Gotta give this in tommorow! Help a noob!

[zartzar]

Ok, basically had my first class of php yesterday. Got this as part of an assignment:

 

"For this part of the exercise you must use the rand() function to generate a number

between 10 and 30. You should then print the words “This line is in font‐size x px” where x

is the number generated by the rand() function. The words should be displayed in the

relevant font size."

 

I can get my sentence to display the random number, but for the life of me i cant change the actual font size. Whats wrong with my code?:

 

$font=rand(1,30);

 

 

print "<div style='font-size:$fontpx';>This line is in font size $font px</div>";

 

 

[premiso]

encapsulate $font in the string with {$font} and it will work.  The string thinks you are trying to call "$fontpx" is is not a variable.

 

Also note, that if it is 10 to 30, you have a 1 instead of 10 in your rand function.

[rhodesa]

PHP will evaluate variables inside double quoted strings, but it sees your first variable as $fontpx not $font followed by the characters p and x. You need to either stop/start the string:

print "<div style='font-size:".$font."px';>This line is in font size $font px</div>";

or use curly braces to show php where the variable start/stops:

print "<div style='font-size:{$font}px';>This line is in font size $font px</div>";

 

read up on it here:

http://us.php.net/manual/en/language.types.string.php#language.types.string.parsing

[zartzar]

encapsulate $font in the string with {$font} and it will work.  The string thinks you are trying to call "$fontpx" is is not a variable.

 

Also note, that if it is 10 to 30, you have a 1 instead of 10 in your rand function.

 

I think, i think i love you!

 

I tried normal brackets, but didn't think of {}

 

Thx!!

[killah]

echo '<div style="font-size: '.rand(10,30).'px;">This font size is undefined?</div>';