Question

我有一个非常简单的课程

structure(list(soc_sec = c("AA2105480", "AA2105480", "AB4378973", 
"AB4990257", "AB7777777", "AB7777777", "AB7777777", "AC4285291", 
"AC4285291", "AC6039874", "AC6039874", "AC6039874", "AC6039874", 
"AC6547645", "AC6547645", "AC6547645", "AD1418577", "AD1418577", 
"AD1418577", "AD1418577"), group_count = c(5L, 7L, 1L, 2L, 5L, 
6L, 5L, 2L, 1L, 9L, 7L, 8L, 7L, 7L, 6L, 1L, 7L, 8L, 6L, 7L), 
    total_creds = c(14, 17, 0, 1, 12, 15, 15, 3, 3, 17.5, 16, 
    12.5, 13.5, 18, 17, 2, 13, 16, 15, 13), group_start = structure(c(12792, 
    12656, 12438, 16314, 12438, 12656, 12792, 16314, 16447, 14615, 
    14979, 14852, 15217, 12792, 12656, 12893, 15714, 15945, 16078, 
    16673), class = "Date"), group_end = structure(c(12919, 12762, 
    12545, 16418, 12864, 12762, 12915, 16418, 16540, 14735, 15093, 
    14964, 15329, 12915, 12762, 12935, 15842, 16052, 16195, 16784
    ), class = "Date")), class = c("tbl_df", "data.frame"), row.names = c(NA, 
-20L), .Names = c("soc_sec", "group_count", "total_creds", "group_start", 
"group_end"))

我从服务器获取此类的元素并将json转换为此类：

export class Foo {

    name: string;

    index: number;

    toFullString(): string {
        return `${this.name} | ${this.index}`;
    }
}

现在发生的事情是，在我调用.map(response => response.json() as Foo)方法后，它失败了，因为这不是＆＃34;真正的＆＃34; {C}的toFullString()对象。

那么实现真正的Foo对象的正确方法是什么，最好不需要编写自己的构造函数等等。

TypeScript中的类型安全有时真的是个笑话。

Answer 1

您可以创建Object并为其指定JSON值。

.map(res => this.onResponse(res))

-

function onResponse(res:any){
    this.foo = new Foo();
    this.foo.name = res['name'];
    ...
}

或者将JSON值赋给Foo的构造函数

this.foo = new Foo(res['name'], ...);

Answer 2

要么您必须编写自己的构造函数，要么必须使用返回的obejcts并使用# Get the License SkuId from a template user that we want to apply to the new user $licensedUser = Get-AzureADUser -ObjectId "TemplateUser@contoso.com" # Get the new User we want to apply the license too $user = Get-AzureADUser -ObjectId "newuser@contoso.com" # Create the new License object $license = New-Object -TypeName Microsoft.Open.AzureAD.Model.AssignedLicense $license.SkuId = $licensedUser.AssignedLicenses.SkuId # Create the Licenses Table and add the license from above $licenses = New-Object -TypeName Microsoft.Open.AzureAD.Model.AssignedLicenses $licenses.AddLicenses = $license # Apply the license to the new user Set-AzureADUserLicense -ObjectId $user.ObjectId -AssignedLicenses $licenses附加函数（技术上合法的JS）。一个你不想要的，另一个有点粗略。

您已经看过的构造函数方法，但我将在此重复：

for(let obj of responseObjects) { obj['toFullString'] = function ... }

你也可以在地图上做同样的事情。

constructor(public name, public index) {
    this.name = name;
    this.index = index;
}

自动映射仅适用于数据对象，这是一个限制，因为JS是它的基础语言。

这是一个示例将该函数附加到Plain Old JavaScript Object的plunker： http://plnkr.co/edit/BBOthl0rzjfEq3UjD68I?p=preview

如何在打字稿中从JSON中获取正确的类？

2 个答案: