在我正在构建的软件中,我正在使用网络摄像头拍照,然后将其上传到 PHP 服务器供以后使用。我已经使用表单成功地将图像从我的计算机上传到服务器,但是由于我要上传的图像已经预先确定,所以我一直在研究使用 AJAX。我从网络摄像头拍摄快照的代码如下:
var video = document.getElementById('video');
// Get access to the camera!
if (navigator.mediaDevices && navigator.mediaDevices.getUserMedia) {
// Not adding `{ audio: true }` since we only want video now
navigator.mediaDevices.getUserMedia({video: true}).then(function (stream) {
video.src = window.URL.createObjectURL(stream);
video.play();
});
}
// Elements for taking the snapshot
var canvas = document.getElementById('canvas');
var context = canvas.getContext('2d');
var video = document.getElementById('video');
// Trigger photo take
document.getElementById("snap").addEventListener("click", function () {
var img = context.drawImage(video, 0, 0, 640, 480);
});
document.getElementById("capture").addEventListener("click", function () {
context.drawImage(video, 0, 0, 640, 480);
//var capture = canvas.toDataURL();
});
这是我保存这张图片的 php 代码:
<?php
if (isset($_POST['upload'])) {
$image_name = $_FILES['image']['name'];
$image_name = $_FILES['image']['type'];
$image_name = $_FILES['image']['size'];
$image_tmp_name = $_FILES['image']['tmp_name'];
if ($image_name == '') {
echo "<script>alert('Please select an image.')</script>";
exit();
} else {
move_uploaded_file($image_tmp_name, "pictures/$image_name");
echo "Image uploaded succesfully.";
echo "<img src= 'pictures/$image_name'>";
}
}
?>
我确信这两段代码都能正常工作,因为我已经分别对它们进行了测试。如前所述,我正在尝试使用 AJAX 将它们连接在一起。我为此找到的示例代码是:
function saveImage() {
var capture = canvas.toDataURL("image/png");
var xmlHttpReq = false;
if (window.XMLHttpRequest) {
ajax = new XMLHttpRequest();
}
else if (window.ActiveXObject) {
ajax = new ActiveXObject("Microsoft.XMLHTTP");
}
ajax.open('POST', 'picture.php', false);
ajax.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
ajax.onreadystatechange = function () {
console.log(ajax.responseText);
}
ajax.send("imgData=" + capture);
}
我在按钮中使用了 onClick = "saveImage()"来尝试调用该函数并将图像上传到服务器。当我按下按钮时,该文件似乎根本无法访问服务器。有什么我想念的吗?
最佳答案
您没有将其作为文件发送,而是将其作为一次使用发送 Blob或 File对象
var blob = new Blob([capture ], {type: 'application/octet-binary'});
var form = new FormData();
var fileName = 'snap.png'; //filename
form.append('something', blob, fileName);
在服务器上解码并保存
$dest = 'uploads/'.$_FILES['something']['name'];
move_uploaded_file($_FILES['something']['tmp_name'],$dest);
$file = file_get_contents($dest);
$tmp = explode(',',$file); //remove data: header
file_put_contents($dest,base64_decode($tmp[1])); //decode base64 n save
关于javascript - 使用 AJAX 将图像从 Canvas 上传到 PHP 服务器,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40372352/