PHP - Problems calling a function with parameters

Question

I need to call a function, in PHP, that accepts 3 parameters, RGB Values. This function converts RGB color values to HSL values, so (R,G,B) is needed in the parenthesis.

This is my function:

function RGBtoHSL($red, $green, $blue) {
  // convert colors
}

Which, if I make a test call of the following, it works just fine:

RGBtoHSL(255,0,0);

and also works like this:

RGBtoHSL(255,000,000);

Now, further down my page I have a variable $displayRGB which holds the current pixels RGB values in this format xxx,xxx,xxx. I've echoed this variable to test the format matches my requirements and it does, but when I try and add this variable to my function caller, it fails with the error "Missing argument 2, Missing argument 3" and points to this line:

RGBtoHSL($displayRGB);

I'm still teething in PHP (come from ASP), can somebody please help point me in the right direction and pass me my dummy?

@David - That's not how upvotes work - People upvote answers based on their merit, not when they were posted. — nickb
– nickb, Commented Jul 25, 2012 at 21:44
@David: consider yourself downvoted for being cheeky. That's not how it works. — TheCarver
– TheCarver, Commented Jul 25, 2012 at 21:49

Mekswoll · Accepted Answer · 2012-07-25 21:29:45Z

4

Your $displayRGB is a single variable (of type string, I presume). What you can do is split this string into an array:

$rgbArray = explode(',', $displayRGB);

Then pass it to your function

RGBtoHSL($rgbArray[0], $rgbArray[1], $rgbArray[2]);

answered Jul 25, 2012 at 21:29

Mekswoll

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Comments

Matt · Accepted Answer · 2012-07-25 21:28:47Z

3

You can't pass in an array (I assume $displayRGB is an array) as "all three arguments" in PHP. Try

RGBtoHSL($displayRGB[0], $displayRGB[1], $displayRGB[2]);

or modify your function to accept an array.

If $displayRGB is a string of "xxx,yyy,zzz" you can run an explode on it

$colors = explode(",", $displayRGB);

and it will set $colors as an array with indices containing xxx, yyy and zzz.

Then pass it as I mentioned above.

answered Jul 25, 2012 at 21:28

Matt

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3 Comments

Hamish Over a year ago

I suspect that OP is providing a string, but yeah, the error is pretty clear.

Matt Over a year ago

Edited to reflect these comments.

TheCarver Over a year ago

Thanks Matt, it worked fine. It was a string so thanks for the edit too.

j0k · Accepted Answer · 2012-07-26 14:56:53Z

1

try this instead of eval

call_user_func_array('RGBtoHSL', explode(',', $displayRGB));

edited Jul 26, 2012 at 14:56

j0k

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answered Jul 25, 2012 at 21:34

w43L

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Comments

Jerzy Zawadzki · Accepted Answer · 2012-07-25 21:30:16Z

0

what's your $displayRGB value?

if it's "255,0,0" you should first do "explode"

e.g.

<?php

list($r,$g,$b)=explode(',',$displayRGB);
RGBtoHSL($r,$g,$b);

answered Jul 25, 2012 at 21:30

Jerzy Zawadzki

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1 Comment

Matt Over a year ago

Nice usage of list(). I didn't think of that.

Tim Withers · Accepted Answer · 2012-07-25 21:31:28Z

-2

I will probably get flamed for this, but this would definitely work:

eval("RGBtoHSL($displayRGB);");

Don't do it. It will work... but don't do it.

answered Jul 25, 2012 at 21:31

Tim Withers

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1 Comment

TheCarver Over a year ago

That's funny. "Don't do it. It will work... but don't do it" totally put me off your answer. I said to myself "Don't try it". Nice one though.

David · Accepted Answer · 2012-07-25 22:46:45Z

-2

You can't put a string in a function and expect it to explode the string for you.

You need to go like this:

$string = '255,0,0';
$array = explode(',', $string);
RGBtoHSL($array[0], $array[1], $array[2]);

edited Jul 25, 2012 at 22:46

answered Jul 25, 2012 at 21:29

David

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1 Comment

ShadowStorm Over a year ago

-1 Should be explode(',', $string); with a comma. This will throw syntax error.

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