在一个简单的函数中包装取回

async function post_www_url_encdoded(url, data) {
    const body = new URLSearchParams();
    for (let key in data) {
        body.append(key, data[key]);
    }
    return await fetch(url, { method: "POST", body });
}

const response = await post_www_url_encdoded("https://example.com/login", {
    "name":"ali",
    "password": "1234"});
if (response.ok){ console.log("posted!"); }

在最初的示例中，您有一个transformRequest函数，它将对象转换为Form Encoded数据。

在修改后的示例中，您已将其替换为JSON。stringify将对象转换为JSON。

在这两种情况下，你都有'Content-Type': 'application/x-www-form-urlencoded;charset=UTF-8'，所以在这两种情况下，您都声称要发送表单编码的数据。

使用你的表单编码函数，而不是JSON.stringify。

重新更新:

在第一个获取示例中，将主体设置为JSON值。

现在您已经创建了一个Form Encoded版本，但是您没有将主体设置为该值，而是创建了一个新对象，并将Form Encoded数据设置为该对象的属性。

不要创建额外的对象。把你的值赋给身体。

var details = {
    'userName': 'test@gmail.com',
    'password': 'Password!',
    'grant_type': 'password'
};

var formBody = [];
for (var property in details) {
  var encodedKey = encodeURIComponent(property);
  var encodedValue = encodeURIComponent(details[property]);
  formBody.push(encodedKey + "=" + encodedValue);
}
formBody = formBody.join("&");

fetch('http://identity.azurewebsites.net' + '/token', {
  method: 'POST',
  headers: {
    'Accept': 'application/json',
    'Content-Type': 'application/x-www-form-urlencoded'
  },
  body: formBody
})

它对我很有帮助，而且没有任何错误

参考资料:https://gist.github.com/milon87/f391e54e64e32e1626235d4dc4d16dc8

根据规范，使用encodeURIComponent不会给你一个符合要求的查询字符串。州:

Control names and values are escaped. Space characters are replaced by +, and then reserved characters are escaped as described in [RFC1738], section 2.2: Non-alphanumeric characters are replaced by %HH, a percent sign and two hexadecimal digits representing the ASCII code of the character. Line breaks are represented as "CR LF" pairs (i.e., %0D%0A). The control names/values are listed in the order they appear in the document. The name is separated from the value by = and name/value pairs are separated from each other by &.

问题是，encodeURIComponent将空格编码为%20，而不是+。

表单主体应该使用其他答案中显示的encodeURIComponent方法的变体进行编码。

const formUrlEncode = str => {
  return str.replace(/[^\d\w]/g, char => {
    return char === " " 
      ? "+" 
      : encodeURIComponent(char);
  })
}

const data = {foo: "bar߃©˙∑  baz", boom: "pow"};

const dataPairs = Object.keys(data).map( key => {
  const val = data[key];
  return (formUrlEncode(key) + "=" + formUrlEncode(val));
}).join("&");

// dataPairs is "foo=bar%C3%9F%C6%92%C2%A9%CB%99%E2%88%91++baz&boom=pow"

只是这样做，UrlSearchParams做的把戏 这是我的代码，如果能帮到别人的话

import 'url-search-params-polyfill';
const userLogsInOptions = (username, password) => {



// const formData = new FormData();
  const formData = new URLSearchParams();
  formData.append('grant_type', 'password');
  formData.append('client_id', 'XXXX-app');
  formData.append('username', username);
  formData.append('password', password);
  return (
    {
      method: 'POST',
      headers: {
        // "Content-Type": "application/json; charset=utf-8",
        "Content-Type": "application/x-www-form-urlencoded",
    },
      body: formData.toString(),
    json: true,
  }
  );
};


const getUserUnlockToken = async (username, password) => {
  const userLoginUri = `${scheme}://${host}/auth/realms/${realm}/protocol/openid-connect/token`;
  const response = await fetch(
    userLoginUri,
    userLogsInOptions(username, password),
  );
  const responseJson = await response.json();
  console.log('acces_token ', responseJson.access_token);
  if (responseJson.error) {
    console.error('error ', responseJson.error);
  }
  console.log('json ', responseJson);
  return responseJson.access_token;
};

在一个简单的函数中包装取回

async function post_www_url_encdoded(url, data) {
    const body = new URLSearchParams();
    for (let key in data) {
        body.append(key, data[key]);
    }
    return await fetch(url, { method: "POST", body });
}

const response = await post_www_url_encdoded("https://example.com/login", {
    "name":"ali",
    "password": "1234"});
if (response.ok){ console.log("posted!"); }