weixin_33708432 2017-01-26 10:57 采纳率: 0%
浏览 9

Ajax无法正常工作。

i am trying to register customer and want live checking of email is registered with us or not. mobile number is registered with us, i am not getting the required result in both the event it give me same result if its registered of not. can you please help me. result in both event is ///"Already Registered with this E-mail";

if(isset($_POST['name'])){
$sellerEmail =$_POST['name'];

    $query_verify_mob= new ParseQuery("Vyapari");
    $query_verify_mob->EqualTo("sellerEmail", $sellerEmail);
    $result=count($query_verify_mob);




if($result ==1)
{
    echo "<span style='color:red;'>Already Registered with this E-mail...</span>";


}
else
{     
    echo "<span style='color:green;'>Email ID is Available...</span>"; 
}}.
  • 写回答

1条回答 默认 最新

  • weixin_33693070 2017-01-26 11:54
    关注

    0 input in what ParseQuery is, aniway u must just check the count of emails with that input email.

    Instead of $result == 1 u should use $result > 0.

    评论

报告相同问题?

悬赏问题

  • ¥100 set_link_state
  • ¥15 虚幻5 UE美术毛发渲染
  • ¥15 CVRP 图论 物流运输优化
  • ¥15 Tableau online 嵌入ppt失败
  • ¥100 支付宝网页转账系统不识别账号
  • ¥15 基于单片机的靶位控制系统
  • ¥15 真我手机蓝牙传输进度消息被关闭了,怎么打开?(关键词-消息通知)
  • ¥15 装 pytorch 的时候出了好多问题,遇到这种情况怎么处理?
  • ¥20 IOS游览器某宝手机网页版自动立即购买JavaScript脚本
  • ¥15 手机接入宽带网线,如何释放宽带全部速度