小编use*_*491的帖子

我使用contentEditable div,我想在用户修改它之后访问它的内容(onBlur).如何访问DOM元素的innerHtml或textContent？任何的想法 ？

这是一个简单的例子:

import Html exposing (div, text)
import Html.App exposing (beginnerProgram)
import Html.Events exposing (onBlur)
import Html.Attributes exposing (contenteditable)


main =
  beginnerProgram { model = "Hello", view = view, update = update }


view model =
  div [contenteditable True, onBlur (SaveInput)] [ text model ]


type Msg = SaveInput


update msg model =
  case msg of
    SaveInput ->
      "ok"

我希望使用textContent将公式替换为"ok" 以获取用户输入的值.我正在使用elm v 0.17.

我或python是否与以下代码混淆？我希望__le__被召唤a <= ab,而不是__ge__:

#!/usr/bin/env python2

class B(object):
    def __ge__(self, other):
        print("__ge__ unexpectedly called")

class A(object):
    def __le__(self, other):
        print("__le__ called")

class AB(A, B):
    pass

a = A()
ab = AB()

a <= ab # --> __ge__ unexpectedly called
ab <= a # --> __le__ called

我得到了与python 2.7,3.2和pypy 1.9相同的行为.

我该怎么办才能被__le__召唤而不是__ge__？

我编译了" Hello world "程序

import Html exposing (text)

main =
  text "Hello, World!"

在Windows上使用elm 0.17,没有编译错误:

elm make hello.elm --output index.html

当我在index.html上打开Chrome时,我会看到一个空白页面.Chrome的控制台在index.html中显示2个错误:

Uncaught ReferenceError: _elm_lang$virtual_dom$VirtualDom$text is not defined
Uncaught ReferenceError: Elm is not defined

如果我在该文件上运行elm-reactor,我也会得到一个白页,并在控制台中出现类似的错误.

不知何故,路径丢失了...这是elm-package.json文件:

{
"version": "1.0.0",
"summary": "helpful summary of your project, less than 80 characters",
"repository": "https://github.com/user/project.git",
"license": "BSD3",
"source-directories": [
    "."
],
"exposed-modules": [],
"dependencies": {
    "elm-lang/core": "4.0.0 <= v < 5.0.0",
    "elm-lang/html": "1.0.0 <= v < 2.0.0"
},
"elm-version": "0.17.0 <= v < 0.18.0"
} …

我想将程序的视图和更新部分分离为单独的源文件,但是,如何共享消息和订阅声明？

我已将todomvc 示例onenter中的代码改编为 create ，但它不起作用。显然，没有传递给 Elm。那么，我怎样才能检测到shift-Enter呢？onShiftEntershiftKey

onShiftEnter : Msg -> Attribute Msg
onShiftEnter msg =
  let
    tagger (code, shift) =
      if code == 13 && shift then msg else NoOp
  in
    on "keydown" 
       (Json.Decode.map tagger 
          ( Json.Decode.tuple2 (,)
              (Json.Decode.at ["keyCode"] Json.Decode.int)
              (Json.Decode.at ["shiftKey"] Json.Decode.bool)
          )
       )

我对此非常陌生,我正在尝试创建一个基于文本的扫雷.我希望玩家决定他希望网格有多大.我的问题是,应该确保用户输入1到10之间的数字的if语句不起作用.请看一看.

scanf ("%i/%i",&x,&y);
if (0 < x < 11 && 0 < y < 11)
{   
printf ("you have selected %i by %i\n",x,y);
for (i = 0; i < x; i++) 
    {
    for (j = 0; j < y; j++) 
        {
        grid[x][y] = 'O';
        printf ("%c ", grid[x][y]);
        }
    printf ("\n");
    }
}
else
printf ("Wrong gridsize");