imagearc

(PHP 4, PHP 5, PHP 7, PHP 8)

imagearc — Draws an arc

Description

imagearc(
    GdImage $image,
    int $center_x,
    int $center_y,
    int $width,
    int $height,
    int $start_angle,
    int $end_angle,
    int $color
): bool

imagearc() draws an arc of circle centered at the given coordinates.

Parameters

image: A GdImage object, returned by one of the image creation functions, such as imagecreatetruecolor().
center_x: x-coordinate of the center.
center_y: y-coordinate of the center.
width: The arc width.
height: The arc height.
start_angle: The arc start angle, in degrees.
end_angle: The arc end angle, in degrees. 0° is located at the three-o'clock position, and the arc is drawn clockwise.
color: A color identifier created with imagecolorallocate().

Return Values

Returns true on success or false on failure.

Changelog

Version	Description
8.0.0	`image` expects a GdImage instance now; previously, a valid `gd` resource was expected.

Examples

Example #1 Drawing a circle with imagearc()

<?php

// create a 200*200 image
$img = imagecreatetruecolor(200, 200);

// allocate some colors
$white = imagecolorallocate($img, 255, 255, 255);
$red   = imagecolorallocate($img, 255,   0,   0);
$green = imagecolorallocate($img,   0, 255,   0);
$blue  = imagecolorallocate($img,   0,   0, 255);

// draw the head
imagearc($img, 100, 100, 200, 200,  0, 360, $white);
// mouth
imagearc($img, 100, 100, 150, 150, 25, 155, $red);
// left and then the right eye
imagearc($img,  60,  75,  50,  50,  0, 360, $green);
imagearc($img, 140,  75,  50,  50,  0, 360, $blue);

// output image in the browser
header("Content-type: image/png");
imagepng($img);

// free memory
imagedestroy($img);

?>

The above example will output something similar to:

Output of example : Drawing a circle with imagearc()

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User Contributed Notes 20 notes

down

eamon at hostelworld dot com ¶

20 years ago

Right...
possibly the easiest way of drawing a filled circle:
Loop through the imagearc function incrementing the diameter by one pixel:
<?
 // --- code fragment --- //

for($i=1; $i<$Diameter; $i++){
  imagearc($Image, $CenterX, $CenterY, $i, $i, $Start, $End, $Color);
}

// --------------------- //

?>

This works great for circles with diameters up to about 60 or 70 pixels wide. After that, you start to get pixle gaps.

down

chandlerklebs at gmail dot com ¶

12 years ago

This is an example script I wrote for myself to help me learn how to used the imagearc functions. Maybe if will also help others.

<?php
//example PHP script of imagearc functions
$image_width=360;$image_height=360;
$img = imagecreatetruecolor($image_width,$image_height); //make image variable

//create a background color by making a filled rectangle
$color = imagecolorallocate($img,255,255,255);
imagefilledrectangle($img,0,0,$image_width,$image_height,$color);

$r=$image_width/2 - $image_width/32 ; //radius
$cx=$image_width/2;
$cy=$image_height/2;

$color = imagecolorallocate($img,0,0,0);
imagearc($img, $cx, $cy, $r*2, $r*2,  0, 360, $color); //regular outlines arc

imagefilledarc($img, $cx, $cy, $r*1, $r*1,  0, 90, $color,IMG_ARC_CHORD); //filled triangle with chord of circle
imagefilledarc($img, $cx, $cy, $r*1, $r*1,  180, 270, $color,IMG_ARC_PIE); //pie slice

$font_number=5; //can use built in fonts numbered 1 to 5
$string="Hello world!";
imagestring($img, $font_number, $cx-(imagefontwidth($font_number)*strlen($string))/2, $cy-120, $string, $color);

header("Content-type: image/png");
imagepng($img);// output image in the browser

$filename="imagearc";
imagepng($img,"./frames/$filename.png",9); //make highly compressed png

imagedestroy($img);
?>

down

jinny at 263 dot net ¶

22 years ago

imagesetstyle() sets the style to be used by all line drawing functions when drawing with the special color .

Here goes a example of drawing a dashed-line circle.enjoy!

