I have a following PHP file TimeRotateImage.php which prints an image
?php
$curr_time = (localtime());
if ($curr_time[2] = 6 and $curr_time[2] = 17) {
PRINT IMAGE_1;
}
else {
PRINT IMAGE_2;
}
?
I want to use this script in another file to display the image. the tag
should be
img
Deepak Dhake wrote:
But i am not getting any output if i follow the above procedure. Can you
tell me how to do it? I have to have the script TimeRotateImage.php
which calculates which image to print accoring to local time and i want
to embed the file name in html tag to print it on screen.
It's
PRINT is nothing but...
TimeRotateImage.php
?php
$curr_time = (localtime());
if ($curr_time[2] = 6 and $curr_time[2] = 17) {
print img src='image1.jpg';
}
else {
print img src='image2.jpg';
}
?
this script (something like this) should be called from another script like,
?PHP
print img
Deepak Dhake wrote:
?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
Klaus Reimer wrote:
This can't work. You browser tries to download an image with the name
'img src=c.jpeg'.
Ah, I'm talking nonsense. I meant the browser tries to DISPLAY an image
with the CONTENT 'img src=c.jpeg'. The browser can't do this.
That's why you don't see anything.
--
Bye, K
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