Ajax函数返回未定义？

Question

async function login(username, password) {
  await $.ajax({
    url: "/api/login",
    type: "POST",
    data: `{"username": "${username}", "password": "${password}"}`,
    dataType: "text",

    success: async function(response) {
      var a = await JSON.parse(response)
      console.log(a.token.access_token)
      return a.token.access_token
    },

    error: function(response) {
      return "false"
    }
  })
}

eyJhbGciOiJIUzI1NiIsInR5cCI6IkpXVCJ9.eyJhdXRob3JpemVkIjp0cnVlLCJleHAiOjE2MTgzODIzNzUsInVzZXJuYW1lIjoic2sxMTIyIn0.e16nHpJUsNmMSAcPgaHkmJgwrnBkp4jB7KHe_ylAvKc // console.log() from login()

undefined // console.log() in another function

console.log(await login(questions[0].value, questions[2].value))

Answer 1

• 您需要使用$.ajax()的Promise语义，success回调对返回值无效。
• 您需要返回来自$.ajax()调用的promise。
• 您的login()函数中绝对不需要异步或await。

正确代码：

function login(username, password) {
  return $.ajax({
    url: "/api/login",
    type: "POST",
    data: {
      username: username,
      password: password
    },
    dataType: "text"
  }).then(function (response) {
    var data = JSON.parse(response);
    return data.token.access_token;
  }).fail(function (jqXhr, status, error) {
    return false;
  });
}

和

console.log(await login(questions[0].value, questions[2].value));

function login(username, password) {
  return $.post("/api/login", {username: username, password: password})
    .then((data) => data.token.access_token)
    .fail((jqXhr, status, error) => false);
}

function login(username, password) {
  try {
    const data = await $.post("/api/login", {username: username, password: password});
    return data.token.access_token;
  } catch (err) {
    return false;
  }
}