<?php

header("Content-type: image/jpeg");
$im = imagecreate(100,100);

$b   = imagecolorallocate ($im, 0, 0, 0);
$w   = imagecolorallocate ($im, 255, 255, 255);

$style = array ($b,$b,$b,$b,$b,$w,$w,$w,$w,$w);

imagesetstyle ($im, $style);

imagearc($im,50,50,100,100,0,360,IMG_COLOR_STYLED);

imagejpeg($im);
imagedestroy($im);
?>

down

cleverphp ¶

6 years ago

imagearc example works not well,as it lacks of this line
"$white = imagecolorallocate($img, 255, 255, 255);
    imagefill($img,0,0,$white);
    $black = imagecolorallocate($img, 0, 0, 0);
"

down

marc at resiteit dot com ¶

23 years ago

Round cornered anti-aliased dynamically sized button.



$w=40;

$h=20;

$im = ImageCreate($w,$h);

$white=ImageColorAllocate($im,255,255,255);

ImageFilledRectangle($im,0,0,$w,$h,$white);

imagecolortransparent ($im, $white);

ImageTTFText ($im, $h+ceil($h/3)+1, 0, -1, $h-1, $col1, "arialbd.ttf", "O");

ImageTTFText ($im, $h+ceil($h/3)+1, 0, $w-$h, $h-1, $col1, "arialbd.ttf", "O");

ImageTTFText ($im, $h+ceil($h/3)+1, 0, 1, $h-1, $col1, "arialbd.ttf", "O");

ImageTTFText ($im, $h+ceil($h/3)+1, 0, $w-$h-2, $h-1, $col1, "arialbd.ttf", "O");

$points=array(

    1,round($h/2), 

    round($h/4),$h-round($h/4), 

    round($h/2),$h, 

    $w-(round($h/2)),$h, 

    $w-(round($h/4)),$h-round($h/4), 

    $w-2,round($h/2), 

    $w-round($h/4),round($h/4), 

    $w-round($h/2),0, 

    round($h/2),0, 

    round($h/4),round($h/4)

);

imagefilledpolygon ($im, $points, 10, $col1);



header("content-type: image/gif");

header("Content-Disposition: filename=name.gif");

ImageGif($im);

ImageDestroy($im);

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jerryscript at aol dot com ¶

20 years ago

[note-Apache/1.3.29 (Win32) PHP/4.3.4]

The imagearc (and imageellipse) functions do not accept line thicknesses when drawn from 0 to 360 degrees.

Drawing from 0 to 359 and again from 359 to 360 does create an ellipse with the current line thickness.

Jerry

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foripepe at yahoo dot com ¶

23 years ago

To fill an arc (DiameterX != DiameterY):



<?

function imagefilledarc($Image, $CenterX, $CenterY, $DiameterX, $DiameterY, $Start, $End, $Color) {

    // To draw the arc

    imagearc($Image, $CenterX, $CenterY, $DiameterX, $DiameterY, $Start, $End, $Color);

    // To close the arc with 2 lines between the center and the 2 limits of the arc

    $x = $CenterX + (cos(deg2rad($Start))*($DiameterX/2));

    $y = $CenterY + (sin(deg2rad($Start))*($DiameterY/2));

    imageline($Image, $x, $y, $CenterX, $CenterY, $Color);

    $x = $CenterX + (cos(deg2rad($End))*($DiameterX/2));

    $y = $CenterY + (sin(deg2rad($End))*($DiameterY/2));

    imageline($Image, $x, $y, $CenterX, $CenterY, $Color);

    // To fill the arc, the starting point is a point in the middle of the closed space

    $x = $CenterX + (cos(deg2rad(($Start+$End)/2))*($DiameterX/4));

    $y = $CenterY + (sin(deg2rad(($Start+$End)/2))*($DiameterY/4));

    imagefilltoborder($Image, $x, $y, $Color, $Color);

}

?>



To close the arc with 2 lines (DiameterX != DiameterY):



<?

function imagenofilledarc($Image, $CenterX, $CenterY, $DiameterX, $DiameterY, $Start, $End, $Color) {

    // To draw the arc

    imagearc($Image, $CenterX, $CenterY, $DiameterX, $DiameterY, $Start, $End, $Color);

    // To close the arc with 2 lines between the center and the 2 limits of the arc

    $x = $CenterX + (cos(deg2rad($Start))*($DiameterX/2));

    $y = $CenterY + (sin(deg2rad($Start))*($DiameterY/2));

    imageline($Image, $x, $y, $CenterX, $CenterY, $Color);

    $x = $CenterX + (cos(deg2rad($End))*($DiameterX/2));

    $y = $CenterY + (sin(deg2rad($End))*($DiameterY/2));

    imageline($Image, $x, $y, $CenterX, $CenterY, $Color);

}

?>



An example:

<?

