如何发送表单而不刷新页面？ [英] How to send a form without refreshing the page?

问题描述

我不知道PHP。

我不希望用户转到 http://www.example.com /feedback-thanks.html 发送表单后。我想要的是发送按钮下方的文本或任何不刷新页面。

我删除了 header（Location ：http://www.example.com/feedback-thanks.html）; 但我没有收到电子邮件，用户被重定向到feedback.php ... :(

html

 < form method =postaction =feedback.php> 
< input name =emailtype =email/> 
< textarea name =messageid =feedback -textareaautofocus =autofocusrequired =required>< / textarea> 
 < / form> 
  

feedback.php

 <？php 
 $ email = $ _REQUEST ['email']; 
 $ message = $ _REQUEST [消息']; 
 
 mail（[e-mail]，Subject Here，
 $ message，From：$ email）; 
 header（Location ：http：//www.example.com/fee dback-thanks.html）; 
？> 
  

解决方案

您必须使用 $ .ajax（）为此

• http://net.tutsplus.com/tutorials/javascript-ajax/5-ways-


• http://www.sitepoint.com/ajax-jquery/

这是我在我的一个应用程序

  $（'＃submitSearch'）。click（function（）{
 
 var formData = {
 searchData：$（'＃searchData'）.val（），
}; 
 
 $ .ajax（{
 url：siteUrl +' fetch / search'，
 type：POST，
 data：formData，
 cache：true，
 beforeSend：function（）{
 jQuery（'＃main '）的.html（' < d iv class =loading>< img src ='+ siteUrl +'resources / imgs / ajax-loader.gifalt =正在加载.../>< / div>'）; 
}，
 success：function（data）{
 jQuery（'＃main'）。empty（）; 
 jQuery（'＃main'）。append（data）; 
}，
 error：function（x，e）{
 if（x.status == 0）{
 alert（'您正在离线!! \\\
请检查你的网络。'）; 
} else if（x.status == 404）{
 alert（'Requested URL not found。'）; 
} else if（x.status == 500）{
 alert（'Internel Server Error。'）; 
} else if（e =='parsererror'）{
 alert（'Error.\\\
Parsing JSON Request failed。'）; 
} else if（e =='timeout'）{
 alert（'Request Time out。'）; 
} else {
 alert（'Unknow Error.\\\
'+ x.responseText）; 
} 
} 
}）; 
返回false; 
}）; 
  

I don't know PHP.

I don't want the user to go to http://www.example.com/feedback-thanks.html after sending the form. What I want is text below the sent button or anything without refreshing the page.

I removed header( "Location: http://www.example.com/feedback-thanks.html" ); but i don't receive the e-mail and the user is redirected to feedback.php... :(

html

<form method="post" action="feedback.php">
<input name="email" type="email"/>
<textarea name="message" id="feedback-textarea" autofocus="autofocus" required="required" ></textarea>
<button type="submit" id="feedback-button-send">send</button>
</form>

feedback.php

<?php
  $email = $_REQUEST['email'] ;
  $message = $_REQUEST['message'] ;

  mail( "[e-mail]", "Subject Here",
    $message, "From: $email" );
  header( "Location: http://www.example.com/feedback-thanks.html" );
?>

解决方案

You will have to use $.ajax() for that

• http://net.tutsplus.com/tutorials/javascript-ajax/5-ways-to-make-ajax-calls-with-jquery/
• http://www.sitepoint.com/ajax-jquery/

this is what I use in one of my apps

$('#submitSearch').click(function() {

        var formData = {
            searchData: $('#searchData').val(),
        };

        $.ajax({
            url: siteUrl + 'fetch/search',
            type: "POST",
            data: formData,
            cache: true,            
            beforeSend:function(){
                jQuery('#main').html('<div class="loading"><img src="' + siteUrl + 'resources/imgs/ajax-loader.gif" alt="Loading..." /></div>');
            },
            success: function(data) {
                jQuery('#main').empty();
                jQuery('#main').append(data);
            },
            error:function(x,e){
                if(x.status==0){
                    alert('You are offline!!\n Please Check Your Network.');
                }else if(x.status==404){
                    alert('Requested URL not found.');
                }else if(x.status==500){
                    alert('Internel Server Error.');
                }else if(e=='parsererror'){
                    alert('Error.\nParsing JSON Request failed.');
                }else if(e=='timeout'){
                    alert('Request Time out.');
                }else {
                    alert('Unknow Error.\n'+x.responseText);
                }
            }
        });
        return false;
    });

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