更新：
这是整个代码，我几乎都复制并粘贴了。
<!DOCTYPE HTML>
<head>
<script type="text/javascript" src="http://code.jquery.com/jquery-latest.js"></script>
<script language="JavaScript" type="text/javascript"></script>
<style>
</style>
</head>
<body>
<script>
Webcam.set({
width: 320,
height: 240,
dest_width: 640,
dest_height: 480,
image_format: 'jpeg',
jpeg_quality: 90,
force_flash: false
});
</script>
<div id="web_camera" style="width:320px; height:240px;"></div>
<div id="cam_result"></div>
<script type="text/javascript" src="webcam.js"></script>
<script language="JavaScript">
document.addEventListener("DOMContentLoaded", function(event) {
Webcam.set({
width: 320,
height: 240,
image_format: 'jpeg',
jpeg_quality: 90
});
Webcam.attach( '#web_camera' );
function take_snapshot() {
// take snapshot and get image data
Webcam.snap( function(data_uri) {
// display results in page
document.getElementById('cam_results').innerHTML =
'<h2>Here is your image:</h2>' +
'<img src="'+data_uri+'"/>';
Webcam.upload( data_uri, 'upload.php', function(code, text) {
// Upload complete!
// 'code' will be the HTTP response code from the server, e.g. 200
// 'text' will be the raw response content
});
} );
}
});
</script>
<a href="javascript:void(take_snapshot())">Take Snapshot</a>
</body>
我正在使用此链接
http://mycodingtricks.com/javascript/webcam-api/
这个看起来好多了但可能是同样的事情
http://www.html5rocks.com/en/tutorials/getusermedia/intro/
我关注的是data_uri，也是网址上传
所以摄像头工作，显示我的脸，无论如何，但我推这个
<a href="javascript:void(take_snapshot())">Take Snapshot</a>
没有任何反应。我看到左下方的小灰框说javascript：void（take_snapshot（））我想知道我是否应该放一个参数......
可能有几个问题，我正在使用域映射，文件夹可能指向错误，或者可能是文件权限问题，我用www-data
这是建议的upload.php，或者由第一个链接
给出<?php
// be aware of file / directory permissions on your server
move_uploaded_file($_FILES['webcam']['tmp_name'], '/tabdater/uploads/webcam'.md5(time()).rand(383,1000).'.jpg');
?>
我很感激任何帮助。
答案 0 :(得分：5)
这应该是一个很好的起点。我希望这会对你有所帮助。
<!doctype html>
<html lang="en">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<title>Cam Snap</title>
<script type="text/javascript" src="webcam.js"></script>
<script language="JavaScript">
function take_snapshot() {
Webcam.snap(function(data_uri) {
document.getElementById('results').innerHTML = '<img id="base64image" src="'+data_uri+'"/><button onclick="SaveSnap();">Save Snap</button>';
});
}
function ShowCam(){
Webcam.set({
width: 320,
height: 240,
image_format: 'jpeg',
jpeg_quality: 100
});
Webcam.attach('#my_camera');
}
function SaveSnap(){
document.getElementById("loading").innerHTML="Saving, please wait...";
var file = document.getElementById("base64image").src;
var formdata = new FormData();
formdata.append("base64image", file);
var ajax = new XMLHttpRequest();
ajax.addEventListener("load", function(event) { uploadcomplete(event);}, false);
ajax.open("POST", "upload.php");
ajax.send(formdata);
}
function uploadcomplete(event){
document.getElementById("loading").innerHTML="";
var image_return=event.target.responseText;
var showup=document.getElementById("uploaded").src=image_return;
}
window.onload= ShowCam;
</script>
<style type="text/css">
.container{display:inline-block;width:320px;}
#Cam{background:rgb(255,255,215);}#Prev{background:rgb(255,255,155);}#Saved{background:rgb(255,255,55);}
</style>
</head>
<body>
<div class="container" id="Cam"><b>Webcam Preview...</b>
<div id="my_camera"></div><form><input type="button" value="Snap It" onClick="take_snapshot()"></form>
</div>
<div class="container" id="Prev">
<b>Snap Preview...</b><div id="results"></div>
</div>
<div class="container" id="Saved">
<b>Saved</b><span id="loading"></span><img id="uploaded" src=""/>
</div>
</body>
</html>
PHP（必须有上传目录） -
<?php
define('UPLOAD_DIR', 'uploads/');
$img = $_POST['base64image'];
$img = str_replace('data:image/jpeg;base64,', '', $img);
$img = str_replace(' ', '+', $img);
$data = base64_decode($img);
$file = UPLOAD_DIR . uniqid() . '.png';
$success = file_put_contents($file, $data);
print $success ? $file : 'Unable to save the file.';
?>
归功于This Blog的php！
答案 1 :(得分：1)
它不起作用，因为您在take_snapshot()函数中包含了document.addEventListener("DOMContentLoaded", function(event) {函数。
请按照教程...
您的代码应如下所示：
<!DOCTYPE html>
<html>
<head>
<title>Javascript Webcam Demo - <MyCodingTricks/></title>
<link href="http://mycodingtricks.com/demo/style.css" rel="stylesheet"/>
</head>
<body>
<h3>Demonstrates simple 320x240 capture & display</h3>
<div id="my_camera"></div>
<!-- A button for taking snaps -->
<form>
<input type=button class="btn btn-success" value="Take Snapshot" onClick="take_snapshot()">
</form>
<div id="results" class="well">Your captured image will appear here...</div>
<script src="http://mycodingtricks.com/demo/script.js"></script>
<!-- First, include the Webcam.js JavaScript Library -->
<script type="text/javascript" src="webcam.min.js"></script>
<!-- Configure a few settings and attach camera -->
<script language="JavaScript">
Webcam.set({
width: 320,
height: 240,
image_format: 'jpeg',
jpeg_quality: 90
});
Webcam.attach( '#my_camera' );
function take_snapshot() {
// take snapshot and get image data
Webcam.snap( function(data_uri) {
// display results in page
document.getElementById('results').innerHTML =
'<h2>Here is your image:</h2>' +
'<img src="'+data_uri+'"/>';
Webcam.upload( data_uri, 'upload.php', function(code, text) {
// Upload complete!
// 'code' will be the HTTP response code from the server, e.g. 200
// 'text' will be the raw response content
});
} );
}
</script>
</body>
</html>