doing math displaying + or -

[sonny]

Hi

I am trying to display a point differential result from two db fields

example
10-minus 9, and display +1
Is that possible? or do I need to add the + sign to the result manually

Thanks
Sonny

[mlseim]

Maybe with a "sign specifier"?
http://php.net/manual/en/function.sprintf.php

Or it might be just easier to do it manually?

[sonny]

I almost got it,

PHP Code:

$dada = $num1 += (-$num2); 


Sonny

[Fou-Lu]

Wut? This simply assigns $dada the new value of $num1 which is equal to the negative value of $num2 added to $num1. This still is a number and will not show the sign unless its negative. This will also adjust the value of $num1.
Did you read the link provided? Use a print formatter to keep it as a number and add a sign:

PHP Code:

$var = 15;
printf('%+d', $var); 


Will show +15.

[sonny]

Originally Posted by Fou-Lu

Wut? This simply assigns $dada the new value of $num1 which is equal to the negative value of $num2 added to $num1. This still is a number and will not show the sign unless its negative. This will also adjust the value of $num1.
Did you read the link provided? Use a print formatter to keep it as a number and add a sign:

PHP Code:

$var = 15;
printf('%+d', $var); 


Will show +15.

Ok I got it, the problem I had was I needed to put that into a variable.
It seems to work, but if anyone sees something wrong anyway
please let me know.

PHP Code:

$var=10;
$num=@sprintf('%+d', $var);

//Example
echo "<td>" . $num . "</td>; 


[jmj001]

why not this?

PHP Code:

$var=10;
echo "<td>" . printf('%+d', $var) . "</td>; 


just use a print formatter at the point where you output the value

adding the extra line in there seems redundant

[sonny]

Originally Posted by jmj001

why not this?

PHP Code:

$var=10;
echo "<td>" . printf('%+d', $var) . "</td>; 


just use a print formatter at the point where you output the value

adding the extra line in there seems redundant

Your right, I used something like that before, don't know why I did not
this time. their must have been a reason, just forgot it.

Sonny

[Fou-Lu]

If you are straight out printing it, just use a printf:

PHP Code:

printf('<td>%+d</td>', $var); 


Much faster.
You may have wanted to store it as a string for later uses. Not really necessary since printf is a construct so you'll see almost no performance loss by using it.

Edit:
Actually, printf is not considered a construct; how odd. Still, I can't see tremendous overhead in its use.