我想为我的所有图像制作一个弹出式模式按钮,但只有第一个有效,我知道我必须添加一些foreach代码,但我不知道如何。<div id="myModal" class="modal"> <span class="close">×</span> <img class="modal-content" id="img01"></div><div class="gallery"> <div class="containerPhotos"> <?php require_once 'classes/dbh.php'; if (isset($_GET['entry_id'])) { $entry_id = $_GET['entry_id']; $query = mysqli_query($conn, "SELECT * FROM images WHERE entry_id='$entry_id'"); if ($query->num_rows > 0) { while ($row = $query->fetch_assoc()) { $dirname = 'uploads/' . $row["file_name"]; echo '<img class="photo" onclick="showImage(this)" id="myImg" src="' . $dirname . '" />'; } } } ?> </div></div> <script> 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; } var span = document.getElementsByClassName("close")[0]; // When the user clicks on <span> (x), close the modal span.onclick = function () { modal.style.display = "none"; } </script>我希望使用foreach代码为所有人工作,但我不知道如何。
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