我有一些参数,我想POST表单编码到我的服务器:

{
    'userName': 'test@gmail.com',
    'password': 'Password!',
    'grant_type': 'password'
}

我像这样发送我的请求(目前没有参数)

var obj = {
  method: 'POST',
  headers: {
    'Content-Type': 'application/x-www-form-urlencoded; charset=UTF-8',
  },
};
fetch('https://example.com/login', obj)
  .then(function(res) {
    // Do stuff with result
  }); 

如何在请求中包含表单编码的参数?


当前回答

不需要使用jQuery、querystring或手动组装有效负载。URLSearchParams是一种方法,这里是一个最简洁的答案与完整的请求示例:

fetch('https://example.com/login', {
  method: 'POST',
  body: new URLSearchParams({
    param: 'Some value',
    anotherParam: 'Another value'
  })
})
  .then(response => {
    // Do stuff with the response
  });

同样的技术使用async / await。

const login = async () => {
  const response = await fetch('https://example.com/login', {
    method: 'POST',
    body: new URLSearchParams({
      param: 'Some value',
      anotherParam: 'Another value'
    })
  })

  // Do stuff with the response
}

是的,您可以使用Axios或任何其他HTTP客户端库来代替本机获取。

其他回答

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!"); }

只需将主体设置为如下所示

var reqBody = "username="+username+"&password="+password+"&grant_type=password";

then

fetch('url', {
      method: 'POST',
      headers: {
          //'Authorization': 'Bearer token',
          'Content-Type': 'application/x-www-form-urlencoded; charset=UTF-8'
      },
      body: reqBody
  }).then((response) => response.json())
      .then((responseData) => {
          console.log(JSON.stringify(responseData));
      }).catch(err=>{console.log(err)})

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;
};
*/ import this statement */
import qs from 'querystring'

fetch("*your url*", {
            method: 'POST',
            headers: {'Content-Type': 'application/x-www-form-urlencoded;charset=UTF-8'},
            body: qs.stringify({ 
                username: "akshita",
                password: "123456",
            })
    }).then((response) => response.json())
      .then((responseData) => {
         alert(JSON.stringify(responseData))
    })

在使用npm i querystring后,保存它的工作很好。

您可以使用react-native-easy-app,更容易发送http请求和制定拦截请求。

import { XHttp } from 'react-native-easy-app';

* Synchronous request
const params = {name:'rufeng',age:20}
const response = await XHttp().url(url).param(params).formEncoded().execute('GET');
const {success, json, message, status} = response;


* Asynchronous requests
XHttp().url(url).param(params).formEncoded().get((success, json, message, status)=>{
    if (success){
       this.setState({content: JSON.stringify(json)});
    } else {
       showToast(msg);
    }
});