Getting PHP to change a randomly generated CSS background color

Question

I realize that there have already been several threads on this type of question, but none that seemed to match mine. I have code that is correctly displaying a series of 6 numbers for a CSS background. Those 6 numbers are then stored into an array and the contents of that array are concatenated with a "#" sign in front of it. That, then is stored as another variable. Here:

$v = 0;
$array = array();
$tot;
$other0 = null;
$other1 = null;
$other2 = null;
$other3 = null;
$other4 = null;
$other5 = null;
$color;

for ($i=0;$i<6;$i++){ //loops 6 times to create 5 numbers
    $tot = rand(0, 15);

    if (($tot>9) && ($tot<16)){ //assigns 10 to "a" in hex
        if ($tot == 10){
            $tot = $other0;
            $other0 = "a";
            //echo $other0;
            $array[$v++] = $other0;
        }
        elseif ($tot == 11){ //assigns 11 to "b" in hex
            $tot = $other1;
            $other1 = "b";
            //echo $other1;
            $array[$v++] = $other1;
        }
        elseif ($tot == 12){ //assigns 12 to "b" in hex
            $tot = $other2;
            $other2 = "c";
            //echo $other2;
            $array[$v++] = $other2;
        }
        elseif ($tot == 13){ //assigns 13 to "b" in hex
            $tot = $other3;
            $other3 = "d";
            //echo $other3;
            $array[$v++] = $other3;
        }
        elseif ($tot == 14){ //assigns 14 to "b" in hex
            $tot = $other4;
            $other4 = "e";
            //echo $other4;
            $array[$v++] = $other4;
        }
        elseif ($tot == 15) { //assigns 15 to "b" in hex
            $tot = $other5;
            $other5 = "f";
            //echo $other5;
            $array[$v++] = $other5;
        }

    }
    else {
        //echo $tot;
        $array[$v++] = $tot; //if not, then just assign to array
    }
}
$var = "";
$color = "";
for ($t=0; $t<6;$t++){
    $var .= (string) $array[$t]; //displays the 6 numbers as one string
}
//var_dump($var); //for more visual reference, uncomment the var_dump
$color = "#" . $var;
echo $color;

That above is in PHP.

How do I get that variable to be displayed using CSS? Is it like (in the same PHP file): echo "<style>color:$color</style>";

or do I have to make a style.php for it to be referenced? If I do, how do I do that?

Many thanks!

you would need to run the php before the body tag in the html. you could then do an inline style in the page for body { background-color:<?PHP echo $color; ?> } — bretterer
– bretterer, Commented Jul 12, 2012 at 15:43
You really need to learn hexadecimal :) check out: php.net/manual/en/function.dechex.php — Mahn
– Mahn, Commented Jul 12, 2012 at 15:44
@ChaseYuan Remember this: whatever you want to do there is a function for that in PHP. — PeeHaa
– PeeHaa, Commented Jul 12, 2012 at 15:52
I'd suggest using 3 randomly generated values between 0 and 255, for instance: $blub = "#".dechex(mt_rand(0, 255)).dechex(mt_rand(0, 255)).dechex(mt_rand(0, 255)); — Mahn
– Mahn, Commented Jul 12, 2012 at 16:02

SMacFadyen · Accepted Answer · 2012-07-12 15:44:47Z

1

Well it looks like "<style>color:$color</style>"; will output broken html/css.

Try amending

echo '<style type="text/css"> body { background: '.$color.' } </style>';

answered Jul 12, 2012 at 15:44

SMacFadyen

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1 Comment

SMacFadyen Over a year ago

Yeah, I hit submit by accident. [Edited]

Doug Molineux · Accepted Answer · 2012-07-12 15:44:15Z

1

It should be

<style>
body {
   background-color: "<?php echo $color; ?>";
}
</style>

This needs to be within your <head> element.

answered Jul 12, 2012 at 15:44

Doug Molineux

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4 Comments

ParaChase Over a year ago

Hm. This doesn't seem to work because $color is called before it is defined.

Doug Molineux Over a year ago

Where are you defining $color? The PHP should all be executed before the HTML is generated. But $color will need to be defined before you echo it

ParaChase Over a year ago

I have my html generated before the php is called. Should I move it after the function?

Doug Molineux Over a year ago

If you have a PHP file that contains HTML and PHP, all the PHP will execute first, it doesn't matter whether or not the HTML appears first in the file. It will execute all PHP first. The PHP that exists (aside from the HTML) will run sequentially however. So the function / code that generates the $color variable, will need exist above where you echo it. With all that being said, you maybe better off using a different suggestion (like Mahn's) to generate the color var, since it sounds like it may not actually be a color anyway. Hopefully this helps.

1321941 · Accepted Answer · 2012-07-12 15:44:36Z

1

<body style="background-color:<?php echo $color; ?>;">


content


</body>

the style attribute allows you to define custom css to elements, the css attribute, background-color basically defines the color of the background for that element... self explanatory, I know!

answered Jul 12, 2012 at 15:44

1321941

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Comments

Community · Accepted Answer · 2017-05-23 12:12:28Z

1

Why dont you take a look at these links.

Change background color of a page using php

So instead of doing:

if blablabla {
    echo '<body style="background-color:white">'; 
}
else {  
    echo '<body style="background-color:orange">'; 
}

Do this:

echo '<body style="background-color:(variable)">';

edited May 23, 2017 at 12:12

CommunityBot

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answered Jul 12, 2012 at 15:48

Eric Sauer

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Comments

aurechabnv · Accepted Answer · 2013-03-21 11:40:17Z

<?php
    $alpha = ["a","b","c","d","e","f","0","1","2","3","4","5","6","7","8","9"];
    $color = "#";
    for ($i=0; $i<6; $i++) {
        $index = rand(0,count($alpha)-1);
        $color .= $alpha[$index];
    }
?>
<html>
    <body style="background:<?php echo $color ?>">
        <?php echo $color ?>
    </body>
</html>

It seems you've made things a little more complicated than it could be. Here's a sample with html integration, as you asked for in the first place. :)

You declare all the elements you need in an array, then you loop 6 times to: generate a random index, take the corresponding element from the array and concatenate it to your color variable.

Finally you echo it anywhere you need in your html file, as said earlier.

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