现在我需要添加一个新的菜单项及其对应的asp文件,我在项目的菜单管理信息中添加了内容,然后通过ftp上传了该asp文件,但是它不起作用,请问我在后台sql中做了什么吗?但这是用C写的,我不知道C ...
这是main.asp中的菜单部分
function recursionMenu(parent, flag) {
var menus;
$.ajax({
type : 'get',
url : "/action/getusermenu",
data :{},
dataType : 'json',
async : 'true',
success : function(data) {
menus = data;
var topMenus = [];
var parentMenus = [];
//获取根节点
for (var i = 0; i < menus.length; i++) {
var menu = menus[i];
var code = menu.f_menu_code;
var parent = menu.f_parent_menucode;
if(parent == ""){
topMenus.push(menus[i]);
}else{
parentMenus.push(menus[i]);
}
}
for(var j = 0;j < topMenus.length;j++){
var child = [];
var ulParent = $('<ul class="sub-menu">');
for(var k = 0;k < parentMenus.length;k++){
if (topMenus[j].f_menu_code == parentMenus[k].f_parent_menucode){
child.push(parentMenus[k]);
var url = parentMenus[k].f_data_url+".asp";
var liPparent = $("<li class='nav-item'><a target='"+url+"' title='"+topMenus[j].f_menu_name+"|"+parentMenus[k].f_menu_name+"'class='nav-link'><span class='title'>"+parentMenus[k].f_menu_name+"</span></a></li>'");
ulParent.append(liPparent);
}
}
这是getusermenu动作
int total;
char ct[10], data[65536], *json, *role_key, *menucode, *menus;
memset(data, 0, 2048);
role_key = websGetSessionVar(wp, "roleid", "");
if (opendb()){
sprintf(sqlstr, "select * from tscu_menu m where f_menu_code in (select f_menu_code from tscu_role_menu where f_role_key = '%s') order by f_sequence", role_key);
getDataBySQL(sqlstr);
menus = createInnerJson();
websSetSessionVar(wp, "menus", menus);
sendMsgInfo(wp, menus);
}
else{
sendDBOpenError(wp);
}
closedb();
}