ajax走完后台不走success也不走error

BinZor 2017-07-10 06:30:09
最近遇到一个问题,求各位大神帮忙
我先把代码贴出来
<span >
<input type="hidden" id="buttonValue" name="buttonValue" value="po_o_submit">
<button name="commit" class="Whitetext" onclick="reqoperate('po_o_submit')" value="提交">
<img src="<%=request.getContextPath() %>/billow/ico45.gif" width="16" height="16" align="absmiddle"/>提交</button>
</span>


function reqoperate(op,type) {
switch(op) {
case "po_o_submit":{ //提交
if(confirm("确定要提交吗?")){
allFolw();
}
break;
}


function allFolw(){
var val = "&val=businessId:" + examineId +"^processInstanceLifeBean:budgetAdjustProcessInstanceLife&business=yearAdjustType";
//提交后的返回值,如果不为空说明已经进入工作流状态
jQuery("val").val(val);
if ("" != taskId){
tosubmit('10',taskId);
}else{ //如果为空,提交进入工作流状态
document.getElementById("val").value = val;
tosubmit('10',taskId,val);
}
}



/**
* 流程提交
**/

function tosubmit(processId, taskId) {
var url = document.getElementById("basePath").value + "/requestTransition.do?taskId=" + taskId + "&pdId=" + processId;
AjaxSubmit(url);
}

function tosubmit(processId, taskId, val) {
var url = document.getElementById("basePath").value + "/requestTransition.do?taskId=" + taskId + "&pdId=" + processId;
if (undefined != val) {
url += val;
}
AjaxSubmit(url);
}
function AjaxSubmit(url) {------------------------------------------------------------------------------------------------------------------------------
jQuery.ajax({
type: "POST",
url: url,
data: '',
success: function(msg){
callback(msg);
}
});
}
function callback(data) {}





	@SuppressWarnings("unchecked")
@Override
public ActionForward execute(ActionMapping mapping, ActionForm form,
HttpServletRequest request, HttpServletResponse response) throws IOException {
System.out.println("----------------------------------");
User su = (User) request.getSession().getAttribute("User");
String taskId = request.getParameter("taskId");
String val = request.getParameter("val");
String business = request.getParameter("business");
Writer writer = response.getWriter();
Long pdIdL = 0l;
String responseXML = "";
try {

if (null == taskId || "".equals(taskId.trim()) || "null".equals(taskId.trim())) {
pdIdL = (Long) workFlowTemplateService.findWorkFlow(business, su.getSysUser().getOrgId());
if (pdIdL == 0) {
writer.write("error:没有为该业务表单指定相关流程");
return null;
}
}
Map variables = new HashMap();
if (null != val && !"".equals(val.trim())) {


String[] vals = val.split("\\^");
for (int i = 0; i < vals.length; i++) {
String[] temp = vals[i].split(":");
if (temp.length == 2) {
//处理预算责任部门
if (WorkFlowTemplate.COUNTERSIGNED_DEPARTMENT.equals(temp[0])) {
variables.put(WorkFlowTemplate.COUNTERSIGNED_DEPARTMENT, temp[1]);
variables.put(WorkFlowTemplate.COUNTERSIGNED_DEPARTMENT_LENGTH, temp[1].split(",").length);
} else if (WorkFlowTemplate.BUSINESS_ID.equals(temp[0])) { //处理业务表单ID
variables.put(temp[0], temp[1]);
} else if ("totalCost".equals(temp[0]) || "totalFund".equals(temp[0])) {
variables.put(temp[0], Double.parseDouble(temp[1].replace(",", "")));
} else {
variables.put(temp[0], temp[1]);
}
}
}
}
List transitionInfo = null;
if (null == taskId || "".equals(taskId.trim())) {
variables.put(WorkFlowTemplate.DRAFTER, su.getSysUser().getUserId());
variables.put(WorkFlowTemplate.DRAFTER_NAME, su.getSysUser().getUsername());
variables.put(WorkFlowTemplate.DRAFTER_ORG, su.getSysUser().getOrgId());
variables.put(WorkFlowTemplate.DRAFTER_ORG_NAME, su.getSysUser().getOrgname());
transitionInfo = templateParseTransition.parseDrafterTransition(pdIdL, variables, su);
} else {
transitionInfo = templateParseTransition.parseTransition(taskId, su);
}
if (null != transitionInfo && transitionInfo.size() > 0) {
responseXML = templateParseTransition.parseXMLD(transitionInfo);
} else {
responseXML = "error:没有可选路径";
}
System.out.println(responseXML);
} catch (Exception e) {
responseXML = e.getMessage();
e.printStackTrace();
} finally {
if (null != writer) {
try {
writer.write(responseXML);
writer.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
return null;
}


问题是这样子的,jquery1.11.1的版本,然后如果我在java方法上打断点,就会执行success里的方法,如果不打,就只会刷新页面。很奇怪的问题。我也不知道为什么会这样,f12,也没有报错,求大神给指点指点
...全文
397 2 打赏 收藏 转发到动态 举报
写回复
用AI写文章
2 条回复
切换为时间正序
请发表友善的回复…
发表回复
2017-07-10
  • 打赏
  • 举报
回复
javascript没有重载 jQuery("val").val(val); => jQuery("#val").val(val);
孟子E章 2017-07-10
  • 打赏
  • 举报
回复
<button 要加type=button,默认是submit,提交页面了

87,910

社区成员

发帖
与我相关
我的任务
社区描述
Web 开发 JavaScript
社区管理员
  • JavaScript
  • 无·法
加入社区
  • 近7日
  • 近30日
  • 至今
社区公告
暂无公告

试试用AI创作助手写篇文章吧