我正在使用$cordovaFileTransfer.upload()
将图片从设备上传到我的服务器。在文档中,它说它不包括这些参数:server, filePath, options
。我需要传递一些额外的自定义参数:例如user_id
。有谁知道是否可以这样做?谢谢。
答案 0 :(得分:0)
您可以在options对象中发送这些自定义参数,执行以下操作:
var options = {};
options.headers = {
Connection: "close"
}
options.chunkedMode = false;
options.userName = 'Mati';
$cordovaFileTransfer.upload(server, filePath, options)
答案 1 :(得分:0)
You can add "params" to options object, for example :
var options = {
fileKey: "name",
fileName: "image.png",
chunkedMode: false,
mimeType: "image/png",
httpMethod: "post", //here the methed (get, post,..)
params: $scope.Data //here you can submit the data
};
答案 2 :(得分:0)
您需要使用'params'属性,如下所示:
var _options = {
fileKey: "image_name",
fileName: "image.jpg",
chunkedMode: false,
mimeType: "image/jpeg",
httpMethod: "POST",
headers: { },
params: {
"extra_param_name": "Jamilson",
"extra_param_last_name": "Junior"
}
};
希望这有帮助。
答案 3 :(得分:0)
var options = new FileUploadOptions();
options.fileKey = "image";
options.fileName="image.jpg";
options.mimeType = "image/jpeg";
options.httpMethod="POST";
options.headers = {
Connection: "close"
};
options.chunkedMode = false;
var params = {};
params.user_id ="123";
params.username="username";
options.params = params;