我想<% f.submit %>在新对象表单上调用一个确认弹出窗口.有没有办法在没有javascript的情况下执行此操作?
device当从rfid部分view/cabinet/show页面导航到新表单时,如何获取值以预填充新表单? devicehas_one rfid.
来自橱柜/展示的链接:
<%= link_to "Create New Device (non-functional)", new_device_path({:id => @rfid.id, cabinet_id: @cabinet.id}), :class => "btn btn-primary" %>
Run Code Online (Sandbox Code Playgroud)
的devices_controller,我想有create方法工作的时候是0或2 PARAMS传递:
def create(options)
if options[:cabinet_id] and options[:id]
@rfid = Rfid.find(params[:id])
@device = Device.new(params[:device])
@device.rfid = @rfid
@device.cabinet_id = :cabinet_id
else
@device = Device.new(params[:device])
@cabinet = Cabinet.find(params[:device][:cabinet_id])
@device.row_id = @cabinet.row_id
end
respond_to do |format|
if @device.save and @cabinet.refresh_capacity_used and @cabinet.refresh_consumption
format.html { redirect_to @device, notice: 'Device was successfully created.' }
format.json …Run Code Online (Sandbox Code Playgroud) 我正在尝试对表单的表单字段进行泛化.我想要有不同的parsley.js验证要求.html中简单字段的示例可能是:required 有2个非标准属性.
我尝试了以下方法,但得到的错误是:attr定义> 1次:
<input type=text required data-parsley-range=[1,100] />
Run Code Online (Sandbox Code Playgroud)
如何在th:attr中定义> 1个属性,只是插值?
奖金,我如何有条件地放置attrs,例如data-parsley-range,如果我可以避免它,我根本不想打印.范围也一样.
谢谢!
无论我如何格式化此请求的原始部分,我都无法避免下面的解析错误.
我有一个带有传递规范的create方法的Rails API,以说明控制器消息是合理的:
describe "POST power_up" do
let!(:client) { FactoryGirl.create :client, name: "joe", auth_token: "asdfasdfasdfasdfasdfasdf" }
it "should create and save a new Power Up" do
expect { post :create, format: :json, power_up: FactoryGirl.attributes_for(:power_up) }.to change(V1::PowerUp, :count).by(1)
end
end
Run Code Online (Sandbox Code Playgroud)
我正在使用Postman尝试POST.无论我尝试什么,我都会收到错误:
Started POST "/v1/power_ups.json" for 127.0.0.1 at 2014-08-30 18:05:29 -0400
Error occurred while parsing request parameters.
Contents:
{
'name': 'foo',
'description': 'bar'
}
ActionDispatch::ParamsParser::ParseError (795: unexpected token at '{
'name': 'foo',
'description': 'bar'
}
Run Code Online (Sandbox Code Playgroud)
邮差要求设置:

我也尝试过:
{
'power_up': {
'name': 'foo', …Run Code Online (Sandbox Code Playgroud) 文件a.txt看起来像:
ABC
Run Code Online (Sandbox Code Playgroud)
文件d.txt看起来像:
DEF
Run Code Online (Sandbox Code Playgroud)
我正试图拿"DEF"并将其附加到"ABC",所以a.txt看起来像
ABC
DEF
Run Code Online (Sandbox Code Playgroud)
我尝试过的方法总是完全覆盖第一个条目,所以我总是最终得到:
DEF
Run Code Online (Sandbox Code Playgroud)
以下是我尝试过的两种方法:
FileChannel src = new FileInputStream(dFilePath).getChannel();
FileChannel dest = new FileOutputStream(aFilePath).getChannel();
src.transferTo(dest.size(), src.size(), dest);
Run Code Online (Sandbox Code Playgroud)
......我试过了
FileChannel src = new FileInputStream(dFilePath).getChannel();
FileChannel dest = new FileOutputStream(aFilePath).getChannel();
dest.transferFrom(src, dest.size(), src.size());
Run Code Online (Sandbox Code Playgroud)
API不清楚transferTo和transferFrom param描述:
谢谢你的任何想法.
我有测试确保时间戳不会改变。他们在我的本地环境中通过,但在构建中失败。它与“updated_at”返回的值有关,即将第二个小数的最后 3 位数字更改为 0。以下是失败的示例:
RSpec::Expectations::ExpectationNotMetError:
expected: 2015-01-06 10:16:29.948655841 -0500
got: 2015-01-06 10:16:29.948655000 -0500
RSpec::Expectations::ExpectationNotMetError:
expected: 2015-01-06 10:16:31.236196108 -0500
got: 2015-01-06 10:16:31.236196000 -0500
Run Code Online (Sandbox Code Playgroud)
如何构建断言以忽略最后 3 位数字?是否有一个时间函数只比较时间,比如分数的百位?
谢谢。
在UNIX中,我读到将shell脚本移动到/ usr/local/bin将允许您通过只需键入"[scriptname] .sh"并按Enter键从任何位置执行脚本.