    $destImage = imagecreate( 216, 152 );

    $c0 = imagecolorallocate( $destImage, 0, 255, 255 );

    $c1 = imagecolorallocate( $destImage, 0, 0, 0 );

    $c2 = imagecolorallocate( $destImage, 255, 0, 0 );

    ImageFilledRectangle ( $destImage, 0, 0, 216, 152, $c0 );

    imagefilledarc( $destImage, 108, 76, 180, 80, 0, 130, $c1 );

    imagenofilledarc( $destImage, 108, 76, 180, 80, 0, 130, $c2 );

    header("content-type: image/PNG");

    ImagePNG( $destImage );

    ImageDestroy( $destImage );

?>

down

travis at duluth dot com ¶

24 years ago

The wierd thing is that the first two integers tell where to place the "circle".

So for example I first create the "pallet" to place the circle on.

$image = imagecreate(500, 500);  

(this makes a huge 500x500 gif :) )

$colorBody = imagecolorallocate($image, 0, 0, 0);

(make the default color of the "pallet" black

$circleColor = imagecolorallocate($image, 255, 0, 255);

(going to make the circle an ugly pink color)

imagearc($image, 250, 250, 300, 300, 0, 360, $circleColor);

Places the image in the center (250,250) and the circle is 300 pixels in diameter.



Hope this helps.

 

Travis Kent Beste

down

joe dot tym at gmail dot com ¶

12 years ago

I didn't have much luck with the other two functions, one of them makes circles that look like they've been printed on a dot-matrix printer. This simple function builds a border out of circles, seems to work nicely.



<?php

function imagearcunfilled($image,$x,$y,$width,$height,$border_thickness, $color) {



        imagesetthickness($image, 1);



        $x_radius = $width / 2;

        $y_radius = $height / 2;



        for ($i = 0; $i < 360; $i++) {

                if (TRUE) {

                        $x2 = $x + cos($i) * $x_radius;

                        $y2 = $y + sin($i) * $y_radius;

                        imagefilledarc($image,$x2,$y2,$border_thickness,$border_thickness,0,360,$color,IMG_ARC_PIE);

                }

        }

}

?>

down

mojiro at awmn dot net ¶

18 years ago

A previous for the Rotated (Filled)Ellipse note from(nojer2 at yahoo dot com, 02-Apr-2001 12:06) has a mistake, at the second arc. Replace them with the following listing.

if ($filled) {
    triangle($im, $cx, $cy, $cx+$px, $cy+$py, $cx+$x, $cy+$y, $colour);
    triangle($im, $cx, $cy, $cx-$px, $cy-$py, $cx-$x, $cy-$y, $colour);
} else {
    imageline($im, $cx+$px, $cy+$py, $cx+$x, $cy+$y, $colour);
    imageline($im, $cx-$px, $cy-$py, $cx-$x, $cy-$y, $colour);
}

down

Anonymous ¶

21 years ago

Please note that in order to draw a complete circle or ellipse (without using the imageellipse) you mustn't use 0? for both s and e. If you do this you will get, umm, nothing. Instead set s to 0? and e to 360? to get a complete circle or ellipse.

down

nojer2 at yahoo dot com ¶

23 years ago

Here's the function to draw rotated ellipses again. This time I've optimised it a bit, fixed the no-fill bug, and used a 'squishratio' rather than a 'radiusmodifier', to make the curves perfect, so ignore my previous version.



