如何从mysql数据库中的图像创建图像幻灯片

Question

I have a mysqli database and form which allows me to store an id, name and photo. The path of the photo is set to an "images" folder on the server. I have a query which can

SELECT * FROM images WHERE name = $pagetitle.

This works absolutely fine, outside of the javascript slideshow. When i put a php command in the javascript where it is looking for which images to display, the js only shows 1 image, and not ALL images.

Any help would be appreciated, thanks.

The section of the code in question is below...

index.php

            <!-- 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 -->

</div>

Answer 1

Try this:

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