IntegriTivity
01-06-2014, 06:42 PM
I am using Ultimate Fade-in Slideshow v 2.0 and need to substitute my own image array but have not found the correct syntax to make it work.
I have a directory of images that I'm reading with PHP and then splitting those images into two arrays (for the Left side and the Right side because I have two slideshows on the same page).
The code below is only for the left slideshow; I know how to make the other one separate.
I have not included the PHP code for getting the filenames ($banner) because it already works. I have included the creation of the JS array in case anyone else would like to use this code.
However, my problem is simply using my JS array as a variable for the imagearray. I have tried all kinds of syntax, to no avail.
I currently have a document.write turned on to show the results of the array variable I want to use and the test page is http://www.roseritchieacademy.com/indexTEST.php.
The variable array is this:
["images/banner/!headerLeft.jpg", "", "", ""],["images/banner/1.jpg", "", "", ""],["images/banner/11.jpg", "", "", ""],["images/banner/13.jpg", "", "", ""],["images/banner/15.jpg", "", "", ""],["images/banner/3.jpg", "", "", ""],["images/banner/5.jpg", "", "", ""],["images/banner/7.jpg", "", "", ""],["images/banner/9.jpg", "", "", ""]
<script language="JavaScript" type="text/JavaScript">
// Find out how many photos are in the array
var BannerCount = <?php echo count($banner); ?>;
// Put PHP array into a Javascript Array
var aBanner= new Array(BannerCount);
<?php
for($i=0; $i < count($banner); $i++){
$MyJava = "aBanner[".$i."] = '" . $banner[$i] . "';"; // develop the JS
echo $MyJava; // write the JS to browser
}
?>
var imagearrayL = new Array(); // Set up array needed for slideshow
var imagearrayR = new Array(); // Set up array needed for slideshow
// Add each photo to one of the arrays
for(i=0; i<=BannerCount-1; i++) // For each record in banner array
{
var $BannerPic = '["images/banner/' + aBanner[i] + '"]'; // put banner pic into JS variable
if(i % 2 == 0) { //If pic number is even (modulus remainder 0)
imagearrayL.push($BannerPic); // put it in the Left Array
}
else { //If pic number is odd, put it in the Right Array
imagearrayR.push($BannerPic); // put it in the Left Array
}
}
var mygallery=new fadeSlideShow({
wrapperid: "fadeshow1", //ID of blank DIV on page to house Slideshow
dimensions: [266, 200], //width/height of gallery in pixels. Should reflect dimensions of largest image
imagearray: imagearrayL, // THIS IS THE LINE THAT IS NOT WORKING ***************************************
displaymode: {type:'auto', pause:2500, cycles:0, wraparound:false},
persist: false, //remember last viewed slide and recall within same session?
fadeduration: 600, //transition duration (milliseconds)
descreveal: "ondemand",
togglerid: ""
})
alert ('Javascript is finished and was not broken.');
</script>
I have a directory of images that I'm reading with PHP and then splitting those images into two arrays (for the Left side and the Right side because I have two slideshows on the same page).
The code below is only for the left slideshow; I know how to make the other one separate.
I have not included the PHP code for getting the filenames ($banner) because it already works. I have included the creation of the JS array in case anyone else would like to use this code.
However, my problem is simply using my JS array as a variable for the imagearray. I have tried all kinds of syntax, to no avail.
I currently have a document.write turned on to show the results of the array variable I want to use and the test page is http://www.roseritchieacademy.com/indexTEST.php.
The variable array is this:
["images/banner/!headerLeft.jpg", "", "", ""],["images/banner/1.jpg", "", "", ""],["images/banner/11.jpg", "", "", ""],["images/banner/13.jpg", "", "", ""],["images/banner/15.jpg", "", "", ""],["images/banner/3.jpg", "", "", ""],["images/banner/5.jpg", "", "", ""],["images/banner/7.jpg", "", "", ""],["images/banner/9.jpg", "", "", ""]
<script language="JavaScript" type="text/JavaScript">
// Find out how many photos are in the array
var BannerCount = <?php echo count($banner); ?>;
// Put PHP array into a Javascript Array
var aBanner= new Array(BannerCount);
<?php
for($i=0; $i < count($banner); $i++){
$MyJava = "aBanner[".$i."] = '" . $banner[$i] . "';"; // develop the JS
echo $MyJava; // write the JS to browser
}
?>
var imagearrayL = new Array(); // Set up array needed for slideshow
var imagearrayR = new Array(); // Set up array needed for slideshow
// Add each photo to one of the arrays
for(i=0; i<=BannerCount-1; i++) // For each record in banner array
{
var $BannerPic = '["images/banner/' + aBanner[i] + '"]'; // put banner pic into JS variable
if(i % 2 == 0) { //If pic number is even (modulus remainder 0)
imagearrayL.push($BannerPic); // put it in the Left Array
}
else { //If pic number is odd, put it in the Right Array
imagearrayR.push($BannerPic); // put it in the Left Array
}
}
var mygallery=new fadeSlideShow({
wrapperid: "fadeshow1", //ID of blank DIV on page to house Slideshow
dimensions: [266, 200], //width/height of gallery in pixels. Should reflect dimensions of largest image
imagearray: imagearrayL, // THIS IS THE LINE THAT IS NOT WORKING ***************************************
displaymode: {type:'auto', pause:2500, cycles:0, wraparound:false},
persist: false, //remember last viewed slide and recall within same session?
fadeduration: 600, //transition duration (milliseconds)
descreveal: "ondemand",
togglerid: ""
})
alert ('Javascript is finished and was not broken.');
</script>