ajax提交
$("#subm").click(function(){
var formData = new FormData();
formData.append("appFile", document.getElementById("appFile").files[0]);
alert("8888888888888888888888888");
$.ajax({
url: '${ctx}/appresources/fileUpload.shtml',
type: "POST",
data: formData,
dataType: "json",
contentType: false,
processData: false,
success: function (data) {
alert("上传成功");
$("#versionCode").val(data.versionCode);
$("#appVersion").val(data.appVersion);
$("#appPackageName").val(data.appPackageName);
},
error: function () {
alert("上传失败！");
}
});
});
后台处理：
@ResponseBody
@RequestMapping("fileUpload")
public JSONObject fileUpload(MultipartFile appFile,Model model,HttpServletRequest request,HttpServletResponse response) {
ServletContext
//........................省略
String infos=ApkUtil.getApkInfo(needPath+"source"+File.separator+"apk"+File.separator+fileName);
String[] info=infos.split(",");
String versionCode=info[0];//版本号
String versionName=info[1];//版本名
String packageName=info[2];//包名
String appName=fileName;
AppResourcesFormMap appResourcesFormMap = getFormMap(AppResourcesFormMap.class);
appResourcesFormMap.put("appName",fileName);
appResourcesFormMap.put("appVersion",versionName);
appResourcesFormMap.put("appPackageName",packageName);
appResourcesFormMap.put("versionCode", versionCode);
System.out.println(appResourcesFormMap);
//model.addAttribute("appresources", appResourcesFormMap);
JSONObject fromObject = JSONObject.fromObject(appResourcesFormMap);
//String string = fromObject.toString();
return fromObject;
}
更多关于ajax相关内容感兴趣的读者可查看本站专题：《jquery中Ajax用法总结》、《JavaScript中ajax操作技巧总结》、《PHP+ajax技巧与应用小结》及《asp.net ajax技巧总结专题》
希望本文所述对大家ajax程序设计有所帮助。