<?php

function rotatedellipse($im, $cx, $cy, $width, $height, $rotateangle, $colour, $filled=true) {

   $step=2;

   $cosangle=cos(deg2rad($rotateangle));

   $sinangle=sin(deg2rad($rotateangle));



   $squishratio = $height/$width;

   $nopreviouspoint = true;

   for ($angle=0; $angle<=(180+$step); $angle+=$step) {

       

      $ox = ($width * cos(deg2rad($angle)));

      $oy = ($width * sin(deg2rad($angle))) * $squishratio;



      $x =  + (($ox * $cosangle) - ($oy * $sinangle));

      $y = $centrey + (($ox * $sinangle) + ($oy * $cosangle));

  

      if ($nopreviouspoint) {

        $px=$x;

        $py=$y;

        $nopreviouspoint=false;

      }



      if ($filled) {

         triangle($im, $cx, $cy, $cx+$px, $cy+$py, $cx+$x, $cy+$y, $colour);

         triangle($im, $cx, $cy, $cx-$px, $cx-$py, $cx-$x, $cy-$y, $colour);

      } else {

         imageline($im, $cx+$px, $cy+$py, $cx+$x, $cy+$y, $colour); 

         imageline($im, $cx-$px, $cx-$py, $cx-$x, $cy-$y, $colour); 

      }

      $px=$x;

      $py=$y;

   } 

}



function triangle($im, $x1,$y1, $x2,$y2, $x3,$y3, $colour) {

   $coords = array($x1,$y1, $x2,$y2, $x3,$y3);

   imagefilledpolygon($im, $coords, 3, $colour); 

}

?>

down

timothyhouck at yahoo dot com ¶

24 years ago

To do filled arcs, try something like this:



<?php

$diameter = 50;

imagearc($image, 25, 25, $diameter, $diameter, $start, $end, $color);

while($diameter > 0) {

  imagearc($image, 25, 25, $diameter, $diameter, $start, $start + 1, $color);

  imagearc($image, 25, 25, $diameter, $diameter, $end - 1, $end, $color);

  $diameter--;

}

?>



...well you get the point.  It's a kludge, and *very* slow, but it's free.

down

nojer2 at yahoo dot com ¶

23 years ago

Here's a dashed circle function:



<?php

function dashedcircle($im, $cx, $cy, $radius, $colour, $dashsize=5) {



   $dash=false;

   for ($angle=0; $angle<=(180+$dashsize); $angle+=$dashsize) {

      $x = ($radius * cos(deg2rad($angle)));

      $y = ($radius * sin(deg2rad($angle)));



      if ($dash) {

         imageline($im, $cx+$px, $cy+$py, $cx+$x, $cy+$y, $colour); 

         imageline($im, $cx-$px, $cx-$py, $cx-$x, $cy-$y, $colour); 

      }

      $dash=!$dash;

      $px=$x;

      $py=$y;

   } 

}

?>

down

anton dot vandeghinste at telenet dot be ¶

14 years ago

I needed an arc with a thick border and i didn't like to use 359.9 as end angle so i made a function that works pretty well:

<?php
function imagearcthick($image, $x, $y, $w, $h, $s, $e, $color, $thick = 1)
{
    if($thick == 1)
    {
        return imagearc($image, $x, $y, $w, $h, $s, $e, $color);
    }
    for($i = 1;$i<($thick+1);$i++)
    {
        imagearc($image, $x, $y, $w-($i/5), $h-($i/5),$s,$e,$color);
        imagearc($image, $x, $y, $w+($i/5), $h+($i/5), $s, $e, $color);
    }
}
?>

down

lucas dot delmas at live dot fr ¶

11 years ago

The imagearc function has a precision of one degree. The function truncates the $start and $end values to the inferior degree.

For example if the starting angle you calculated is : -178.62450462172°
and the ending angle is : -152.78056427917°
imagearc will draw a curve from -178° to -152°.

If you need accurate curves drawing, you need to use a loop to draw little step-by-step lines. By creating a large number of short enough lines, you will create the impression of a curve with accuracy.

down

ajim1417 at gmail dot com ¶

14 years ago

I wrote a simple function that can draws an arc counter-clockwisekly. Here it is :

<?php

function imagearcCC(&$im, $cx, $cy, $w, $h, $s, $e, $c) {
$start = 360 - $e;
$end = 360 - $s;
return imagearc($im, $cx, $cy, $w, $h, $start, $end, $c);
}

?>

The params of this function is exactly the same as the usual imagearc function.

down

logang at deltatee dot com ¶

21 years ago

Heres a function to make a curve between two points... This will be a downward curve but it wouldn't be hard to make a similar function to make an upward curve. The first point has to be to the left of the second point ($x1 < $x2), and height is actually backwards. The larger height is the less of a crest the curve has. I imagine with a few modifications this functions could make upward curves as well.

