如果你正在使用JQuery，这也是有效的。

fetch(url, {
      method: 'POST', 
      body: $.param(data),
      headers:{
        'Content-Type': 'application/x-www-form-urlencoded'
      }
})

对于上传表单编码的POST请求，我建议使用FormData对象。

示例代码:

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

var formData = new FormData();

for (var k in params) {
    formData.append(k, params[k]);
}

var request = {
    method: 'POST',
    headers: headers,
    body: formData
};

fetch(url, request);

更简单:

fetch('https://example.com/login', {
    method: 'POST',
    headers:{
      'Content-Type': 'application/x-www-form-urlencoded'
    },    
    body: new URLSearchParams({
        'userName': 'test@gmail.com',
        'password': 'Password!',
        'grant_type': 'password'
    })
});

文档:https://developer.mozilla.org/en-US/docs/Web/API/WindowOrWorkerGlobalScope/fetch

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;
};

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"

你必须自己把x-www-form-urlencoded有效负载放在一起，就像这样:

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('https://example.com/login', {
  method: 'POST',
  headers: {
    'Content-Type': 'application/x-www-form-urlencoded;charset=UTF-8'
  },
  body: formBody
})

注意，如果你在一个(足够现代的)浏览器中使用fetch，而不是React Native，你可以创建一个URLSearchParams对象并使用它作为主体，因为fetch标准声明，如果主体是一个URLSearchParams对象，那么它应该被序列化为application/x-www-form-urlencoded。然而，你不能在React Native中这样做，因为React Native没有实现URLSearchParams。