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Javascript 在typescript中访问类属性_Javascript_Node.js_Typescript - Fatal编程技术网

Javascript 在typescript中访问类属性

Javascript 在typescript中访问类属性,javascript,node.js,typescript,Javascript,Node.js,Typescript,我有一个typescript类,其构建如下 export default class myClass{ myns: any; async initializing() { … this.myns = na.map(function (item) { return item["name"]; }); // here we have value console.log(this.myns); } search(test, input) { input = input || "";

我有一个typescript类,其构建如下

export default class myClass{

myns: any;

async initializing() {
…
this.myns = na.map(function (item) {
  return item["name"];
});

// here we have value 
console.log(this.myns);

}


search(test, input) {

input = input || "";
return new Promise(function (resolve) {
    let fuz= fuzzy.filter(input, this.myns);  //here I want to access myns but here it comes undefined in debug
    resolve(
      fuz.map(function (el) {
        return el.original;
      })
    );
});



}

}
我想访问函数
search
中的
myns
(在函数中search是非加密的,但在init中它有数据)search我怎么做

不仅
myns
未定义
也未定义
尝试执行
(resolve)=>{
而不是
函数(resolve){
,以便将其绑定到回调

编辑:

运行此代码对我很有用:

class myClass {
  myns: any;

  async initializing() {
    this.myns = [{ name: 'test1' }, { name: 'test2' }].map(function (item) {
      return item["name"];
    });

    console.log(this.myns);
  }

  search(test, input) {
    input = input || "";

    return new Promise((resolve) => {
      console.log('test');
      console.log(this.myns);

      resolve();
    });
  }
}

const test = new myClass();

test.initializing();
test.search('lala', 'lala2');
正如预期的那样,产出是:

[ 'test1', 'test2' ]
test
[ 'test1', 'test2' ]
你正在使用的模糊库是什么