Reputation: 151
I have a problem in my if else condition. My Screen Width is:1366 My code should show the if part but always it is showing the else part. I don't know WHY.. First of all i run the if condition without casting but problem was same... Then later on i casted into (int) but still the problem is same
$screenWidth = "<script>document.writeln(screen.width);</script>";
echo "Screen Width is:" . $screenWidth;
if( (int)$screenWidth > 500)
{
echo "if part:" . $screenWidth;
}
else
{
echo "else part:" . $screenWidth;
}
Upvotes: 0
Views: 57
Reputation: 3813
You really can't do this with PHP. Check out this question and answer. However, you can use Javascript to accomplish what you're after. It just requires a destination div to print the results.
<div id="output">
</div>
<script>
var screenWidth = screen.width;
if(screenWidth > 500){
document.getElementById('output').innerHTML = "if part:" + screenWidth;
} else {
document.getElementById('output').innerHTML = "else part:" + screenWidth;
}
</script>
Upvotes: 1