function ImageCurveDown ($image, $x1, $y1, $x2, $y2, $height, $color) {
    $presicion = 1;

    for ($left = ($x1-$x2); $left < 0; $left++){
        if ($y1 < $y2) {
            $cy = $y2 + $height;
            $cx = $x1 - $left;
        } else {
            $cy = $y1 + $height;
            $cx = $x2 + $left;
        }
        $nx1 = abs($x1 - $cx);
        $ny1 = abs($y1 - $cy);
        $nx2 = abs($x2 - $cx);
        $ny2 = abs($y2 - $cy);

        if ($y1 < $y2) {
            if ($nx2 == 0 || $ny1 == 0) continue;
            $angle1 = atan($height/$nx2);
            $A1 = $nx2/cos ($angle1);
            $B1 = $ny2/sin ($angle1);
            $angle2 = pi()/2 +atan($left/$ny1);
            $A2 = $nx1/cos ($angle2);
            $B2 = $ny1/sin ($angle2);
        } else {
            if ($ny2 == 0 || $nx1 == 0) continue;
            $angle1 = atan($ny2/$nx2);
            $A1 = abs($nx2/cos ($angle1));
            $B1 = abs($ny2/sin ($angle1));
            $angle2 = atan($height/$nx1);
            $A2 = abs ($nx1/cos ($angle2));
            $B2 = abs($ny1/sin ($angle2));
        }

        if (abs($A1 - $A2) < $presicion && abs ($B1 - $B2) < $presicion) {
            ImageArc($image, $cx, $cy, $A1*2, $B1*2, 180+rad2deg($angle2), 360-rad2deg($angle1), $color);
        }
    }
}

down

arve at skogvold dot as ¶

23 years ago

I found a better way for drawing a pie chart:



header ("Content-type: image/png");

$diameter = 100;

$radius = $diameter / 2;

$centerX = $radius;

$centerY = $radius;



$im = @ImageCreate ($diameter, $diameter)

    or die ("Cannot Initialize new GD image stream");



$background = ImageColorAllocate ($im, 0, 0, 0);

$red = ImageColorAllocate ($im, 176, 0, 0);



function fill_arc($start, $end, $color) {

    global $diameter, $centerX, $centerY, $im, $radius;

    imagearc($im, $centerX, $centerY, $diameter, $diameter, $start, $end, $color); 

    imageline($im, $centerX, $centerY, $centerX + cos(deg2rad($start)) * $radius, $centerY + sin(deg2rad($start)) * $radius, $color);

    imageline($im, $centerX, $centerY, $centerX + cos(deg2rad($end)) * $radius, $centerY + sin(deg2rad($end)) * $radius, $color);

    imagefill ($im,$centerX + $radius * 0.5 *cos(deg2rad($start+($end-$start)/2)), $centerY + $radius * 0.5 * sin(deg2rad($start+($end-$start)/2)), $color);

    }





fill_arc(0,30,$red);

// Will make a red filled arc, starting at 0 degrees, ending at 30 degrees



ImagePng ($im);

down

-1

cbriou at orange-art dot fr ¶

24 years ago

There is another way to fill an arc :



<?php

// To draw the arc

$Color = imagecolorallocate($Image, $Red, $Green, $Blue);

imagearc($Image, $CenterX, $CenterY, $Diameter, $Diameter, $Start, $End, $Color);

// To close the arc with 2 lines between the center and the 2 limits of the arc

$x = $CenterX + (cos(deg2rad($Start))*($Diameter/2));

$y = $CenterY + (sin(deg2rad($Start))*($Diameter/2));

imageline($Image, $x, $y, $CenterX, $CenterY, $Color);

$x = $CenterX + (cos(deg2rad($End))*($Diameter/2));

$y = $CenterY + (sin(deg2rad($End))*($Diameter/2));

imageline($Image, $x, $y, $CenterX, $CenterY, $Color);

// To fill the arc, the starting point is a point in the middle of the closed space

$x = $CenterX + (cos(deg2rad(($Start+$End)/2))*($Diameter/4));

$y = $CenterY + (sin(deg2rad(($Start+$End)/2))*($Diameter/4));

imagefilltoborder($Image, $x, $y, $Color, $Color);

?>

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imagearc

Description

Parameters

Return Values

Changelog

Examples

See Also

Found A Problem?

User Contributed Notes 20 notes