Rank: Member
Groups: Member
Joined: 11/16/2010(UTC)
Posts: 11
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Hi, I am using the aurigma image uploader 6.0. It is displaying the image uploader well in my page. When i hit that page through ajax it is not displaying and error occurs. I called the below function and assign the image uploader to the div through ajax. When i run the file in the browser it is running but when i assign it in the div html is assigning but the uploader is not running and showing the syntax error as Code:'syntax error
[Break on this error] if (ImageUploader1_w.getControlType() == "Java")\n'
Please help me... Code:<script>
function funTest()
{
$.ajax({
type : "POST",
url : g_site_path+"/galleryupload/gallery/",
data : 'rand='+Math.random(),
dataType : 'script',
success: function(script){
alert(script);
alert(document.getElementById('popDiv'));
document.getElementById('popDiv').innerHTML = script;
alert(document.getElementById('popDiv').innerHTML);
}
});
}
</script>
Edited by moderator Friday, January 7, 2011 7:07:58 AM(UTC)
| Reason: Not specified
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Rank: Advanced Member
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Joined: 6/16/2009(UTC)
Posts: 134
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Hello! If you want to insert Image Uploader into the popup div you should use getHtml method: Code:
var iu = new ImageUploaderWriter("ImageUploader1", 650, 400);
...
document.getElementById('popDiv').innerHTML = iu.getHtml();
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Rank: Member
Groups: Member
Joined: 11/16/2010(UTC)
Posts: 11
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andreym wrote:Hello! If you want to insert Image Uploader into the popup div you should use getHtml method: Code:
var iu = new ImageUploaderWriter("ImageUploader1", 650, 400);
...
document.getElementById('popDiv').innerHTML = iu.getHtml();
Hi, The Url [g_site_path+"/galleryupload/gallery/"] displays the aurigma image uploder 6.0 and also write the image uploader in the php function. I want to use the same uploader function in the popup div. So i call the url through ajax and assign the aurigma image uploader 6.0 in the div through the below function. But it is not assigning the aurigma image uploader to the div. Please help me. <script> function funTest() { $.ajax({ type : "POST", url : g_site_path+"/galleryupload/gallery/", data : 'rand='+Math.random(), dataType : 'script', success: function(script){ alert(script); alert(document.getElementById('popDiv')); document.getElementById('popDiv').innerHTML = script; alert(document.getElementById('popDiv').innerHTML); } }); } </script> Thanks in advance, Regards, Vimala
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Rank: Advanced Member
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Joined: 5/29/2010(UTC)
Posts: 1,310
Thanks: 8 times
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Hello Vimala,
Unfortunately, the way you want to implement is impossible. You can use the method offered by Andrey in the previous post. |
Best regards,
Dmitry Obukhov
Technical Support. Aurigma, Inc.
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Rank: Newbie
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Joined: 1/24/2011(UTC)
Posts: 1
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