fetch要求参数传递，遇到请求无法正常获取数据，网上其他很多版本类似这样：
???fetch(url ,{ ???????method: ‘POST‘, ???????headers:{ ???????????????????‘Accept‘: ‘application/json, text/plain, */*‘, ???????????????????‘Content-Type‘: ‘application/json‘ ???????}, ????????body: JSON.stringify({a:1,b:2}) ???}).then(function(response){ ???????return response.json(); ???}).then(function(data){ ????????console.log(data); ???});经过改进和测试，如下：
var ur = ‘xxx‘,params = {page:1,rows:10},param=‘‘;for(var key in params){ ?param += key + ‘=‘ + params[key] + ‘&‘;}if(param) param = param.substring(0,param.length-1);var requestConfig = { ?method: ‘POST‘, ?credentials: ‘include‘, ?headers: { ???‘Accept‘:‘application/json, text/plain, */*‘, ???‘Content-Type‘: ‘application/x-www-form-urlencoded‘ ????????????????} ?????};Object.defineProperty(requestConfig,‘body‘,{ ???value: param});fetch(url,requestConfig).then(function(res){ ??return res.json(); }).then(function(json){ ??console.log(json.data);});fetch获取json的正确姿势
原文地址：http://www.cnblogs.com/xtreme/p/7447059.html