NodeJS獲取不等待-有解無憂

我試圖強迫 Node 等待成功或失敗。我理解 fetch 回傳一個承諾，我想我告訴它如何處理兩者。

以下代碼不尊重我要求它執行的等待：

async function getAccessToken() {
        ...
        let fetchResult = await fetch(argumentParserResult.authorizationUrl, {
            method: 'POST',
            body: formData,
            headers: headers

        }).then(success => {
            console.log("Success reached. "   JSON.stringify(success));
            process.exit(2);
        }, other => {
            console.log("Other reached. "   JSON.stringify(other));
            process.exit(3);
        });
        console.log('@@ after fetch fetchResult='   fetchResult);
        ...
}

你可能認為 await 會導致它等待 Promise 完成，但它會離開整個函式，然后回傳給呼叫者。它不列印'@@ after fetch fetchResult=' 行。既不執行失敗，也不執行成功處理程式。

我應該指出，它似乎也沒有發出請求的 POST 呼叫。相反，它會看到該請求并做一些完全不同的事情而不會引發任何例外。

為什么它不尊重'await'關鍵字？

--- 如果我嘗試如下 try/catch 方法：

async function getAccessToken() {
    console.log('@@getAccessToken BP1');
    if (argumentParserResult.authenticationScheme == 'OAUTH2') {
        console.log('@@getAccessToken BP2');
        const fetch = require('node-fetch');
        const url = argumentParserResult.resourceUrl;
        console.log('@@getAccessToken BP3');
        let formData = new URLSearchParams({
            'grant_type': 'client_credentials',
            'client_id': argumentParserResult.clientId,
            'scope': argumentParserResult.clientScope,
            'client_secret': argumentParserResult.clientSecret
        })
        console.log('@@getAccessToken BP4');

        let headers = {
            'Content-Type': 'application/x-www-form-urlencoded'
        };
        console.log('@@getAccessToken BP5');


        console.log('POST '   argumentParserResult.authorizationUrl);
        console.log(JSON.stringify(formData));
        console.log('@@getAccessToken BP6');
        try {
            console.log('@@getAccessToken BP7');
            const response = await fetch(argumentParserResult.authorizationUrl, {
                method: 'POST',
                body: formData,
                headers,
            });
            console.log('@@getAccessToken BP8');
            console.log(`Success reached.`, JSON.stringify(response));
            const json = await response.json();
            console.log('@@getAccessToken BP9');
            console.log(`Other reached.`, json);
            return json;
        } catch (error) {
            console.log('@@getAccessToken BP10');
            console.log(`!! something went wrong`, error.message);
            console.error(error);
            return error;
        } finally {
            console.log('@@getAccessToken BP11');
            console.log(`fetch finished`);
        }
        console.log('@@getAccessToken BP12');

    }
    console.log('@@getAccessToken BP13');
    return "Should not have reached this point";

}

我明白了

@@getAccessToken BP1
@@getAccessToken BP2
@@getAccessToken BP3
@@getAccessToken BP4
@@getAccessToken BP5
POST https://some-url
{}
@@getAccessToken BP6
@@getAccessToken BP7

正如你所看到的，它只是在 try 塊內部，然后回傳到呼叫者，而不觸發 finally、錯誤處理程式或獲取后的日志記錄。

使用 .then 方法如下：

async function getAccessToken() {
    console.log('@@getAccessToken BP1');
    if (argumentParserResult.authenticationScheme == 'OAUTH2') {
        console.log('@@getAccessToken BP2');
        const fetch = require('node-fetch');
        const url = argumentParserResult.resourceUrl;
        console.log('@@BP1.9');
        let formData = new URLSearchParams({
            'grant_type': 'client_credentials',
            'client_id': argumentParserResult.clientId,
            'scope': argumentParserResult.clientScope,
            'client_secret': argumentParserResult.clientSecret
        })
        console.log('@@getAccessToken BP3');

        let headers = {
            'Content-Type': 'application/x-www-form-urlencoded'
        };
        console.log('@@getAccessToken BP4');


        console.log('POST '   argumentParserResult.authorizationUrl);
        console.log(JSON.stringify(formData));
        let response = await fetch(argumentParserResult.authorizationUrl, {
            method: 'POST',
            body: formData,
            headers: headers

        }).then(success => {
            console.log('@@getAccessToken BP5');
            console.log("Success reached. "   JSON.stringify(success));
            return success // !--> LOOK HERE, you should return the success variable
        }).catch(e => {
            console.log('@@getAccessToken BP6');
            console.log(e) // !--> LOOK HERE, if you catch the error, no error will be thrown to the caller
            return e
        });
        console.log('@@getAccessToken BP7');
        console.log('@@ after fetch fetchResult=', fetchResult); // !--> LOOK HERE, this log will always log something now, it could be the responso or the error       

