3
Gentlemen, I’m no expert in PHP, I’m starting now, and would like a direction on how to display the video of my IP Camera - P2P
snapshot.php
<?php
$img="http://usuario:senha@ipddns:portacamip/mjpeg.cgi?user=usuario&password=senha&channel=0&.mjpg";
readfile($img);
?>
index php.
<img src="http://meudominio.com/snapshot.php" width="640" height="380" name="refresh">
<script language="JavaScript" type="text/javascript">
image = "http://meudominio.com/snapshot.php"
function Start() {
tmp = new Date();
tmp = "?"+tmp.getTime()
document.images["refresh"].src = image+tmp
setTimeout("Start()", 300)
}
Start();
</script>
What am I doing wrong? Could you please explain to me?
Thank you.
The path of the video is correct?
– Ramon Ruan
If the file
snapshot.php
is at the root of the site and the domain is correct, yes– Lucas
Ah, ball show then, as soon as you get home, I’ll be
– Ramon Ruan
Do you have any quick port forwarding tips? how it works?
– Ramon Ruan
I don’t understand that
– Lucas
what treatment I could do to verify the existence of this variable or file?
$img="http://usuario:senha@ipddns:portacamip/mjpeg.cgi?user=usuario&password=senha&channel=0&.mjpg";
– Ramon Ruan
That would be another question :)
– Lucas