帶有彈出圖像的圖像 Bootstrap 4 測試版

Question

我改為使用模態。

我把這個網站改成了 BS4 測試版。 在此頁面上，當您單擊圖像時，它應該會打開一個帶有圖像的彈出窗口或模式。

 <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>

我在結束正文標記之前在頁面底部列出了 BS .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>

我究竟做錯了什么？ 這曾經在 Bootstrap 3 時有效。

Answer 1

您的方法彈出窗口似乎在全局范圍內不可用。

從我的控制台查看錯誤：

當我單擊 Roswell Toyota 等項目時出現此錯誤。

另外你在這個腳本上有一個錯誤：

    $(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);
        }
      });
    });

您應該在“)};”處添加

Answer 2

我的朋友，分號有一個小錯誤.....

$(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);
  }
});

});

在trigger("reset");后的倒數第二行看到trigger("reset");

並且請在 jquery 之后直接添加 BS4 js 這是一個很好的做法，它可能會導致意外錯誤......