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[英]Passing data from ASP.NET View to a Controller Async using JS in .NET core
[英]passing data from controller to javascript in .Net Core
我想将数据从 controller 传递到 javascript 文件。 这是我的 model 字段
[Key]
public int id_field { get; set; }
public string name { get; set;
public int field_type { get; set; }
这是我的游戏模型
public class GameModel
{
public List<Board> board { get; set; }
public List<Player> player_list { get; set; }
public List<Dice> dices_value { get; set; }
public List<Field> field_list { get; set; }
}
这是 javascript 代码:
<script>
function setup() {
createCanvas(880, 880);
background(255);
for (var i = 0; i < 11; i++) {
var posX = map(i, 0, 11, 0, width);
var posY = map(i, 0, 11, 0, height);
var posX2 = map(i, 0, 11, 0, width);
var posY2 = map(i, 0, 11, height, 0);
var tileRowUp = new Tile(posX, 0, 80, 80);
if (i == 1) {
tileRowUp.nameOfField=@Html.Raw(Json.Serialize(Model.field_list[1].name));
}
if (i == 2) {
tileRowUp.nameOfField=@Html.Raw(Json.Serialize(Model.field_list[2].name));
}
tileRowUp.show();
var tileColLeft = new Tile(0, posY, 80, 80);
if (i == 0) {
tileColLeft.nameOfField=@Html.Raw(Json.Serialize(Model.field_list[0].name));
}
tileColLeft.show();
var tileRowdown = new Tile(posX2, height - 80, 80, 80);
tileRowdown.show();
var tileColRight = new Tile(width - 80, posY2, 80, 80);
tileColRight.show();
}
var mysteriousCard1 = new Tile(170, 190, 100, 100);
mysteriousCard1.show();
var mysteriousCard2 = new Tile(570, 490, 100, 100);
mysteriousCard2.show();
var dice = new Tile(390, 390, 50, 50);
dice.show();
}
class Tile {
constructor(x, y, lar, alt, id_Field,nameOfField,TypeOfField) {
this.x = x;
this.y = y;
this.lar = lar;
this.alt = alt;
this.id_Field = id_Field;
this.nameOfField = nameOfField;
this.TypeOfField = TypeOfField;
}
show() {
//noStroke();
rect(this.x, this.y, this.lar, this.alt);
text(this.nameOfField, this.x + 10, this.y + 10);
}
}
使用tileColLeft.nameOfField=@Html.Raw(Json.Serialize(Model.field_list[0].name));
我将数据分配给 object 名称 tileColLeft。 在这里,我将 js 代码和 HTML 混合在一起,我不想这样做。 这项工作,它向我显示来自数据库的数据。 这是结果https://imgur.com/FoeZZKL
此外,脚本 javascript 现在位于 my.cshtml 文件中。 我必须使这个 js 代码成为外部文件。 我考虑过使用 JsonResult 但我不明白我该怎么做。 有人有什么建议吗?
您可以在您的 cshtml 中使用以下内容-
@section Scripts{
<script>
var objModel=@Html.Raw(Json.Serialize(Model));
setup(objModel);
</script>
}
现在您可以在外部 JS 文件中定义设置 function 并且您需要进行必要的更改以使用 function 参数,而不是在 ZC1C425268E68385D1AB5074 中使用 @Html.Raw 。
function setup(modelObj) {
createCanvas(880, 880);
background(255);
for (var i = 0; i < 11; i++) {
var posX = map(i, 0, 11, 0, width);
var posY = map(i, 0, 11, 0, height);
var posX2 = map(i, 0, 11, 0, width);
var posY2 = map(i, 0, 11, height, 0);
var tileRowUp = new Tile(posX, 0, 80, 80);
if (i == 1) {
tileRowUp.nameOfField =modelObj.field_list[1].name;
}
if (i == 2) {
tileRowUp.nameOfField =modelObj.field_list[2].name;
}
tileRowUp.show();
var tileColLeft = new Tile(0, posY, 80, 80);
if (i == 0) {
tileColLeft.nameOfField =modelObj.field_list[0].name;
}
tileColLeft.show();
var tileRowdown = new Tile(posX2, height - 80, 80, 80);
tileRowdown.show();
var tileColRight = new Tile(width - 80, posY2, 80, 80);
tileColRight.show();
}
var mysteriousCard1 = new Tile(170, 190, 100, 100);
