Getting an undefined variable error for variable defined within function in php

Question

I have the following code:

$SpeedA = 5; 
$SpeedB = 5; 
$Distance = 20; 

function CalDistance ($SpeedA, $SpeedB, 
$Distance)
{
    $DistanceA = (($SpeedA / $SpeedB) * 
    $Distance) / (1 + ($SpeedA / $SpeedB));

    Return $DistanceA;

}
echo $DistanceA;

I get this error:

Notice: undefined variable $DistanceA

Why $DistanceA is regarded as undefined and how to fix it?

Side note: I'd of ordinarily have closed this as an existing undefined index/variable duplicate, but I applaud your use of the "undefined-variable" tag. You can consult the answer that was given below; it seems to be correct to me. — Funk Forty Niner
– Funk Forty Niner, Commented Oct 1, 2018 at 1:23

Geshode · Accepted Answer · 2018-10-01 01:06:17Z

2

You never call the function CalDistance, before your echo. So, you try to echo an undefined $DistanceA.

So, you could do something like this:

$SpeedA = 5; 
$SpeedB = 5; 
$Distance = 20; 

function CalDistance ($SpeedA, $SpeedB, 
$Distance)
{
    $DistanceA = (($SpeedA / $SpeedB) * 
    $Distance) / (1 + ($SpeedA / $SpeedB));

    Return $DistanceA;

}

$call = CalDistance($SpeedA, $SpeedB, $Distance);
echo $call;

answered Oct 1, 2018 at 1:06

Geshode

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2 Comments

Caryn May Over a year ago

I see. That makes sense. I'm new and some times I just look at it too long and can see the wood for the trees. Thanks!

Geshode Over a year ago

Your welcome. If it solved your problem, then please accept the answer. So, that future readers, see that your problem was solved.

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