怎么才可以让返回成功的函数：success写到它的参数内部，
function success(data){
console.log(data);
}
jsonp({
url:'https://sp0.baidu.com/5a1Fazu8AA54nxGko9WTAnF6hhy/su',
type:'get',
data:{
wd:'jsonp'
},
callback :'cb',
success:success
});
function jsonp(options){
var url = options.url;
var data = options.data;
format(data,options,function(str,callback){
var oBody = document.getElementsByTagName('body')[0];
var oScript = document.createElement('script');
oScript.setAttribute('src',url+'?'+str + options.callback+'='+callback);
oBody.appendChild(oScript);
});
return options.success;
};
function format(data,options,callback){
var callbackName = '';
var str = '';
for(var p in data){//格式化get提交的参数
str += p+'='+data[p]+'&';
}
for(var p in options){
if(options[p] == options.success){//取出要返回的函数名
callbackName = p;
callback && callback(str,callbackName);
}
}
}
如果是这么写success:function(data){conosle.log(data);}
会报一个success is undefined错误
