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获取form表单数据

var modelObj = {};
var modelFieldsArray = $('#AddMusic').serializeArray();
$.each(modelFieldsArray, function () {
modelObj[this.name] = this.value;
});

 

modelObj.MusicUrl = musicApp.commonKit.upLoadMng.fileMusicUrl;
modelObj.ImgUrl = musicApp.commonKit.upLoadMng.fileImgUrl;
var modelStr = JSON.stringify(modelObj);
var requestModel = {};
requestModel.method = "MusicCreate";
requestModel.modelstr = modelStr;

posted @ 2018-07-17 15:04  刘奇云  阅读(698)  评论(0编辑  收藏  举报