Promise 中如何 sleep(不是实现 sleep)

我的函数原型如下

const sleep = (ms) => new Promise((resolve) => setTimeout(resolve, ms));


class GversionsBulkUpdate extends React.Component {

  ......

  update = async () => {
    const { handleLoadingClose, handleLoading, appStore } = this.props;
    handleLoading();

    let bulkMax = 10;   // 最大并发值
    const fails = [];

    await new Promise((resolve) => {
      const isFinish = finish(this.state.list.length, resolve);                  // finish mark
      this.state.list.forEach((pkginfo) => {
        while (bulkMax <= 0) await sleep(300); // 限制并发数量
        bulkMax -= 1;
        const body = { address: pkginfo.url, gversion: pkginfo.gversion, ftype: pkginfo.ftype };
        goGameRequest.createPfile(appStore.user, pkginfo.pkg.package_id, body,
          () => { console.log('finish one'); bulkMax += 1; isFinish.next(); },  // error callback
          (msg) => {                                                            // success callback
            console.log(msg);
            bulkMax += 1;
            isFinish.next();
            fails.push({ pkg: pkginfo.pkg, msg });
          });
      });
    });
    console.log(fails)
    const msg = fails.length === 0 ? '批量更新执行完毕' : '部分包更新失败';
    handleLoadingClose(msg);
    this.setState({ parameters: Object.assign({}, BaseParameters),
      list: Object.assign([], baseList),
      packages: Object.assign([], this.pacakges) });
  };
  
  ......

createPfile 是个 http 请求

想用 while (bulkMax <= 0) await sleep(300); 限制并发请求数量

但是语法错误....Promise 中需要等待要如何写..有点懵比了