@ type / mongoose无法实现方法时该怎么办

在猫鼬文档中,我找到了有用的方法<nav class="nav navbar-expand-sm navbar-dark"> <a class="navbar-brand" href="" id="homeOverride"> <span class="overrideFonts"> Home </span> </a> <button class="navbar-toggler" type="button" data-toggle="collapse" data-target="#navbarSupportedContent" aria-controls="navbarSupportedContent" aria-expanded="false" aria-label="Toggle navigation"> <span class="navbar-toggler-icon"></span> </button> <div class="collapse navbar-collapse" id="navbarSupportedContent"> <ul class="navbar-nav mr-auto"> <li class="nav-item"> <a class="nav-link" href=""> <span class="overrideFonts"> Projects </span> </a> </li> <li class="nav-item"> <a class="nav-link" href=""> <span class="overrideFonts"> Contact </span> </a> </li> </ul> </div> </nav> https://mongoosejs.com/docs/subdocs.html#finding-a-subdocument

arrayOfSubdocuments.id

这是import { connect,Schema,model,Document } from 'mongoose'; connect('mongodb://db-dev/shop',{ useNewUrlParser: true }); // typescript types type MChild = Document & { name: String,} type MParent = Document & { children: MChild[],} // mongoose schemas const childSchema = new Schema({ name: String }); const parentSchema = new Schema({ children: childSchema[],}); const main = async () => { const Child = model<MChild>('Child',childSchema); const Parent = model<MParent>('Child',parentSchema); const temp = await new Parent({ children: [{ name: 'alex',},{ name: 'bob',}],}).save(); temp.children.id(someId); // property 'id' doesn't exist on type MChild[] }; main(); 错误还是我错误地确定了类型@types/mongoose

ccnc402 回答:@ type / mongoose无法实现方法时该怎么办

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