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
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"
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!"); }
使用URLSearchParams
https://developer.mozilla.org/en-US/docs/Web/API/URLSearchParams
var data = new URLSearchParams();
data.append('userName', 'test@gmail.com');
data.append('password', 'Password');
data.append('grant_type', 'password');
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);
}
});
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。
