Question

我使用以下代码和Ethan Marcotte的书来检查浏览器是否支持某些DOM功能。

// Is this browser sufficiently modern to continue?
if ( !( "querySelector" in document
  && "addEventListener" in window
  && "getComputedStyle" in window) ) {
    return;
}
window.document.documentElement.className += " enhanced";

var nav = document.querySelector( ".nav ul" ),
navToggle = document.querySelector( ".nav .skip" );
if ( navToggle ) {
  navToggle.addEventListener( "click",   function( e ) {
    if ( nav.className == "open" ) {
      nav.className = "";
    } else {
    nav.className = "open";
    }
    e.preventDefault();
  }, false );
}

开发人员工具报告以下错误;

Uncaught SyntaxError：非法返回语句

我已尝试将返回值更改为false，但这并不起作用。基本上，如果找到这些DOM功能，它应该将增强类添加到html标记中。想知道是否有人可以在我出错的地方提供帮助。

感谢。

Answer 1

我建议你在加载函数中插入所有代码：

window.onload = function () {
  if (!("querySelector" in document
        && "addEventListener" in window
        && "getComputedStyle" in window)) {
    return;
  }
  window.document.documentElement.className += " enhanced";

  var nav = document.querySelector(".nav ul"),
      navToggle = document.querySelector(".nav .skip");
  if (navToggle) {
    navToggle.addEventListener("click", function (e) {
      if (nav.className == "open") {
        nav.className = "close";
      } else {
        nav.className = "open";
      }
      e.preventDefault();
    }, false);
  }
}

<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" integrity="sha384-1q8mTJOASx8j1Au+a5WDVnPi2lkFfwwEAa8hDDdjZlpLegxhjVME1fgjWPGmkzs7" crossorigin="anonymous">
<script src="http://code.jquery.com/jquery-1.12.1.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/js/bootstrap.min.js" integrity="sha384-0mSbJDEHialfmuBBQP6A4Qrprq5OVfW37PRR3j5ELqxss1yVqOtnepnHVP9aJ7xS" crossorigin="anonymous"></script>


<ul class="nav nav-pills nav-stacked">
    <li class="active"><a href="#">Home</a></li>
    <li class="dropdown">
        <a class="skip dropdown-toggle" data-toggle="dropdown" href="#">Menu 1 <span class="caret"></span></a>
        <ul class="dropdown-menu">
            <li><a href="#">Submenu 1-1</a></li>
            <li><a href="#">Submenu 1-2</a></li>
            <li><a href="#">Submenu 1-3</a></li>
        </ul>
    </li>
    <li><a href="#">Menu 2</a></li>
    <li><a href="#">Menu 3</a></li>
</ul>

检查浏览器支持时出现非法返回声明

1 个答案: