如何显示来自 mysql 数据库的图像幻灯片
[英]How to display a slideshow of images from a mysql database
问题描述
我有一个名为images的表，字段为id, name, photo ， photo字段为varchar因为图片使用 php 表单存储到服务器上。 在我的代码只显示表格中的 1 张图像的那一刻，请有人帮助我，以便它使用查询对所有图像进行幻灯片放映：
SELECT * FROM images WHERE name='$pagetitle'
$pagetitle在脚本的其他地方定义。
问题出在 index.php 文件中
索引.php
<!-- Image Slide Show Start --> <div style="display: flex; justify-content: center;"> <img align="middle" src="" name="slide" border=0 width=300 height=375> <?php require('mysqli.php'); $data = $conn->query("SELECT * FROM images WHERE name= '$pagetitle'"); $image = mysqli_fetch_array( $data ); $directory = ("/images/".$image['photo'] . ""); $conn->close(); ?> <script> //configure the paths of the images, plus corresponding target links slideshowimages("<?php echo $directory ?>") //configure the speed of the slideshow, in miliseconds var slideshowspeed=2000 var whichlink=0 var whichimage=0 function slideit(){ if (!document.images) return document.images.slide.src=slideimages[whichimage].src whichlink=whichimage if (whichimage<slideimages.length-1) whichimage++ else whichimage=0 setTimeout("slideit()",slideshowspeed) } slideit() </script> </div><br><br> <!-- Image Slide Show End -->
2 个解决方案
解决方案1
2 已采纳 2016-03-10 15:33:47
解决的代码如下：
<?php // Connect to the database require('mysqli.php'); // Query for a list of all existing files $sql = "SELECT * FROM images WHERE name= '$pagetitle'"; $result = $conn->query($sql); $directory = ''; while( $image = $result->fetch_assoc() ) $directory .= ($directory != '' ? "," : '') . ('"/images/'.$image["photo"] . '"'); // Check if it was successfull if($directory != '') { // if there are images for this page, run the javascript ?><script> //configure the paths of the images, plus corresponding target links slideshowimages(<?php print $directory ?>) //configure the speed of the slideshow, in miliseconds var slideshowspeed=2000 var whichlink=0 var whichimage=0 function slideit(){ if (!document.images) return document.images.slide.src=slideimages[whichimage].src whichlink=whichimage if (whichimage<slideimages.length-1) whichimage++ else whichimage=0 setTimeout("slideit()",slideshowspeed) } slideit() </script> <? } else { // If there are not any images for this page, leave the space blank echo ""; } // Close the mysql connection $conn->close(); ?>
解决方案2
0 2020-03-20 12:34:41
<?php
$servername = "";
$username = "";
$password = "";
$dbname = "";
$conn = new mysqli($servername, $username, $password, $dbname);
//Make connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//SQL query
$sql = "SELECT * FROM TABLE_NAME";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$dir= "" . $row["field"]. "";
echo" <div class='content section' style='max-width:500px'>
<img class='slides' src='$dir' style=width:100%></div>";
}
} else {
echo "Empty Gallery";
}
?>
<html>
<body>
<script>
var index = 0;
slideshow();
function slideshow() {
var i;
var x = document.getElementsByClassName("slides");
for (i = 0; i < x.length; i++) {
x[i].style.display = "none";
}
index++;
if (index > x.length) {index = 1}
x[index-1].style.display = "block";
setTimeout(slideshow, 3000); // Change slides every 3 seconds
}
</script>
</body>
</html>
如何从mysql数据库中的图像创建图像幻灯片
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