javascript - 原生js封装jsonp函数

Question

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

Answer 1

https://jsfiddle.net/hsfzxjy/...

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