谷歌浏览器chrome console 发送POST/GET请求

打开chrome 开发者工具
登录你访问的系统,复制你后台系统提供的url
在Console中输入如下代码 可发送http post/get请求

POST请求:

//方法一:
var url = "/dict/test";
var params = {advertiserUid: 1232131, advertiserWeiboNickname: "18"};
var xhr = new XMLHttpRequest();
xhr.open("POST", url, true);
xhr.setRequestHeader("Content-Type", "application/json");
xhr.onload = function (e) {
  if (xhr.readyState === 4) {
    if (xhr.status === 200) {
      console.log(xhr.responseText);
    } else {
      console.error(xhr.statusText);
    }
  }
};
xhr.onerror = function (e) {
  console.error(xhr.statusText);
};
xhr.send(JSON.stringify(params));

//方法二:
var url = "/dict/test";
var params = "score=5&abc=6";
var xhr = new XMLHttpRequest();
xhr.open("POST", url, true);
xhr.setRequestHeader("Content-type","application/x-www-form-urlencoded"); 
xhr.onload = function (e) {
  if (xhr.readyState === 4) {
    if (xhr.status === 200) {
      console.log(xhr.responseText);
    } else {
      console.error(xhr.statusText);
    }
  }
};
xhr.onerror = function (e) {
  console.error(xhr.statusText);
};
xhr.send(params);

 

GET请求:

 1 var url = "/v1/query/listDicts?types=userType,userStatus";
 2 var xhr = new XMLHttpRequest();
 3 xhr.open("GET", url, true);
 4 xhr.onload = function (e) {
 5   if (xhr.readyState === 4) {
 6     if (xhr.status === 200) {
 7       console.log(xhr.responseText);
 8     } else {
 9       console.error(xhr.statusText);
10     }
11   }
12 };
13 xhr.onerror = function (e) {
14   console.error(xhr.statusText);
15 };
16 xhr.send(null);

 

posted @ 2019-07-15 23:48  经管时评  阅读(8288)  评论(0编辑  收藏  举报