我在python中有以下请求
import requests, json, io
cookie = {}
payload = {"Name":"abc"}
url = "/test"
file = "out/test.json"
fi = {'file': ('file', open(file) )}
r = requests.post("http://192.168.1.1:8080" + url, data=payload, files=fi, cookies=cookie)
print(r.text)
发送文件,并将表单字段发送到后端。如何使用Angular $ http执行相同的操作(发送文件+表单字段)。目前,我喜欢这样,但不知道如何发送文件。
var payload = {"Name":"abc"};
$http.post('/test', payload)
.success(function (res) {
//success
});
答案 0 :(得分:127)
我必须同时上传文件并发送用户令牌信息时遇到类似问题。 transformRequest
FormData
以及 $http({
method: 'POST',
url: '/upload-file',
headers: {
'Content-Type': 'multipart/form-data'
},
data: {
email: Utils.getUserInfo().email,
token: Utils.getUserInfo().token,
upload: $scope.file
},
transformRequest: function (data, headersGetter) {
var formData = new FormData();
angular.forEach(data, function (value, key) {
formData.append(key, value);
});
var headers = headersGetter();
delete headers['Content-Type'];
return formData;
}
})
.success(function (data) {
})
.error(function (data, status) {
});
帮助:
$scope.file
为了获取文件app.directive('file', function () {
return {
scope: {
file: '='
},
link: function (scope, el, attrs) {
el.bind('change', function (event) {
var file = event.target.files[0];
scope.file = file ? file : undefined;
scope.$apply();
});
}
};
});
,我使用了自定义指令:
<input type="file" file="file" required />
HTML:
{{1}}
答案 1 :(得分:18)
我无法让Pavel的回答与发布到Web.Api应用程序时一样。
问题似乎在于删除标题。
headersGetter();
delete headers['Content-Type'];
为了确保允许浏览器将Content-Type与边界参数一起默认,我需要将Content-Type设置为undefined。使用Pavel的示例,从未设置边界,从而导致400 HTTP异常。
关键是删除删除上面标题的代码,并手动将标题内容类型设置为null。从而允许浏览器设置属性。
headers: {'Content-Type': undefined}
这是一个完整的例子。
$scope.Submit = form => {
$http({
method: 'POST',
url: 'api/FileTest',
headers: {'Content-Type': undefined},
data: {
FullName: $scope.FullName,
Email: $scope.Email,
File1: $scope.file
},
transformRequest: function (data, headersGetter) {
var formData = new FormData();
angular.forEach(data, function (value, key) {
formData.append(key, value);
});
return formData;
}
})
.success(function (data) {
})
.error(function (data, status) {
});
return false;
}
答案 2 :(得分:7)
我最近编写了一个支持本机多文件上传的指令。我创建的解决方案依赖于服务来填补您使用$ http服务识别的差距。我还添加了一个指令,它为角度模块提供了一个简单的API,用于发布文件和数据。
使用示例:
<lvl-file-upload
auto-upload='false'
choose-file-button-text='Choose files'
upload-file-button-text='Upload files'
upload-url='http://localhost:3000/files'
max-files='10'
max-file-size-mb='5'
get-additional-data='getData(files)'
on-done='done(files, data)'
on-progress='progress(percentDone)'
on-error='error(files, type, msg)'/>
您可以找到code on github以及my blog
上的文档由您来处理Web框架中的文件,但我创建的解决方案提供了将数据提供给服务器的角度接口。您需要编写的角度代码是响应上传事件
angular
.module('app', ['lvl.directives.fileupload'])
.controller('ctl', ['$scope', function($scope) {
$scope.done = function(files,data} { /*do something when the upload completes*/ };
$scope.progress = function(percentDone) { /*do something when progress is reported*/ };
$scope.error = function(file, type, msg) { /*do something if an error occurs*/ };
$scope.getAdditionalData = function() { /* return additional data to be posted to the server*/ };
});
答案 3 :(得分:2)
您也可以使用HTML5上传。您可以使用此AJAX uploader。
JS代码基本上是:
$scope.doPhotoUpload = function () {
// ..
var myUploader = new uploader(document.getElementById('file_upload_element_id'), options);
myUploader.send();
// ..
}
从HTML输入元素中读取
<input id="file_upload_element_id" type="file" onchange="angular.element(this).scope().doPhotoUpload()">
答案 4 :(得分:2)
这是我的解决方案:
// Controller
$scope.uploadImg = function( files ) {
$scope.data.avatar = files[0];
}
$scope.update = function() {
var formData = new FormData();
formData.append('desc', data.desc);
formData.append('avatar', data.avatar);
SomeService.upload( formData );
}
// Service
upload: function( formData ) {
var deferred = $q.defer();
var url = "/upload" ;
var request = {
"url": url,
"method": "POST",
"data": formData,
"headers": {
'Content-Type' : undefined // important
}
};
console.log(request);
$http(request).success(function(data){
deferred.resolve(data);
}).error(function(error){
deferred.reject(error);
});
return deferred.promise;
}
// backend use express and multer
// a part of the code
var multer = require('multer');
var storage = multer.diskStorage({
destination: function (req, file, cb) {
cb(null, '../public/img')
},
filename: function (req, file, cb) {
cb(null, file.fieldname + '-' + Date.now() + '.jpg');
}
})
var upload = multer({ storage: storage })
app.post('/upload', upload.single('avatar'), function(req, res, next) {
// do something
console.log(req.body);
res.send(req.body);
});
<div>
<input type="file" accept="image/*" onchange="angular.element( this ).scope().uploadImg( this.files )">
<textarea ng-model="data.desc" />
<button type="button" ng-click="update()">Update</button>
</div>
答案 5 :(得分:0)
答案 6 :(得分:0)
在我的解决方案中,我有
$scope.uploadVideo = function(){
var uploadUrl = "/api/uploadEvent";
//obj with data, that can be one input or form
file = $scope.video;
var fd = new FormData();
//check file form on being
for (var obj in file) {
if (file[obj] || file[obj] == 0) {
fd.append(obj, file[obj]);
}
}
//open XHR request
var xhr = new XMLHttpRequest();
// $apply to rendering progress bar for any chunking update
xhr.upload.onprogress = function(event) {
$scope.uploadStatus = {
loaded: event.loaded,
total: event.total
};
$scope.$apply();
};
xhr.onload = xhr.onerror = function(e) {
if (this.status == 200 || this.status == 201) {
//sucess
$scope.uploadStatus = {
loaded: 0,
total: 0
};
//this is for my solution
$scope.video = {};
$scope.vm.model.push(JSON.parse(e.currentTarget.response));
$scope.$apply();
} else {
//on else status
}
};
xhr.open("POST", uploadUrl, true);
//token for upload, thit for my solution
xhr.setRequestHeader("Authorization", "JWT " + window.localStorage.token);
//send
xhr.send(fd);
};
}
答案 7 :(得分:0)
请看看我的实施情况。您可以将以下函数包装到服务中:
function(file, url) {
var fd = new FormData();
fd.append('file', file);
return $http.post(url, fd, {
transformRequest: angular.identity,
headers: { 'Content-Type': undefined }
});
}
请注意,file
参数是Blob
。如果您拥有base64
个版本的文件,则可以像以下一样轻松更改为Blob
:
fetch(base64).then(function(response) {
return response.blob();
}).then(console.info).catch(console.error);