我正在创建一个应用程序[托管在Server1上],在其中录制视频并将其上传到服务器[Server2]。我正在使用Ajax上传文件。录制完成后,我将获得文件[videoSrc]的URL作为“ blob:https://server1/a0080679-caaf-41b0-b6a0-ce41d1283476”。我正在使用以下代码使用Ajax上传视频:
function uploadFile()
{
var data = new FormData();
data.append('video', videoSrc);
$.ajax({
type: "POST",
enctype: 'multipart/form-data',
url: "https://server2/upload.php",
data: data,
processData: false,
contentType: false,
cache: false,
timeout: 600000,
success: function (data) {
console.log(data);
},
error: function (e) {
console.log("ERROR : ", e);
}
});
}
在服务器端(PHP)上,我只是在post参数中获取URL。我没有在$ _FILES指令中获取文件。代码:
print_r($_POST);
print_r($_FILES);
输出:
Array
(
[video] => blob:https://server1/a0080679-caaf-41b0-b6a0-ce41d1283476
)
Array
(
)
如何上传录制的视频?请帮助。
我通过反复试验找到了解决方法。我正在发送使用createObjectURL()创建的Blob(记录的输出)的URL。我直接使用Ajax发送了Blob,PHP $ _FILES收到了它。
早期代码:
var blob = new Blob(recordedChunks, {
type: 'video/mp4'
});
//Creating blob url which is only valid for the browser
//where the application is running
var videoSrc = URL.createObjectURL(blob);
var data = new FormData();
data.append('video', videoSrc); //Incorrect: sending the Blob URL
$.ajax({
type: "POST",
enctype: 'multipart/form-data',
url: "https://server2/upload.php",
data: data,
processData: false,
contentType: false,
cache: false,
timeout: 600000,
success: function (data) {
console.log(data);
},
error: function (e) {
console.log("ERROR : ", e);
}
});
更正的代码:
var blob = new Blob(recordedChunks, {
type: 'video/mp4'
});
var data = new FormData();
data.append('video', blob); //Correct: sending the Blob itself
$.ajax({
type: "POST",
enctype: 'multipart/form-data',
url: "https://server2/upload.php",
data: data,
processData: false,
contentType: false,
cache: false,
timeout: 600000,
success: function (data) {
console.log(data);
},
error: function (e) {
console.log("ERROR : ", e);
}
});