Question

如何将id呈现为页面：

下面的代码将从'resources / php / disImg'目录按顺序显示我的所有照片。

我的目标是点击图片并在目录中的图片模式中显示“onclick”图片。

<!-- The Modal -->
<div id="myModal" class="modal">
    <span class="close">&times;</span>
    <img class="modal-content" id="img01">
</div>

<!-- Displays all photos from folder -->
<div class="containerPhotos">
        <?php 
            $dirname = "resources/php/disImg/";
            $images = glob($dirname."*.{jpg,jpeg,png}",GLOB_BRACE);
            natcasesort($images);

            foreach($images as $randomImage) {
                echo '<img id="myImg" src="'.$randomImage.'" class="photo" />';
            }
        ?>
</div>

目前，我可以从php / image目录中使用（id =“myImg”）定位第一个图像，并将其显示在模态中。

点击第一张照片后弹出模态：

<script>
// Get the modal
var modal = document.getElementById('myModal');

var img = document.getElementById('myImg');
var modalImg = document.getElementById("img01");
    img.onclick = function(){
    modal.style.display = "block";
    modalImg.src = this.src;
}

// Get the <span> element that closes the modal
var span = document.getElementsByClassName("close")[0];

// When the user clicks on <span> (x), close the modal
span.onclick = function() { 
    modal.style.display = "none";
}
</script>

任何人都知道如何制作它所以我点击id =“myImg”的图像会弹出并显示在模态中吗？

我无法通过第n次选择获得解决方案，但我认为这样可行。

下面是模态弹出窗口的.css

#myImg {
    border-radius: 5px;
    cursor: pointer;
    transition: 0.3s;
}

#myImg:hover {opacity: 0.7;}

/* The Modal (background) */
.modal {
    display: none; /* Hidden by default */
    position: fixed; /* Stay in place */
    z-index: 1; /* Sit on top */
    padding-top: 100px; /* Location of the box */
    left: 0;
    top: 0;
    width: 100%; /* Full width */
    height: 100%; /* Full height */
    overflow: auto; /* Enable scroll if needed */
    background-color: rgb(0,0,0); /* Fallback color */
    background-color: rgba(0,0,0,0.9); /* Black w/ opacity */
}

/* Modal Content (image) */
.modal-content {
    margin: auto;
    display: block;
    width: 100%;
    max-width: 1000px;
}

/* Add Animation */
.modal-content, #caption {    
    -webkit-animation-name: zoom;
    -webkit-animation-duration: 0.6s;
    animation-name: zoom;
    animation-duration: 0.6s;
}

@-webkit-keyframes zoom {
    from {-webkit-transform:scale(0)} 
    to {-webkit-transform:scale(1)}
}

@keyframes zoom {
    from {transform:scale(0)} 
    to {transform:scale(1)}
}

/* The Close Button */
.close {
    position: absolute;
    top: 15px;
    right: 35px;
    color: #f1f1f1;
    font-size: 40px;
    font-weight: bold;
    transition: 0.3s;
}

.close:hover,
.close:focus {
    color: #bbb;
    text-decoration: none;
    cursor: pointer;
}

/* 100% Image Width on Smaller Screens */
@media only screen and (max-width: 700px){
    .modal-content {
        width: 100%;
    }
}

Answer 1

有几种不同的方法可以实现这一点，但第一个方法就是在PHP foreach循环中为每个img HTML元素添加一个onclick事件。

        foreach($images as $randomImage) {
            echo '<img id="myImg" onclick="showImage(this)" src="'.$randomImage.'" class="photo" />';
        }

然后你需要一个同名的javascript函数，你可以得到这样的源代码。

function showImage(imgElement) { 
   var src = imgElement.getAttribute("src");
   /* Do the stuff with your modal */}

Answer 2

感谢帮助人员，这里有最终的构建工作！

Javascript /

var modal = document.getElementById('myModal');
var modalImg = document.getElementById('img01');

function showImage(imgElement) { 
   var src = imgElement.getAttribute("src");
   modal.style.display = "block";
   modalImg.src = src;
}

// Get the <span> element that closes the modal
var span = document.getElementsByClassName("close")[0];

// When the user clicks on <span> (x), close the modal
span.onclick = function() { 
    modal.style.display = "none";
}

PHP /

foreach($images as $randomImage) {
    echo '<img class="photo" onclick="showImage(this)" id="myImg" src="'.$randomImage.'" />';
}

使用模态弹出查看器显示目录中的PHP图像

2 个答案: