2017-08-17 80 views
-1

嗨我有模型与反馈的形式。模型提交后应该显示另一个感谢模型也需要存储本地存储。帮我。Javascript显示谢谢你模式提交

<div id="myModal2" class="modal fade" role="dialog"> 
<form action="" method="post" onSubmit="popUpToggle();" id="myModal2" > 
<div class="form_fields feedbackfield"> 
<textarea id="txtMessage" name="txtMessage" rows="5" class="form-control contact-input" placeholder="Message *" required></textarea> 
</div> 
<div class="form_fields"> 
<input type="submit" class="contact-submit" name="SUBMIT" value="SUBMIT"> 
</div> 
</form> 
</div> 

<div id="myModal4" class="modal fade" role="dialog"> 
     <div class="modal-dialog">      
       <div class="popupHead3"> 
        Thank you for your feedback  
       </div>  
     </div> 

的Javascript:

setTimeout(function() { 
      if(localStorage.getItem("modalPopup")){$("#myModal2").modal('hide');} 
      else{$('#myModal2').modal('show');} 

     }, 5000); 

     function popUpToggle(){   
      localStorage.setItem("modalPopup", "commit"); 
     } 

回答

0

您可以尝试AJAXformsubmit然后用

$('#myModal4').modal({show: 'true'}); 
+0

我该怎么做..?我给popUpToggle()作为onSubmit – user2172424

0

请尝试以下代码。

$("#myModal2").submit(function(){ 
    event.stopPropagation(); 
    event.preventDefault(); 
    $.ajax({ 
    type: "POST", 
    url: /* submit url or form action url */, 
    data: $(this).serialize(), 
    success: function(d){ 
     /* do here what you want to do */ 

     $("#myModal4").modal({show: "true"}); 
    }, 
    dataType: "JSON" 
    }); 
}); 
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