jquery .ajax()问题 [英] jquery .ajax() issue

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本文介绍了jquery .ajax()问题的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在使用Jquery 1.7.1并且遇到了问题..我正在使用我的脚本中的CRM工作并且正在努力完成页面但我仍然坚持这个问题..

I am using Jquery 1.7.1 and am having issues.. I'm working with a CRM in my script and am working to get the page finished but I'm stuck with this issue..

我的html:

<form class="collector" action="https://www.domain.biz/admin/transact.php" method="POST">
  <input type="hidden" name="method" value="NewProspect">
  <input type="hidden" name="campaignId" value="3">
  <input type="hidden" name="ipAddress" value="<?php echo $_SERVER['REMOTE_ADDR']; ?>">
  <fieldset>
    <div style=" padding-left: 50px">
      <table>
        <tr>
          <td><span style="color:red;">*</span>Your First Name: 
              <span id="rfvFirstName" style="display:none;">*</span>
          </td>
          <td><span style="color:red;">*</span>Your Last Name: 
              <span id="rfvFirstName" style="display:none;">*</span>
          </td>
                                            <td><span style="color:red;">*</span>Your Phone Number: </td>
                                            <td><span style="color:red;">*</span>Primary Email: </td>
                                        </tr>
                                        <tr>
                                            <td>
                                            <input name="firstName" type="text" id="firstName" style="width:150px;" value="">
                                            </td>
                                            <td>
                                            <input name="lastName" type="text" id="lastName" style="width:150px;" value="">
                                            </td>
                                            <td>
                                            <input name="phone" type="text" id="phone" class="" style="width:150px;" value="">
                                            </td>
                                            <td>
                                            <input name="email" type="text" id="email" class="required email" style="width:150px;" value="">
                                            </td>
                                        </tr>
                                    </table>
                                </div>
                                <div class="clear"></div>
                                <center>
                                    <input type="submit" name="continue" id="imgbtnSubmit" class="button" style="background-image: url('<?php echo base_url();?>inc/img/button-check.png');
                                background-repeat: no-repeat; width: 348px; height: 46px; border:none; background-color:transparent;" value="" />
                                </center>
                            </fieldset>
                            <p align="center" style="font-size:12px;">
                                &nbsp;
                            </p>
                        </form>

JS:

$('.collector').submit(function(){
    validate = true;
        $(this).find("input:text:visible:enabled").each(function() {
            if ($(this).attr("value") == "") {
                alert("Please fill in all fields!");
                $(this).focus();
                validate = false;
                return false;
            }
            else if ($(this).hasClass("email") && !$(this).attr("value").match(/@/)) {
                alert("Please enter an email address...");
                $(this).focus();
                validate = false;
                return false;
            }
        });
    if (validate != false)  { 
            $.ajax({
                url: $(this).attr('action'),
                type: 'POST',
                data: $(this).serialize(),
                success: function(response) { 
                    alert(response);
                }
            });
        }
    return false;
});

现在这两件事都有效,而且它们很好地协同工作......问题在于我没有得到任何回应,我不知道为什么。我想这是因为萤火虫说的话...... POST https://www.domain.biz/admin/transact.php 200 OK 1.04s jquery.js(8102行)

Now both of these things work, and they work together fine... the issue comes in that I don't get any response and I'm not sure why. I imagine it is because of what firebug is saying... POST https://www.domain.biz/admin/transact.php 200 OK 1.04s jquery.js (line 8102)

我的firebug中的这一行显示为红色,jquery.js中的第8102行是: xhr.send((s .hasContent&& s.data)|| null);

This line in my firebug is displayed as red, and the line 8102 in jquery.js is: xhr.send( ( s.hasContent && s.data ) || null );

推荐答案

你做跨域ajax请求?我下载了你的代码并做了一个简单的测试:

Do you make an cross domain ajax request ? I downloaded your code and make a simple test:

代码在


  • localhost :8080 / domain1 / a.php


  • localhost:8080 / domain2 / b.php

发生错误

代码


  • localhost:8080 / domain1 / a.php

向页面本身发出ajax请求

Make a ajax request to the page itself


  • (localhost:8080 / domain1 / a.php)

没有错误发生并获得预期的响应。

然后我搜索了[ jquery.ajax跨域请求]的答案,找到一些链接可能会有所帮助:
jQuery AJAX跨域

Then I googled the answer for [jquery.ajax cross domain request],and find some links may helps: jQuery AJAX cross domain

$.ajax({
 url:"testserver.php",
 dataType: 'JSONP', // Notice! JSONP <-- P
 success:function(json){
     // do stuff with json (in this case an array)
     alert("Success");
 },
 error:function(){
     alert("Error");
 },
 });

这篇关于jquery .ajax()问题的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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