获取mysql表中的图像以显示在javascript幻灯片中

Question

I have some images in the table images , the fields are id , name and photo . Where the image exists in photo .

At the minute, the code below not getting any images, although there should be about 4 images that match the query. The images that meet the query should go into the slideshowimages("") variable.

<?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($image) {

// if there are images for this page, run the javascript
?><script>


//configure the paths of the images, plus corresponding target links

        //NEED TO GET ALL RELEVANT IMAGE LOCATIONS INTO LINE BELOW
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> <?php
} else {
    // If there are not any images for this page, leave the space blank
    echo "";
    }

// Close the mysql connection
$conn->close();
?>      

The JavaScript that is in the head

<script language="JavaScript1.1">

var slideimages=new Array()
var slidelinks=new Array()
function slideshowimages(){
for (i=0;i<slideshowimages.arguments.length;i++){
slideimages[i]=new Image()
slideimages[i].src=slideshowimages.arguments[i]
}
}

</script>   

Answer 1

The nice and simply way, is to use an AJAX call, to get your image urls in a JSON array, which you can parse to a javascript array, and iterate and so on. In that case, the added bonus is that you can separate your code to different files by language, and it makes a way nicer code.

But in your current code, you have to iterate your mysqli results with a simple loop. For example:

//...
// 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'] . "'");
//...

In this case, your $directory variable be like something like this:

'/images/image1.jpg','/images/image2.jpg','/images/image3.jpg'

And you hopefully can pass it to the javascript function as an argument list.

I hope I could help.