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Home/ Questions/Q 6153305
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T20:04:23+00:00 2026-05-23T20:04:23+00:00

im having a hard time saving the image captured from the webcam using the

  • 0

im having a hard time saving the image captured from the webcam using the jquery webcam plugin.
here’s the code..

$(document).ready(function(){
    $("#camera").webcam({
        width: 320,
        height: 240,
        mode: "save",
        swffile: "jscam.swf",   
    });

    });

i am using the ‘save’ mode. in the body part..

<div id="camera"></div>
<a href="javascript:webcam.save('upload.php');void(0);">capture</a>

in the upload.php part..

$str = file_get_contents("php://input");
file_put_contents("upload.jpg", pack("H*", $str));

i also tried the callback mode still doesnt work. it seems the blog itself has insufficient examples

http://www.xarg.org/project/jquery-webcam-plugin/

[update]

finally got it working! i can capture the images. i dug up the source code of the page and added an onload eventlistener on my code 😀

now my only problem is how to save the image. the blog doesn’t clearly specify how. it just gave the codes

webcam.save('/upload.php');

which is honestly i don’t know what to do with it, unlike the php code he gave. should i put it in a link? or edit the onCapture part?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T20:04:23+00:00Added an answer on May 23, 2026 at 8:04 pm

    You going to need to do a little PHP Here is a basic upload script, from the JPEGCam project

    <?php
    
    /* JPEGCam Test Script */
    /* Receives JPEG webcam submission and saves to local file. */
    /* Make sure your directory has permission to write files as your web server user! */
    
    $filename = date('YmdHis') . '.jpg';
    $result = file_put_contents( '/path/to/file/store/or/site/' . $filename, 
          file_get_contents('php://input') );
    if (!$result) {
        print "ERROR: Failed to write data to $filename, check permissions\n";
        exit();
    }
    
    $url = 'http://' . $_SERVER['HTTP_HOST'] . dirname($_SERVER['REQUEST_URI']) . '/' 
         . $filename;
    print "$url\n";
    
    ?>
    
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