对于“键盘”对象,我有以下rest结构:
GET /keyboards/ -> Lists user keyboards
POST /keyboards/ -> Creates new keyboard
GET /keyboards/{id} -> Get specific keyboard by id
PUT /keyboards/{id} -> Update specific keyboard by id
DELETE /keyboards/{id} -> Delete specific ke
我已经创建了表示模型,并希望将其(使用AutoMapper)映射到ViewModel中。ViewModel是复合的/因为我使用的是partials,并且我想要重用,例如,KeyboardsViewModel也可以在其他视图/partials上使用。
如何将此演示模型映射(设置映射)到ViewModel中?这是正确的方法吗?
谢谢!
public class MainPresentationModel : BasePresentationModel
{
// Should map into the MainViewModel.Keyboards.Keyboards
public int
我有3个功能,每个功能都与Promise.resolve一起工作,
如何为所有函数使用Promise.resolve?,当我调用所有函数时,这些函数都没有排序
function sendAllText(msg, opts) {
if (locale.keyboards[msg.text].text) {
var i,j,tempstring, promise;
promise = Promise.resolve();
for (i=0,j=locale.keyboards[msg.text].text.length; i<j; i++) {
#function description
def getMoneySpent(keyboards, drives, b):
q = []
for i in range(len(keyboards)):
for j in range(len(drives)):
q.append(keyboards[i] + drives[j])
for m in range(len(q)):
if(q[m] > b):
q.remove
我做了一些搜索,但没有成功,我想知道是否有更好的方法来重写sql查询,因为LEFT JOIN中的这个OR条件会降低性能:(
例如:
SELECT DISTINCT * FROM computers
LEFT JOIN monitors ON computers.brand = monitors.brand
LEFT JOIN keyboards ON computers.type = keyboards.type
LEFT JOIN accessories ON accessories.id = keyboards.id OR accessories.id = monitors.id
GROU
在以下Python代码中:
keyboards = [3, 1]
drivers = [5, 2, 8]
upper_limit = 10
sums = [k + d for k in keyboards for d in drivers if (k + d) <= upper_limit]
我想将k+d的结果存储在列表理解中,以便在列表理解中引用它。在Python3中是可能的吗?
我知道我们可以做到以下几点:
sums = []
for k in keyboards:
for d in drivers:
s = k + d
if s <=
在Hackerrank中,这是一个叫做电子商店的挑战,这是我的代码,它可以在一定程度上进行测试。当输入增加时,编译时间变得更长。所以我需要将一个数组中的每个数字与第二个数组中的其他数字相加。 function getMoneySpent(keyboards, drives, b) {
let purchase = [];
let result = -1;
for(let i = 0; i < keyboards.length; i++) {
for(let j = 0; j < drives.length; j++) {
if((keyboard
如何从电报机的数组中发送有序数据?
例如,我有一个数组中的text1、text2、text3,但是它像text3、text1、text2一样发送它们,所以不按原来的顺序发送。
这是我的密码:
function sendAllText(msg, opts) {
if (locale.keyboards[msg.text].text) {
var i,j,tempstring;
for (i=0,j=locale.keyboards[msg.text].text.length; i<j; i++) {
tempstring = locale.
请帮助解决游戏所需的swtich包
公共静态空main(String[]参数){
Scanner input = new Scanner(System.in);
System.out.print("Please Enter a number");
int day = input.nextInt();
switch(day)
{
case 1: System.out.println("1 Microphone");
break;
case 2: System.out.println("2 Loud Speakers 1 Microphone ")
我有一个文本框,我需要用户选择一个表情符号继续进行。默认情况下,如何触发表情符号键盘?
我知道有一个方法可以得到所有的键盘,但我不知道如何选择一个默认。
NSArray *array = [[NSUserDefaults standardUserDefaults] objectForKey:@"AppleKeyboards"];
NSLog(@"Keyboards: %@", array);
The output of all the installed international keyboards looks as follows:
Keyboards:
在grails文档中:7 Web层-参考文档,部分:绑定到集合和地图;下一个示例显示:
class Album {
String title
static hasMany = [players: Player]
Map players
}
class Player {
String name
}
def bindingMap = [title: 'The Lamb Lies Down On Broadway',
'players[guitar]': [name: 'Steve Hack
这是本的第2部分,非常感谢David的。如果我需要提取由两个关键字限定的日期,该怎么办?
示例:
text = "One 09 Jun 2011 Two 10 Dec 2012 Three 15 Jan 2015 End"
Case 1 bounding keyboards: "One" and "Three"
Result expected: ['09 Jun 2011', '10 Dec 2012']
Case 2 bounding keyboards: "Two" and "End
嗨,我没有通过所有的测试,只有9/16。所以我想知道我的代码问题链接:。
function getMoneySpent(keyboards, drives, b) {
let arr = [];
let lenOfArr1 = keyboards.length;
let lenOfArr2 = drives.length;
let j = drives.length;
arr = keyboards.slice(0);
for (let number of drives) {
arr.push(number);
}
当我制作gui时,我试图使GUI能够打开和关闭机器人。我设法让它打开,但是每次它的运行和我选择选项卡到GUI窗口关闭它,程序就会崩溃。我最好的猜测是,它被困在某个循环中,我不知道如何修复它。这是我的密码
from pynput.keyboard import Key, Controller
import pyautogui
keyboards = Controller()
import keyboard as kb
from threading import Thread
from tkinter import *
from variables import *
import time
#he