如何使打字稿同步执行?

时间:2019-04-24 18:52:48

标签: typescript async-await synchronization

我正在尝试实现上传图片功能,并且有一个如下的类。

class myClass {
  oriData: any;
  name: string;

  constructor(props) {
    this.name = props.name;
    this.oriData = this.readFile(props);
  }

  async readFile(props) {
    return await new Promise((resolve, reject) => {
      let reader = new FileReader();
      reader.readAsDataURL(props);
      reader.onload = () => {
        let result = reader.result;
        resolve(result);
      };
      reader.onerror = reject;
    });
  }
}

private async process(file): Promise<myClass> {
  try {
    let image = await new myClass(file);
    console.log(image.oriData);
    console.log(image.name);
    return Promise.resolve(image);
  }
  catch(err) {
    console.log(err);
  }
}

但是,当我尝试获取image.oriData和image.name时,image.oriData显示为“未定义”,但另一个是正确的。我检查了该步骤,发现进程(文件)仍将实例化myClass,而无需等待reader.onload完成。我相信这应该是一个同步问题。谁能帮我改进这段代码?谢谢!

1 个答案:

答案 0 :(得分:1)

您可以从JavaScript构造函数返回任何文字,因此这是可能模式:

class DontDoThis {
  constructor(props) {
    this.name = props.name;
    const results = this.readFile(props);
    // As a side-effect of the promise completing, set a field
    results.then(data => this.oriData = data);

    // Force this constructor to be treated as async
    // even though we cannot say `async constructor`
    return results.then(() => this, () => this);
  }
}

async function process(...) {
  const complexData = await new DontDoThis(...);
}

也就是说,仅仅拥有一个构造最终值的函数要好得多:

interface ImageData { name: string, oriData: any };
// This can also be a static method, e. g. `ImageData.create(...)`
async function ComplexData(props): Promise<ImageData> {
  return readFile(props).then(data => ({name: props.name, oriData: data}));
}

async function process(props) {
  const complexData = await ComplexData(props);
}

ImageData可以是类而不是接口-要注意的关键是它不能部分构造-ComplexData成功并产生ImageData对象,或者失败而且你一无所有。您永远不会有不安全的使用,因为它是半初始化的ImageData对象。