当使用webcam.js转换base64图像时，我得到损坏的图像
当使用webcam.js转换base64图像时，我得到损坏的图像
提问于 2015-09-15 20:50:50
这是我的html和使用webcam.js的javascript。我只是按照https://github.com/jhuckaby/webcamjs介绍了如何使用现有的表单来实现它。
<html lang="en">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<title>WebcamJS Test Page</title>
<style type="text/css">
body { font-family: Helvetica, sans-serif; }
h2, h3 { margin-top:0; }
form { margin-top: 15px; }
form > input { margin-right: 15px; }
#results { float:right; margin:20px; padding:20px; border:1px solid; background:#ccc; }
</style>
</head>
<body>
<div id="results">Your captured image will appear here...</div>
<h1>WebcamJS Test Page</h1>
<h3>Demonstrates simple 320x240 capture & display</h3>
<div id="my_camera"></div>
<!-- First, include the Webcam.js JavaScript Library -->
<script type="text/javascript" src="../webcam.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' );
Webcam.snap( function(data_uri) {
var raw_image_data = data_uri.replace(/^data\:image\/\w+\;base64\,/, '');
document.getElementById('mydata').value = raw_image_data;
document.getElementById('myform').submit();
} );
</script>
<!-- A button for taking snaps -->
<form id="myform" method="post" action="myscript.php">
<input id="mydata" type="hidden" name="mydata" value=""/>
<input type=button value="Take Snapshot" onClick="take_snapshot()">
<input type="submit" value="submit">
</form>
<!-- Code to handle taking the snapshot and displaying it locally -->
<script language="JavaScript">
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+'"/>';
} );
}
</script>这是保存图像的myscript.php。我成功地将路径保存在数据库中，但我得到了一个损坏的.jpg文件(文件大小始终为7字节)。
<?php
include 'connect.php';
$encoded_data = $_POST['mydata']; // to get the base 64 code image link
$name = base64_decode($encoded_data); // to convert base 64 code
$name = date('YmdHis');
$newname="images/".$name.".jpg";
$file = file_put_contents( $newname, file_get_contents('php://input') );
if (!$file) {
print "Error occured here";
exit();
}
else
{
$sql="INSERT INTO image (images) VALUES('$newname')";
$result=mysqli_query($con,$sql);
$value=mysqli_insert_id($con);
$_SESSION["myvalue"]=$value;
}
$url = 'http://' . $_SERVER['HTTP_HOST'] . dirname($_SERVER['REQUEST_URI']) . '/' . $newname;
print "$url\n";
?>回答 1
Stack Overflow用户
发布于 2016-05-28 13:16:39
var binaryImg = atob(base64string);
var length = binaryImg.length;
var ab = new ArrayBuffer(length);
var ua = new Uint8Array(ab);
for (var i = 0; i < length; i++) {
ua[i] = binaryImg.charCodeAt(i);
}
var blob = new Blob([ab], {
type: "image/jpeg"
});enter code here
var imgFile = new File([blob], 'photo.jpeg', {type: 'image/jpeg'});现在，您可以使用imgFile向远程服务器发送数据。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/32586588
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