我已经移动了具有普通用户和root权限的脚本,但我无法运行它.
剧本:
#! bin/bash
echo "The current date and time is:"
date
echo "The total system uptime is"
uptime
echo "The users currently logged in are:"
who
echo "The current user is:"
who -m
exit 0
Run Code Online (Sandbox Code Playgroud)
当我尝试移动然后运行脚本时会发生这种情况:
[myusername@VDDK13C-6DDE885 ~]$ sudo mv sysinfo.sh /usr/local/bin
[myusername@VDDK13C-6DDE885 ~]$ sysinfo.sh
bash: sysinfo.sh: command not found
Run Code Online (Sandbox Code Playgroud) 我正在做一个简单的教程,无法通过一个非常简单的事情.
我想在页面加载时创建一个元素.我完全遵循了代码.我正在使用Google Chrome.以下不会创建音频元素有什么问题?:
的index.html
<!doctype html>
<html ng-app="myApp">
<head>
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.0.7/angular.min.js"></script>
<script src="js/main.js"></script>
</head>
<body>
<div ng-controller="PlayerController">
<button ng-click="play()" class="button" ng-show="!playing">Play</button>
<button ng-click="stop()" class="button alert" ng-show="playing">Stop</button>
Playing audio: <b>{{ playing }}</b>
</div>
</body>
</html>
Run Code Online (Sandbox Code Playgroud)
main.js(在引用的位置js/main.js中)
var app = angular.module('myApp', []);
app.controller('PlayerController', ['$scope', function($scope) {
$scope.playing = false;
$scope.div = document.createElement("div");
$scope.audio = document.createElement('audio');
$scope.audio.src = 'media/sample_mpeg4.mp4';
}]);
Run Code Online (Sandbox Code Playgroud)
这是源页面检查时的样子:
<!doctype html>
<html ng-app="myApp">
<head>
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.0.7/angular.min.js"></script>
<script src="js/main.js"></script>
</head>
<body>
<div ng-controller="PlayerController">
<button ng-click="play()" class="button" ng-show="!playing">Play</button>
<button ng-click="stop()" class="button alert" ng-show="playing">Stop</button> …Run Code Online (Sandbox Code Playgroud) 测试以下内容需要一些帮助.我正在做关于保护api的RailsCast:http://railscasts.com/episodes/352-securing-an-api?view=asciicast
我有RequestController一个before_filter检查,如果要求有一个令牌:
class RequestsController < ApplicationController
include ActionController::MimeResponds
include ActionController::HttpAuthentication::Token::ControllerMethods
before_filter :restrict_access
respond_to :json
#...
def authenticate
return restrict_access
end
private
def restrict_access
authenticate_or_request_with_http_token do |token, options|
ApiKey.exists?(access_token: token)
end
end
end
Run Code Online (Sandbox Code Playgroud)
我失败的rspec测试看起来像:
it 'responds successfully to generic request because of key protection' do
api_key = ApiKey.create
api_key.save!
get :index
request.headers["token"] = api_key.access_token
expect(response).to be_success # test for the 200 status-code
end
Run Code Online (Sandbox Code Playgroud)
结果: expected success? to return true, got false
我不明白如何将有效的api_key注入到请求中,以便响应将评估为true.有任何想法吗?谢谢.
orderRequestDao.save(new OrderRequest("5000", "body"));成功执行会在 Dynamo 中放置一条记录。任何读取返回的尝试:
[Request processing failed; nested exception is com.amazonaws.services.dynamodbv2.datamodeling.DynamoDBMappingException:
could not invoke null on class
com.cfa.fulfillmentApi.model.OrderRequest
with value 100 of type class java.lang.String] with root cause
Run Code Online (Sandbox Code Playgroud)
(id 为 100 的记录存在)
我正在使用以下 jars (aws.sdk.version: 1.11.86):
<dependency>
<groupId>com.amazonaws</groupId>
<artifactId>aws-java-sdk</artifactId>
<version>${aws.sdk.version}</version>
</dependency>
<dependency>
<groupId>com.amazonaws</groupId>
<artifactId>aws-java-sdk-core</artifactId>
<version>${aws.sdk.version}</version>
</dependency>
<dependency>
<groupId>com.amazonaws</groupId>
<artifactId>aws-java-sdk-dynamodb</artifactId>
<version>${aws.sdk.version}</version>
</dependency>
<dependency>
<groupId>com.github.derjust</groupId>
<artifactId>spring-data-dynamodb</artifactId>
<version>4.4.1</version>
</dependency>
Run Code Online (Sandbox Code Playgroud)
DynamoDb 配置:
主分区键: id (String)
道:
@EnableScan
public interface OrderRequestDao extends CrudRepository<OrderRequest, String> {
OrderRequest findOne(String s);
OrderRequest save(OrderRequest or);
}
Run Code Online (Sandbox Code Playgroud)
域对象: …