为按 F5 会产生浏览器默认对话框，要求发送或取消)。

此页面刷新仅在第二次提交表单(初始 AJAX 调用时从服务器返回的 HTML)后发生。否则似乎没有其他故障。用户输入并通过初始 AJAX 调用发送到服务器的数据也可以。

这就是服务器将 HTML 发送回客户端后出现错误时发生的情况。我已经在下面发布了代码和 HTML，此时任何想法都会受到欢迎，因为我对 JQuery/AJAX 东西真的很陌生!

<!-- found on the contact page -->
<script type="text/javascript">

        $(document).ready( function() {
            $( '#submit' ).click( function() {
                var payload = $( '#form' ).serialize();
                // var name = $('input[name=name]');

                // var data = 'name=' + name.val();

                $.ajax({
                    url: '/contact/post/',
                    type: 'post',
                    data: payload, 
                    cache: false,
                    success: function( html ) {      console.log( html );   
                        $( '#formbox' ).html( html );
                    }
                });
                return false;
            }); // close of #submit

        });

    </script>

<!-- ... ... -->

<div id="formbox">
        <form 
            id="form" 
            method="post" 
            accept-charset="utf-8" 
            action="#">

            <!-- using only one field at the moment (debugging purposes) -->
            <div class="lft"><p><label for="_name">Name</label>&nbsp;</p></div>
            <div class="rgt"><p>&nbsp;<input type="text" id="_name" name="name" size="16" maxlength="32" value="Elizabeth Q" /></p></div>

            <div class="lft"><p>&nbsp;</p></div>
            <div class="rgt"><p>&nbsp;
                <input type="submit" id="submit" name="submit" value="Send Message" />
            </p></div>

        </form>
    </div>




<!-- what is returned from server if no errors with user data -->
<div class="both">
        <p>Thank you for contacting us, please expect a prompt reply shortly.</p>
    </div>


<!-- what is returned from server if there are errors with user data -->
<form 
        id="form" 
        method="post" 
            accept-charset="utf-8" 
            action="#">

            <div class="both">
                <p class="error">Please amend the following error(s) shown below before trying again.</p>
                <br />
                <ul>

                <?php foreach( $this -> get( 'logger' ) -> export() as $error ): ?>
                    <li><p><?php echo $error; ?></p></li>
                <?php endforeach; ?>
            </div>

            <div class="lft"><p><label for="_name">Name</label>&nbsp;</p></div>
            <div class="rgt"><p>&nbsp;<input type="text" id="_name" name="name" size="16" maxlength="32" value="<?php echo $this -> get( 'name' ); ?>" /></p></div>

            <div class="lft"><p>&nbsp;</p></div>
            <div class="rgt"><p>&nbsp;
                <input type="submit" id="submit" name="submit" value="Send Message" />
            </p></div>

        </form>


<!-- the php file that determines user data is valid or not and returns response -->
final class QPage_Handler_Ajax extends QPage_Handler_Validator {
        public function __construct() {
            $this -> initialise();
            $this -> id = 'site'; 
        }

        public function execute( QDataspace_Interface $dataspace ) {
            if( !$this -> validate( $request = QRegistry::get( 'request' ), QRegistry::get( 'logger' )  ) ) {
                // execute this handler as errors found
                $this -> handler -> execute( $dataspace );
            } else { 
                // no errors with data sent to server
                $page = new QPage_View( $request = QRegistry::get( 'request' ) );
                $page -> render( 'contact/post/body.tpl' ); 
            }
        }

        protected function initialise() {
            $this -> forward( new QPage_Handler_Ajax_Success() );

            $this -> addCondition( QValidator::factory()
                -> addCondition( new QValidator_Condition_Required( 'name', 'The field "name" is required.' ) )
                -> addCondition( new QValidator_Condition_Expression( 'name', '/^[a-zA-Z ]+$/', 'The field "name" has illegal character(s).' ) )
                -> addCondition( new QValidator_Condition_Size_Maximum( 'name', 32, 'The field "name" must not exceed 32 characters.' ) )
            );
        }
    }

    final class QPage_Handler_Ajax_Success extends QPage_Handler {
        public function __construct() {
            $this -> id = 'site';
        } 

        public function execute( QDataspace_Interface $dataspace ) { 
            $page = new QPage_View( $request = QRegistry::get( 'request' ) );
            $page -> set( 'logger', QRegistry::get( 'logger' ) );
            $page -> render( 'contact/post/error.tpl' ); 
        }
    }

非常欢迎和赞赏有关如何解决我遇到的问题的任何建议。请记住，我刚开始使用 JQuery，但我已经使用 PHP 多年了。

Answer 1

我想，你应该使用event.preventDefault()

所以，试试这个:

$( '#form' ).submit( function(event) {
                event.preventDefault();
                var payload = $( '#form' ).serialize();
                // var name = $('input[name=name]');

                // var data = 'name=' + name.val();

                $.ajax({
                    url: '/contact/post/',
                    type: 'post',
                    data: payload, 
                    cache: false,
                    success: function( html ) {      console.log( html );   
                        $( '#formbox' ).html( html );
                    }
                });
                return false;
            }); // close of #submit