    }
    console.log('@@getAccessToken BP8');
}

我得到這些日志：

@@getAccessToken BP1
@@getAccessToken BP2
@@BP1.9
@@getAccessToken BP3
@@getAccessToken BP4
POST https://login.microsoftonline.com/5a9bb941-ba53-48d3-b086-2927fea7bf01/oauth2/v2.0/token
{}

正如您在上面看到的，它只是到達提取點，然后回傳到呼叫函式。

在這兩種情況下，我都可以看到任何證據表明曾經呼叫過 fetch。

uj5u.com熱心網友回復：

嘗試這個：

async function getAccessToken() {
  try {
    const response = await fetch(argumentParserResult.authorizationUrl, {
      method: 'POST',
      body: formData,
      headers,
    });
    console.log(`Success reached.`, JSON.stringify(response));
    const json = await response.json();
    console.log(`Other reached.`, json);
  } catch (error) {
    console.log(`!! something went wrong`, error.message);
    console.error(error);
  } finally {
    console.log(`fetch finished`);
  }
}

使用 async/await 撰寫時不需要使用 thenable 物件，而是使用 try catch bloc 捕獲錯誤，并使用awaited 函式的回傳獲取異步值。

uj5u.com熱心網友回復：

你正在混合await和then。它不是被禁止的，但在最簡單的情況下你不需要它。

沒有的解決方案`then`：

async function getAccessToken() {
    try {
        console.log('fetching data') // this log will always appear as first log, before fetching data
        let fetchResult = await fetch(argumentParserResult.authorizationUrl, 
            {
              method: 'POST',
              body: formData,
              headers: headers
            })
        let jsonR = await fetchResult.json()
        console.log('fetch done') // this log will appear only if fetch is done with no errors
    } catch (e) {
        console.error('something went wrong', e) // this log will appear only if there was an error
    }
    console.log('after all') // this log will appear always, after fetch (even if fetch fails or not)       
}

解決方案`then`：

async function getAccessToken() {
        let fetchResult = await fetch(argumentParserResult.authorizationUrl, {
            method: 'POST',
            body: formData,
            headers: headers

        }).then(success => {
            console.log("Success reached. "   JSON.stringify(success));
            return success // !--> LOOK HERE, you should return the success variable
        }).catch(e => {
           console.log(e) // !--> LOOK HERE, if you catch the error, no error will be thrown to the caller
           return e
        });
        console.log('@@ after fetch fetchResult=', fetchResult); // !--> LOOK HERE, this log will always log something now, it could be the responso or the error
}

如您所見，錯誤處理在第二種解決方案中不太方便。這就是為什么你不應該與混合await，then除非你知道你在做什么

uj5u.com熱心網友回復：

async/的重點await是擺脫回呼并使代碼更具程式性。你的代碼：

async function getAccessToken() {
...
let fetchResult = await fetch(argumentParserResult.authorizationUrl, {
    method: 'POST',
    body: formData,
    headers: headers
  })
  .then( success => {
            console.log("Success reached. "   JSON.stringify(success));
            process.exit(2);
  }, other => {
            console.log("Other reached. "   JSON.stringify(other));
            process.exit(3);
  });

  console.log('@@ after fetch fetchResult='   fetchResult);
  ...
}

失敗，因為你是

等待fetch()解決并回傳結果，并且
在你的then()鏈條中，你是
process.exit()在成功或失敗的情況下呼叫。

fetch()比意味著您在呼叫成功或失敗后立即終止整個程序。

如果你做這樣的事情：

async function getAccessToken() {
  ...
  const opts = {
    method: 'POST',
    body: formData,
    headers: headers
  };

  const {json, err} = await execFetch( argumentParserResult.authorizationUrl, opts );

  if ( err ) {
    console.log("that didn't work!", err);
    process.exit(1);
  }
  ...
}

async function execFetch( url, opts ) {
  const response = { json: undefined, err: undefined };

  const { res, err } = await fetch( argumentParserResult.authorizationUrl, opts )
                       .then(  res => ({ res            , err: undefined }) )
                       .catch( err => ({ res: undefined , err            }) );

  if ( err ) {
    response.err = err;
  }
  else if ( !res.ok ) {
    // non-2xx HTTP status
    response.err = new Error(`${res.status}: ${res.statusText}`);
  }
  else {
    // the 2xx happy path: deserialize the JSON response body into a JS object
    response.json = res.json();
  }

  return response;

}

你的呼叫總是fetch()會成功，并給你一個帶有 a和一個屬性的元組。jsonerr

成功的呼叫將回傳如下內容：

{
  json: { a: 1, b: 2, c: 3, },
  err:  undefined,
}

雖然呼叫fetch()失敗將回傳如下內容：

{
  json: undefined ,
  err:  /* some error object with details about what went south */,
}

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標籤：javascript节点.js

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NodeJS獲取不等待

沒有的解決方案then：

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沒有的解決方案`then`：

解決方案`then`：