处理promise中的错误时返回成功

Question

我有一个承诺,处理通过Web API执行的HTTP请求:

promise = promise.then(r => { 
    // ...
  }, error => {
    if (error.status == 404) {
      // Here I can fix an error and continue properly
    } else {
      // Here the error should be propagated further in the promise
    }
}

// later in the code:
promise.catch(r => { /* More error handling */ } );

在代码的后面,这个承诺被链接到更多的错误检查.

在404错误的情况下,我实际上可以"修复"一个问题,我不希望其他处理程序触发.在这种情况下,我宁愿让这个承诺成功.我怎样才能做到这一点？

---

更多代码可以更深入地解释我的案例:

refresh() {
  this.refreshAsync().catch(r => {
     // Show error to the user.
     notifications.showError("Unexpected error happened");
  });
}

async refreshAsync() {
  // Here goes a lot of vue-resource calls to gather the necessary data for a view. They are chained with `await`. Only one of them:
  await this.$http.get(url).then(r => {
    this.data = r.data;
  }, error => {
    // 404 is actually a legit response for API to return, so the user notification above should not be shown
    if (error.status == 404) {
      // handle successfully
    } else {
      // propagate an error, so the handler above could show a warning to the user.
    }

  });

}

Answer 1

您只需返回拒绝/解决即可

if(error.status == 404)
    return Promise.resolve('OK')
else
    return Promise.reject('fail')

我做了一个例子来展示它是如何工作的，仅针对这种情况：

httpRequest = function () {
  return new Promise(function (res, rej) {
    let status = (Math.round(Math.random()) === 1) ? 200 : 404;
    console.log(status)
    if (status === 200)
        return res({ status })
    else
        return rej({ status })
  })
}

let promise =
httpRequest()
    .then(res => Promise.resolve('success'))
    .catch(e => {
        if (e.status === 404)
            return Promise.resolve('success')
        else
            return Promise.reject('failed')
    })

promise
.then(res => {
    console.log(res)
})
.catch(e => {
    console.log('this should not happen')
})