jQuery网络摄像头插件 - 保存图像 [英] jQuery webcam plugin - saving image

问题描述

我很难使用jquery网络摄像头插件保存从网络摄像头捕获的图像。
这里是代码..

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",   
    });

    });

我正在使用'保存'模式。在身体部分..

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>

..

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

我也试过回调模式仍然无法正常工作。似乎博客本身没有足够的例子

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]

终于搞定了！我可以拍摄图像。我挖出了页面的源代码，并在我的代码上添加了一个onload eventlistener：D

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 :D

现在我唯一的问题是如何保存图像。博客没有明确说明如何。它只是给了代码

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');

老实说我不知道​​该怎么做，不像他给的PHP代码。我应该把它放在一个链接中吗？或者编辑onCapture部分？

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?

推荐答案

你需要做一点PHP这里是一个基本上传脚本，来自JPEGCam项目

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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