how to check if a image is loaded with success in php or javascript? - Stack Overflow

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i have some images that get displayed on a page, some of them don't get loaded and i see that non-loading :) small icon.

i was wondering if there is a way to check if the image loaded successful and if not now to display it

this is my image code:

print "<a href=\"".$domain."/picture.php?s=".$s2."&media_path=".$media_path."\">";
    print "<img src=\"".$domain."/".$folder."/".$media_path."\" >";
print "</a>";

what happens is that picture.php looks in the database for the picture name and path and builds the link.

I've looked in the database and the data is there so the images mush be corrupted on the drive, maybe wrong image

So i need some way to check if the image is loaded with success at the end of the script. preferable in php but javascript/jquery is fine also

edit: from the answers i got i did this:

function hideImage()
{
    $('.hideme_image').hide();
}


print "<a class=\"hideme_image\" href=\"".$domain."/picture.php?s=".$s2."&media_path=".$media_path."\">";
    print "<img onerror=\"hideImage()\" src=\"".$domain."/".$folder."/".$media_path."\" >";
print "</a>";

this hides all images, but not all images are corupted.

any ideas?

thanks

i have some images that get displayed on a page, some of them don't get loaded and i see that non-loading :) small icon.

i was wondering if there is a way to check if the image loaded successful and if not now to display it

this is my image code:

print "<a href=\"".$domain."/picture.php?s=".$s2."&media_path=".$media_path."\">";
    print "<img src=\"".$domain."/".$folder."/".$media_path."\" >";
print "</a>";

what happens is that picture.php looks in the database for the picture name and path and builds the link.

I've looked in the database and the data is there so the images mush be corrupted on the drive, maybe wrong image

So i need some way to check if the image is loaded with success at the end of the script. preferable in php but javascript/jquery is fine also

edit: from the answers i got i did this:

function hideImage()
{
    $('.hideme_image').hide();
}


print "<a class=\"hideme_image\" href=\"".$domain."/picture.php?s=".$s2."&media_path=".$media_path."\">";
    print "<img onerror=\"hideImage()\" src=\"".$domain."/".$folder."/".$media_path."\" >";
print "</a>";

this hides all images, but not all images are corupted.

any ideas?

thanks

• php
• javascript
• image

Share Improve this question Follow edited Jan 18, 2012 at 23:35 Patrioticcow asked Jan 18, 2012 at 23:25 PatrioticcowPatrioticcow 27.1k7676 gold badges221221 silver badges339339 bronze badges

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2 Answers 2

Sorted by: Reset to default 7

<img src="image.png" onload="alert('Image loaded');" onerror="alert('Image failed');" />

Something along these lines, is what you're looking for, I think.

You can use onload event on the image tag. Please have a look at this example from W3Schools.

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