我正在尝试使用jQuery post和php上传canvas.todataurl()图像(getUserMedia)到服务器来处理数据,但我遇到了一些问题。我上传的所有图片最终都被破坏了,一半图像丢失了。我还有一个MySQL数据库,我存储与图像相关的数据(标题,文本,日期等)。似乎我拥有的相关数据越多,图像就越破坏。
因此,我想知道这是浏览器限制还是与jQuery帖子有关。我还检查了PHP max_post_size并且它是16mb,所以这应该不是问题。我无权访问服务器设置。我很困惑,我该怎么办?是否可以将canvas.todataurl()划分为多个部分,然后发布?
的JavaScript
window.addEventListener('DOMContentLoaded', function() {
var video = document.getElementById('videoStream');
var canvas = document.getElementById('canvasImage');
var status = document.getElementById('status');
var button = document.getElementById('button');
//var others = document.getElementById('others');
var imageHolder;
document.getElementById('form').style.display = 'none';
var image = null; // kuvan datauri joka lähtee php:lle
window.URL || (window.URL = window.webkitURL || window.mozURL || window.msURL);
navigator.getUserMedia || (navigator.getUserMedia = navigator.mozGetUserMedia || navigator.webkitGetUserMedia || navigator.msGetUserMedia);
// toString : function() {return "video,audio";} canarya varten
if (navigator.getUserMedia) {
navigator.getUserMedia({video: true, audio: false, toString : function() {return "video,audio";}}, onSuccess, onError);
} else {
status.innerText = "getUserMedia is not supported in your browser, sorry :(";
}
function onSuccess(stream) {
var source;
if (window.webkitURL) {
source = window.webkitURL.createObjectURL(stream);
} else {
source = stream; // Opera ja Firefox
}
video.width = 500;
video.height = 375;
video.autoplay = true;
video.src = source;
}
function onError() {
status.innerText = "Please allow access to your webcam.";
}
button.addEventListener('mousedown', function() {
// Poistetaan aikaisempi kuva jos sellaista on
//document.body.removeChild(imageHolder);
// luodaan kuva uudestaan
imageHolder = document.createElement('figure');
imageHolder.id = 'imageHolder';
document.body.appendChild(imageHolder);
img = document.createElement('img');
imageHolder.appendChild(img);
// kuva on yhtäsuuri kuin video
canvas.width = video.width;
canvas.height = video.height;
img.width = 350;
img.height = 225;
// piirretään canvasille kuva videosta
var context = canvas.getContext('2d');
context.drawImage(video, 0, 0, canvas.width, canvas.height);
}, false);
button.addEventListener('mouseup', function() {
// Canvasilta kuvaksi levylle tallentamista varten
canvas.style.display = 'none';
video.style.display = 'none';
button.style.display = 'none';
others.style.display = 'none';
document.getElementById('form').style.display = 'block';
image = canvas.toDataURL('image/png');
img.src = image;
}, false);
// jquery post
$('#send').click(function(){
var image2 = image.replace('data:image/png;base64,', '');
$.post('upload.php',
{
title: $('#title').val(),
blog: $('#blog').val(),
category: $('#category').val(),
author: $('#author').val(),
imagename: image2
});
});
}, false);
PHP upload.php
define('UPLOAD_DIR', 'images/');
$img = $_POST['imagename'];
$img = str_replace(' ','+', $img);
$data = base64_decode($img);
$file = UPLOAD_DIR . uniqid() . '.png';
$success = file_put_contents($file, $data);
print $success ? $file : 'Tiedoston tallennus ei sitten onnistu millään...';
$imagename = $file; // this is the file name for the MySQL database
我的问题是(我认为)image = canvas.toDataURL('image / png');和jQuery帖子。 canvas.toDataUrl()字符串长约70万个字母。
答案 0 :(得分:0)
你可能想试试这个:
<?php
$decoded = "";
for ($i=0; $i < ceil(strlen($encoded)/256); $i++)
$decoded = $decoded . base64_decode(substr($encoded,$i*256,256));
?>
我是从这里得到的:http://www.php.net/manual/en/function.base64-decode.php#92980
代码基本上尝试部分解码base64字符串。我没有测试过这个。我从未处理过与你正在使用的一样大的base64图像。
答案 1 :(得分:0)
拆分它,使用两个变量并在php中合并,工作正常。 ; - )
var resourcelength_all = resource.length;
var resourcelength_split = resourcelength_all / 2;
var resource_part1 = resource.substr(0, resourcelength_split);
var resource_part2 = resource.substr(resourcelength_split, resourcelength_all);