On 14 Apr., 20:18, wyo [EMAIL PROTECTED] wrote:
Okay found the mistake myself just after posting.
$('#files').html('img src='+files[pos]+'
$('#pictures').html('img src='+files[pos]+'
div id=pictures
echo img src=\$files[0]\ class=\picture\;
/div
O. Wyss
On Apr 10, 11:58 pm, Jonathan Sharp [EMAIL PROTECTED] wrote:
Another approach would be to print out the images like:
echo img align=\center\ src=\$f\ style=\display: none;\
class=\photo\brbr;
Doesn't this load all the pictures right away? Since I expect soon a
few hundreds pictures I don't
On 4/11/07, wyo [EMAIL PROTECTED] wrote:
On Apr 10, 11:58 pm, Jonathan Sharp [EMAIL PROTECTED] wrote:
Another approach would be to print out the images like:
echo img align=\center\ src=\$f\ style=\display: none;\
class=\photo\brbr;
Doesn't this load all the pictures right away? Since I
In your foreach loop, instead of printing out the image, place all the
src references into a JavaScript array. Then, you would do something
like the following (this is a high level abstraction):
myImgSrcArray // array containing all your src references
curPos = 0
$(function(){
Another approach would be to print out the images like:
echo img align=\center\ src=\$f\ style=\display: none;\
class=\photo\brbr;
Then the following jQuery code:
$(function(){
$('img.photo:first').show();
$('img.next').bind('click', function() {
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