如何从mysql数据库中的图像创建图像幻灯片

Question

我有一个mysqli数据库和表单，允许我存储ID，名称和照片。照片的路径设置为＆＃34;图像＆＃34;服务器上的文件夹。我有一个可以

的查询

  

SELECT * FROM images WHERE name = $ pagetitle。

这在javascript幻灯片之外非常好用。当我在javascript中放置一个php命令来查找要显示的图像时，js只显示1个图像，而不是所有图像。

任何帮助都将不胜感激，谢谢。

有问题的代码部分如下......

的index.php

＆＃13;

＆＃13;

            <!-- Image Slide Show Start -->
  <div style="display: flex; justify-content: center;">
  <img align="middle" src="" name="slide" border=0 width=300 height=375>
				
<script>

<?php 
require('dbconnect.php');
$data = mysql_query("SELECT * FROM images WHERE name= '$pagetitle'");
$image = mysql_fetch_array( $data );



?>
//configure the paths of the images, plus corresponding target links
slideshowimages("<?php echo "/images/".$image['photo'] . ""?>")

//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 -->

＆＃13;

＆＃13;

＆＃13;

Answer 1

您的更新查询存在语法错误，在字段之间使用,，您还必须在2 '中包含字符串：

$query = "UPDATE page_content SET PageTitle='$pageTitle',
PageContent='$PageContent', PageContent2='$PageContent2' WHERE PageId='$PageId'";

Answer 2

您在查询中的变量和<ul class="slider-thumbs"> <li class="reason-item"><a>Item 1</a></li> <li class="reason-item"><a class="selected">Item 2</a></li> <li class="reason-item"><a>Item 3</a></li> <li class="reason-item"><a>Item 4</a></li> <li class="reason-item"><a>Item 5</a></li> </ul>字段之间错过了,。

''

OR

$sql = "UPDATE page_content SET PageTitle='$pageTitle', 
       PageContent='$PageContent', PageContent2='$PageContent2' 
       WHERE PageId='$PageId'";

// check query executed successfully or get error
$result = mysqli_query($conn,$sql) or die(mysqli_error($conn));

希望它会对你有所帮助:)。

Answer 3

试试这个：

$sql = "SELECT * FROM images WHERE name= '$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 ?>)