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移动返回JSON承诺响应的位置

json javascript

Jimmy · 技术社区 · 7 年前

这是我的密码:

function myfunction(ref) {
  getReport(ref, "queue", "hour", "2018-09-03", "2018-10-04", "pageviews", "page").done(function(r1) {
    getReport(r1.reportID, "get").done(r2=>reportHandler(r1.reportID, r2, 0))
    // HERE IS WHERE I WANT TO GET MY DATA RETURNED
      });
}


function reportHandler(id, r2, retries){
    if(retries >= 3){return}
    if (r2.error == "report_not_ready") {
        console.log("Not ready");
        setTimeout(function() {
          getReport(id, "get").done(r2=>reportHandler(id, r2, retries + 1))
        }, 2000);
      }
      // THIS IS NOT WHERE I WANT MY DATA BUT WHERE IT COMES OUT
      console.log(r2);
}

function getReport(ref, type, granularity, from, to, metric, element) {
  return $.getJSON("report.php", {
    ref: ref,
    type: type,
    granularity: granularity,
    from: from,
    to: to,
    metric: metric,
    element: element,
  });
}

这是我能得到的最接近的:

function myfunction(ref) {
  getReport(ref, "queue", "hour", "2018-09-03", "2018-10-04", "pageviews", "page").done(function(r1) {
    getReport(r1.reportId, "get").done((r2) => {
      reportHandler(r1.reportId, r2, 0);
      console.log(r2);
    });
  });
}

如何将JSON响应的结果从API收集到“myfunction”中,以便在该函数中进行处理。

2 回复 | 直到 7 年前

Tyler Roper 7 年前

你可以用 async / await 为了这个。

你可以使用等待他们的完成。我添加了一个 sleep “等待X毫秒”以替换 setTimeout (通过使用 设置超时 它自己的)。

const sleep = m => new Promise(r => setTimeout(r, m));

async function myfunction(ref) {
  var r1 = await getReport(ref, "queue", "hour", "2018-09-03", "2018-10-04", "pageviews", "page");
  var r2 = await getReport(r1.reportID, "get").then(r2 => reportHandler(r1.reportID, r2, 0));
  
  console.log(r2);
}

async function reportHandler(id, r2, retries) {
  if (retries >= 3 || r2.error != "report_not_ready") return r2;
  
  await sleep(2000);                              //Wait 2 seconds
  var report = await getReport(id, "get");        //Try again
  return reportHandler(id, report, retries + 1);  //Until we get a result
}

function getReport(ref, type, granularity, from, to, metric, element) {
  return $.getJSON("report.php", {
    ref: ref,
    type: type,
    granularity: granularity,
    from: from,
    to: to,
    metric: metric,
    element: element,
  });
}

下面是一些调用JSON端点的临时代码,它模拟了几次重试后才得到成功的结果。

const sleep = m => new Promise(r => setTimeout(r, m));

async function myfunction(ref) {
  var r1 = await getReport();
  var r2 = await getReport().then(r2 => reportHandler(r2, 0));
  console.log(r2);
}

async function reportHandler(r2, retries) {
  if (retries >= 3) return r2;
  console.log(`Attempt #${retries+1} failed. Retrying in 2 seconds.`);
  
  await sleep(2000);
  var report = await getReport();
  return reportHandler(report, retries + 1);
}

function getReport() {
  return $.getJSON('https://jsonplaceholder.typicode.com/todos/1');
}

myfunction();

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Akrion 7 年前

你需要向我保证 reportHandler 像这样:

const myfunction = ref =>
  getReport(ref, "queue", "hour", "2018-09-03", "2018-10-04", "pageviews", "page").then(r1 =>
    getReport(r1.reportId, "get").then(r2 =>
      reportHandler(r1.reportId, r2, 0).then(r3 => {
        console.log(r1,r2,r3);
      })
    );
  );

const reportHandler = (id, r2, retries) =>
    new Promise((resolve, reject) => {
      if(retries >= 3) { resolve() }
      if (r2.error == "report_not_ready") {
        console.log("Not ready");
        setTimeout(function() {
          return getReport(id, "get").then(r2 => reportHandler(id, r2, retries + 1)).then(r => resolve(r))
        }, 2000);
      }
    }

你有 setTimeout 你需要等待。看看这个方法是否适合你。