mysteriousCard1.show();
var mysteriousCard2 = new Tile(570, 490, 100, 100);
mysteriousCard2.show();
var dice = new Tile(390, 390, 50, 50);
dice.show();
}
class Tile {
constructor(x, y, lar, alt, id_Field, nameOfField, TypeOfField) {
this.x = x;
this.y = y;
this.lar = lar;
this.alt = alt;
this.id_Field = id_Field;
this.nameOfField = nameOfField;
this.TypeOfField = TypeOfField;
}
show() {
//noStroke();
rect(this.x, this.y, this.lar, this.alt);
text(this.nameOfField, this.x + 10, this.y + 10);
}
}
我使用 Json 和 ajax 从数据库中获取数据。这是我在 js 中的代码
var fields = [];
for (var i = 0; i < 40; i++) {
fields[i] = { id_Field: 0, nameOfField2: 'what', TypeOfField: 0 }
}
$(document).ready(function () {
//Call EmpDetails jsonResult Method
$.getJSON("Boards/Json",
function (json) {
var tr;
//Append each row to html table
for (var i = 0; i < json.length; i++) {
fields[i].nameOfField2 = json[i].name;
fields[i].id_Field = json[i].id_field;
fields[i].TypeOfField = json[i].field_type;
tr = $('<tr/>');
tr.append("<td>" + fields[i].id_Field + "</td>");
tr.append("<td>" + fields[i].nameOfField2 + "</td>");
tr.append("<td>" + fields[i].TypeOfField + "</td>");
$('table').append(tr);
}
});
});
function setup() {
createCanvas(880, 880);
background(255);
for (var i = 0; i < 11; i++) {
var posX = map(i, 0, 11, 0, width);
var posY = map(i, 0, 11, 0, height);
var posX2 = map(i, 0, 11, 0, width);
var posY2 = map(i, 0, 11, height, 0);
var tileRowUp = new Tile(posX, 0, 80, 80);
if (i == 1) {
tileRowUp.nameOfField = fields[i].nameOfField2;
}
// tileRowUp[2].nameOfField = fields[2].nameOfField2;
tileRowUp.show();
var tileColLeft = new Tile(0, posY, 80, 80);
tileColLeft.show();
var tileRowdown = new Tile(posX2, height - 80, 80, 80);
tileRowdown.show();
var tileColRight = new Tile(width - 80, posY2, 80, 80);
tileColRight.show();
}
var mysteriousCard1 = new Tile(170, 190, 100, 100);
mysteriousCard1.show();
var mysteriousCard2 = new Tile(570, 490, 100, 100);
mysteriousCard2.show();
var dice = new Tile(390, 390, 50, 50);
dice.show();
}
class Tile {
constructor(x, y, lar, alt, id_Field, nameOfField, TypeOfField) {
this.x = x;
this.y = y;
this.lar = lar;
this.alt = alt;
this.id_Field = id_Field;
this.nameOfField = nameOfField;
this.TypeOfField = TypeOfField;
}
show() {
//noStroke();
rect(this.x, this.y, this.lar, this.alt);
text(this.nameOfField, this.x + 10, this.y + 10);
}
}
这段代码
tr = $('<tr/>');
tr.append("<td>" + fields[i].id_Field + "</td>");
tr.append("<td>" + fields[i].nameOfField2 + "</td>");
tr.append("<td>" + fields[i].TypeOfField + "</td>");
$('table').append(tr);
在列和行中显示数据,它可以工作,但这是我现在的问题
tileRowUp.nameOfField = fields[i].nameOfField2;
这没有将数据分配给tileRowUp.nameOfField
,当我显示它时,我看到这个https://imgur.com/rvxgZiu它就像fields[i].nameOfField2
具有相同的默认值......它看起来像 fields[] 值本地的,只有在这里:
//Append each row to html table
for (var i = 0; i < json.length; i++) {
fields[i].nameOfField2 = json[i].name;
fields[i].id_Field = json[i].id_field;
fields[i].TypeOfField = json[i].field_type;
tr = $('<tr/>');
tr.append("<td>" + fields[i].id_Field + "</td>");
tr.append("<td>" + fields[i].nameOfField2 + "</td>");
tr.append("<td>" + fields[i].TypeOfField + "</td>");
$('table').append(tr);
}
有人有一些想法我做错了什么?
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