显示错误不能读取的不确定

Question

function InitMenu() { 
    var bar = menuBar.children 
    for (var i = 0; i < bar.length; i++) { 
     var menu = eval(bar[i].menu) menu.style.visibility = "hidden" 
     bar[i].onmouseover = new Function("ShowMenu(" + bar[i].id + ")") var Items = menu.children 
     for (var j = 0; j < Items.length; j++) { 
      var menuItem = eval(Items[j].id) if (menuItem.menu != null) { 
       menuItem.innerHTML += "<Span Id=" + menuItem.id + "_Arrow class='Arrow'>4</Span>" //var tmp = eval(menuItem.id+"_Arrow") 
        // tmp.style.pixelLeft = menu.getBoundingClientRect().Right //- tmp.offsetWidth - 15 
       FindSubMenu(menuItem.menu) 
      } 
      if (menuItem.cmd != null) { 
       menuItem.onclick = new Function("Do(" + menuItem.id + ")") 
      } 
      menuItem.onmouseover = new Function("highlight(" + Items[j].id + ")") 
     } 
    } 
} 

Answer 1

eval财产“风格”时的数据是JSON对象只能使用。我假设你有一个名为“菜单栏”的菜单内的菜单

var menuBar=document.getElementById("menuBar"); 
function InitMenu() { 
    var bar = menuBar.childNodes; 
    for (var i = 0; i < bar.length; i++) { 
     var menu = bar[i]; 
     menu.style.visibility = "hidden"; 
     bar[i].onmouseover = new Function("ShowMenu(" + bar[i].id + ")") var Items = menu.children 
     for (var j = 0; j < Items.length; j++) { 
      var menuItem = eval(Items[j].id) if (menuItem.menu != null) { 
       menuItem.innerHTML += "<Span Id=" + menuItem.id + "_Arrow class='Arrow'>4</Span>" //var tmp = eval(menuItem.id+"_Arrow") 
        // tmp.style.pixelLeft = menu.getBoundingClientRect().Right //- tmp.offsetWidth - 15 
       FindSubMenu(menuItem.menu) 
      } 
      if (menuItem.cmd != null) { 
       menuItem.onclick = new Function("Do(" + menuItem.id + ")") 
      } 
      menuItem.onmouseover = new Function("highlight(" + Items[j].id + ")") 
     } 
    } 
}