How do I create an tag that look for the image name in 10 different folders and if not found outputs a placeholder image using Jav开发者_Python百科ascript?
Thanks.
I would go with such script:
<script type="text/javascript">
var arrPossibleImageFolders = ["folder1", "folder2", "folder3"];
var strDefaultImage = "nopicture.gif";
function TryDifferentFolder(oImage) {
var folderIndex = parseInt(oImage.getAttribute("folder_index"), 10);
if (isNaN(folderIndex))
folderIndex = 0;
folderIndex++;
if (folderIndex >= arrPossibleImageFolders.length) {
oImage.src = strDefaultImage;
return;
}
var lastSlashIndex = oImage.src.lastIndexOf("/");
var strNewSource = arrPossibleImageFolders[folderIndex] + "/" + oImage.src.substr(lastSlashIndex);
oImage.src = strNewSource;
oImage.setAttribute("folder_index", folderIndex + "");
}
</script>
Then have such image tag to trigger the code:
<img src="folder1/mypicture.gif" onerror="TryDifferentFolder(this);" />
This will search for "mypicture.gif" in folder1
first, then if not found (onerror triggered) will try in folder2 etc, finally loading the default picture.
var img = new Image();
img.onload = function(){ alert("I am loaded"); };
img.onerror = function(){ alert("I am not loaded"); };
img.src = "/foo/bar.gif";
If you get onerror, try setting the source again to the next image in your list.
Really, you should do something like this on the server side, where the files actually are; make a script that looks for the image and sends what it finds.
(Edited to remove my original JS solution, since it was bad, and others have suggested better ones.)
Sounds like you may need to use a bunch of onerror handlers. Try to load some image, catch possible error till you run out of candidates. Load placeholder as the last image.
I'd solve it with Ajax, try to load each image, if you get a 200 that means the image exists, if 404 try next image.
var images = ['/images/1.jpg', '/images/2.jpg', '/images/3.jpg'];
var imageExists = false;
for(var i = 0; i < images.length; i++){
if(!imageExists){
$.ajax({
url: images[i],
async: false,
success: function(){
// use images[i] as your image
imageExists = true;
}
});
}
}
if(!imageExists) {
// use default image
}
Since you are asking about how to do it with a tag i'll show a solution, although i'd never do it myself:
<script>
urlsToTry = ['http://url2.com','http://url3.com','http://you.get.the.picture.com','http://www.google.com/images/logos/ps_logo2.png'];
</script>
<img src="http://url1.com" style="display: none;" onload="this.style.display='block';" onerror="if(urlsToTry.length) { this.src = urlsToTry.shift(); } else { alert('none of the urls worked!'); }">
You can see an example of it here: http://jsfiddle.net/LLQLH/1/
One of the downsides to this approach is that it tries to load the picture serially which can cause quite a wait if they all fail. The parallel approach using proper js would be the way to go imo:
<script>
var loadQueue={};
var urlsToTry = ['http://url1.com','http://url2.com','http://url3.com','http://you.get.the.picture.com'];
function showTag(url) {
var tag = document.getElementById('myimage');
if(url) tag.src = url;
tag.style.display='inline-block';
}
function onLoadEventHandler(url) {
return function() {
delete loadQueue[url];
for(var i = 0, img; img = loadQueue[i]; i++) {
img.onload = null;
}
loadQueue = null;
showTag(this.src);
}
}
function onErrorEventHandler(url) {
return function() {
delete loadQueue[url];
for(var i in loadQueue) return;
showTag(); //we will only get here if loadQueue is empty
}
}
for(var i = 0, url; url = urlsToTry[i]; i++) {
loadQueue[url] = new Image();
loadQueue[url].onload = onLoadEventHandler;
loadQueue[url].onerror = onErrorEventHandler(url);
loadQueue[url].src = url;
}
</script>
<img id="myimage" src="http://www.google.com/images/logos/ps_logo2.png" style="display: none;">
You can see it working here: http://jsfiddle.net/DJ2SH/
lol, this cant be done only with some tag, you should propably make js(jquery/ajax) function to search for that file and then update href attribute of img... or you should explain your intends a little more
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