Question

我已成功使用ajax通过ajax在线刷新联系表单。我正在使用发送到mysql的数据时遇到更多困难。我很欣赏我到目前为止所做的并不打算通过ajax刷新，所以它可能需要一些工作。这就是我所得到的......任何帮助都表示赞赏。

表格

<form name="email_list" action="">

<p><strong>Your Email Address:</strong><br/>
<input type="text" name="email" id="email" size="40">
<input type="hidden" name="sub" id="sub" value="sub">

<p><input type="submit" name="submit" value="Submit Form" class="email_submit"></p>
</form>

JQuery的

$(function() {    
$('.email_submit').submit(function() { 

var email = $("input#email").val();  
            if (name == "") { 
            $("input#email").focus();  
            return false;
        }  
var sub = $("input#sub").val();  
            if (name == "") {  
            $("input#sub").focus();  
            return false;
        }       

var dataString = $(this).serialize();

//alert (dataString);return false;  
/*$.ajax({  
    type: "POST",  
    url: "mailing_list_add2.php",  
    data: dataString,  
    success: function() {  
        $('#display_block')                      
        .hide()  
        .fadeIn(2500, function() {  
            $('#display_block');  
        }); 
    }  
}); 
return false; 
});*/

});

PHP

<?php
// connects the database access information this file
include("mailing_list_include.php");

// the following code relates to mailing list signups only
if (($_POST) && ($_POST["action"] == "sub")) {

if ($_POST["email"] == "") {
        header("Location: mailing_list_add.php");
        exit;
    } else {
        // connect to database
        doDB();

        // filtering out anything that isn't an email address
        if ( filter_var(($_POST["email"]), FILTER_VALIDATE_EMAIL)  == TRUE) {
            echo '';
        } else {
            echo 'Invalid Email Address';
            exit;
        }

        // check that the email is in the database
        emailChecker($_POST["email"]);

        // get number of results and do action
        if (mysqli_num_rows($check_res) < 1) {
            // free result
            mysqli_free_result($check_res); 

            // cleans all input variables at once
            $email = mysqli_real_escape_string($mysqli, $_POST['email']);

            // add record
            $add_sql =  "INSERT INTO subscribers (email) VALUES('$email')";
            $add_res =  mysqli_query($mysqli, $add_sql)
                        or die(mysqli_error($mysqli));
            $display_block = "<p>Thanks for signing up!</p>";

            // close connection to mysql
            mysqli_close($mysqli);
        } else {
            // print failure message
            $display_block = "You're email address - ".$_POST["email"]." - is already subscribed.";
    }
}
}

?>
<html>
<?php echo "$display_block";?>
</html>

**进行更改后更新如上：）

Answer 1

有一些问题，例如：

您需要将javascript附加到事件（表单提交），因此第一部分将是：

$('form').submit(function() {
您需要发送所有（正确的...）表单值，而不仅仅是您的php工作的电子邮件地址：

$('form').submit(function() {
```
   var dataString = $(this).serialize();
```

Answer 2

您错过了服务器中的数据（删除服务器中的HTML标记）：

$.ajax({  
        type: "POST",  
        url: "mailing_list_add.php",  
        data: dataString,  
        success: function(data) {  
            $('#email_add').html("<div id='response'></div>");  
            $('#response').html("<h2>Successfully added to the database.")                      
            .hide()  
            .fadeIn(2500, function() {  
                $('#response').html(data);  
            }); 
        }  
    });

使用ajax提交数据进行刷新

2 个答案: