图像弹出图像引导4测试版

Question

我改为使用模态。

我将此网站更改为BS4测试版。在this page上点击图像时，它应该打开一个弹出图像或模态图像。

<div class="row mb-2"> 
 
    <div class="col-lg-2 col-sm-3 col-xs-4"> 
 
     <a href="javascript:popUp('ltrs/bluewater.htm')"><img src="photos/logo/bluewater.jpg" class="thumbnail img-fluid" alt="Bluewater Resort &amp; Casino"></a> 
 
    </div> 
 
    <div class="col-lg-2 col-sm-3 col-xs-4"> 
 
     <a href="javascript:popUp('ltrs/semhardrock.htm')"><img src="photos/logo/hardrock.gif" class="thumbnail img-fluid" alt="Hard Rock Hotel &amp; Casino" ></a> 
 
    </div> 
 
    <div class="col-lg-2 col-sm-3 col-xs-4"> 
 
     <a href="javascript:popUp('ltrs/trump.htm')"><img src="photos/logo/trump29.jpg" class="thumbnail img-fluid" alt="Trump 29 Casino"></a> 
 
    </div> 
 
    <div class="col-lg-2 col-sm-3 col-xs-4"> 
 
     <a href="javascript:popUp('ltrs/bestwestern.htm')"><img src="photos/logo/bestwestern.gif" class="thumbnail img-fluid" alt="Best Western"></a> 
 
    </div> 
 
    <div class="col-lg-2 col-sm-3 col-xs-4"> 
 
     <a href="javascript:popUp('ltrs/delsol.htm')"><img src="photos/logo/delsol.jpg" class="thumbnail img-fluid" alt="Casino Del Sol"></a> 
 
    </div> 
 
    <div class="col-lg-2 col-sm-3 col-xs-4"> 
 
     <a href="javascript:popUp('ltrs/ag.htm')"><img src="photos/logo/ag.jpg" class="thumbnail img-fluid" alt="Apache Gold Resort Casino"></a> 
 
    </div> 
 
</div>

我在页面的结束标记之前底部列出的BS的.js。

<script src="https://code.jquery.com/jquery-3.2.1.slim.min.js" integrity="sha384-KJ3o2DKtIkvYIK3UENzmM7KCkRr/rE9/Qpg6aAZGJwFDMVNA/GpGFF93hXpG5KkN" crossorigin="anonymous"></script> 
<script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.11.0/umd/popper.min.js" integrity="sha384-b/U6ypiBEHpOf/4+1nzFpr53nxSS+GLCkfwBdFNTxtclqqenISfwAzpKaMNFNmj4" crossorigin="anonymous"></script> 
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/4.0.0-beta/js/bootstrap.min.js" integrity="sha384-h0AbiXch4ZDo7tp9hKZ4TsHbi047NrKGLO3SEJAg45jXxnGIfYzk4Si90RDIqNm1" crossorigin="anonymous"></script> 

我在做什么错？这用于引导程序3时的工作。

Answer 1

看起来您的方法popup在全局范围内不可用。

请参阅错误从我的控制台：

我有当我点击物品例如罗斯威尔丰田这个错误。

另外，你有一个错误在该脚本：

$(document).ready(function() { 
     $("#submit-button").on('click', function(e) { 
     e.preventDefault(); // Prevent form submission 

     // you would need to set the following values from your form fields 
     var name = $("#form_name").val(); 
     var email = $("#form_email").val(); 
     var phone = $("#form_phone").val(); 
     var message = $("#form_message").val(); 
     var dataString = 'surname='+ name + '&email=' + email + '&phone=' + phone + '&message=' + message; 

     $.ajax({ 
     type: "POST", 
     url: "contact.php", 
     data: dataString, 
     success: function() { 
      $(".messages").html("Thank you for contacting me. I will be in touch soon.").css("visibility", "visible"); 
      $("#submit-button").trigger("reset") 
     $('.messages').fadeTo("slow", 1).fadeTo(3000, 0); 
     } 
     }); 
    }); 

您应该添加在 “）};”

Answer 2

我的朋友，有分号的微小误差.....

$(document).ready(function() { 
$("#submit-button").on('click', function(e) { 
e.preventDefault(); // Prevent form submission 

// you would need to set the following values from your form fields 
var name = $("#form_name").val(); 
var email = $("#form_email").val(); 
var phone = $("#form_phone").val(); 
var message = $("#form_message").val(); 
var dataString = 'surname='+ name + '&email=' + email + '&phone=' + phone + '&message=' + message; 

$.ajax({ 
    type: "POST", 
    url: "contact.php", 
    data: dataString, 
    success: function() { 
    $(".messages").html("Thank you for contacting me. I will be in touch soon.").css("visibility", "visible"); 
    $("#submit-button").trigger("reset") 
    $('.messages').fadeTo("slow", 1).fadeTo(3000, 0); 
    } 
}); 

}）;

查看倒数第二行之后trigger("reset");

，并请加上BS4的js jQuery的后直它是很好的做法做，它可能会导致意想不到的错误...