Mongoose：自定义模式类型

Question

我在自定义模式类型上遵循了这个mongoose documentation，创建了一个“大字符串”：

"use strict";

const mongoose = require('mongoose')

let STRING_LARGE = (key, options) => {   
  mongoose.SchemaType.call(this, key, options, 'STRING_LARGE'); 
}; 
STRING_LARGE.prototype = Object.create(mongoose.SchemaType.prototype);

STRING_LARGE.prototype.cast = function(val) {   
  let _val = String(val);   
  if(!/^[a-zA-Z0-9]{0,400}$/.test(_val)){
    throw new Error('STRING_LARGE: ' + val + ' is not a valid STRING_LARGE');   
  }

  return _val; };

module.exports = STRING_LARGE;

我在架构中这样使用它：

"use strict";

const mongoose = require('mongoose');

mongoose.Schema.Types.STRING_LARGE = require('./types/string_large')

const schema = new mongoose.Schema({
  details:     { type: STRING_LARGE, required: true },
  link:        { type: STRING_LARGE, required: true }
});

module.exports = schema;

但我得到了错误：

[路径]\schemas[shema.js]:8
详细信息：{ 类型：STRING_LARGE，必需：true }，

ReferenceError: STRING_LARGE 未定义 在对象。 （[路径]\模式[shema.js]：8:24） ...

-------------- 更新：工作代码 -------------- ------------

使用“函数（）”而不是“（）=>”

"use strict";

const mongoose = require('mongoose')

function STRING_LARGE (key, options) {   
  mongoose.SchemaType.call(this, key, options, 'STRING_LARGE'); 
}; 
STRING_LARGE.prototype = Object.create(mongoose.SchemaType.prototype);

STRING_LARGE.prototype.cast = function(val) {   
  let _val = String(val);   
  if(!/^[a-zA-Z0-9]{0,400}$/.test(_val)){
    throw new Error('STRING_LARGE: ' + val + ' is not a valid STRING_LARGE');   
  }

  return _val; };

使用“mongoose.Schema.Types.LARGE_STRING”而不是“LARGE_STRING”

module.exports = STRING_LARGE;

"use strict";

const mongoose = require('mongoose');

mongoose.Schema.Types.STRING_LARGE = require('./types/string_large')

const schema = new mongoose.Schema({
  details:     { type: mongoose.Schema.Types.STRING_LARGE, required: true },
  link:        { type: mongoose.Schema.Types.STRING_LARGE, required: true }
});

module.exports = schema;

Answer 1

您将您的类型分配给mongoose.Schema.Types.STRING_LARGE，然后使用STRING_LARGE - 这就是您的ReferenceError 被抛出的地方。你必须直接使用你的类型：

const schema = new mongoose.Schema({
  details:     { type: mongoose.Schema.Types.STRING_LARGE, required: true },
  link:        { type: mongoose.Schema.Types.STRING_LARGE, required: true }
});