<?PHP print "<img src='TimeRotateImage.php'>"; ?> did you get what i am saying? please let me know if you have some solution. thanks
No. I must admit, It don't understand it. Let me try: You have a script "a.php" which outputs this static content:
<img src="b.php">
b.php outputs the static content "<img src="c.jpeg">
This can't work. You browser tries to download an image with the name '<img src="c.jpeg">'. I think THAT's your problem and the reason why I did not understand you problem. If you use a PHP script inside an img src attribute then this PHP-Script must output an image. Complete with content-type header and the image as binary data in the body.
Or another solution for a.php:
<html>
<body>
<img src="<?php include('b.php')?>" />
</body>
</html>but I don't see the sense here. You can just do all this in one script:
<?php
$curr_time = (localtime());
if ($curr_time[2] >= 6 and $curr_time[2] <= 17) {
$img = 'a.jpg';
}
else {
$img = 'b.jpg';
}
?>
<html>
<body>
<img src="<?=$img?>" />
</body>
</html>If I still misunderstood your problem then maybe your description was not detailed enough.
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