jquery - 根据 php session 变量生成 jquery 上下文菜单项？

Question

我有this jQuery context-menu ，我需要生成某些菜单项，具体取决于 PHP $_SESSION 变量。例如，如果我是普通用户，我只会看到“新建”和“编辑”选项。但如果我是管理员用户，我也会看到“删除用户”选项。

类似这样的事情:

$(function() {
  $.contextMenu({
    selector: '.context-menu-one',
    autoHide: true,
    callback: function(key, options) {
      if(key=="Edit") myFunction1(this);
      if(key=="New") myFunction2(this);
      if(key=="Del") myFunction3(this);
    },
    items: {
      "New": {name: "New", icon: "fa-plus-circle"},
      "Edit": {name: "Edit",icon: "fa-pencil-square-o"},
      var grade = <?php echo $_SESSION["grade"];?>;     //this doesn't work
      if(grade == "admin"){                             //but this is what I
        "Del": {name: "Delete user", icon: "fa-trash"}, //want
      }
      "sep1": "---------",
      "quit": {name: "Close", icon: function(){
          return 'context-menu-icon context-menu-icon-quit';
        }
      }
    },
  });    
});

关于如何做到这一点有什么想法吗？

编辑:

php $_SESSION["grade"] 没问题，它返回值，所以我对此没有问题。

Answer 1

字典中间不能有语句(特别是变量赋值和 if 语句)。您需要与此类似地提取它

$(function() {
  var options = {
    selector: '.context-menu-one',
    autoHide: true,
    callback: function(key, options) {
      if(key=="Edit") myFunction1(this);
      if(key=="New") myFunction2(this);
      if(key=="Del") myFunction3(this);
    },
    items: {
      "New": {name: "New", icon: "fa-plus-circle"},
      "Edit": {name: "Edit",icon: "fa-pencil-square-o"},
      "sep1": "---------",
      "quit": {name: "Close", icon: function(){
          return 'context-menu-icon context-menu-icon-quit';
        }
      }
    },
  }
  var grade = <?php echo $_SESSION["grade"];?>;     
  if(grade == "admin"){                            
    options.items["Del"] = {name: "Delete user", icon: "fa-trash"}, 
  }
  $.contextMenu(options);    
});)