4

我有以下 HTML:

<button name="darkBlue" onclick="setThemeColor(this.name)">Blue</button>
<button name="black" onclick="setThemeColor(this.name)">Black</button>

这个脚本:

if (localStorage.buttonColor) {
   var themeButtons = document.querySelectorAll(".theme");
   for (var button in themeButtons) {
      themeButtons[buttons].removeAttribute("disabled");
   }
   // I need here to disable the button with the name that matches localstorage name
}

我已经有了从所有按钮中删除禁用的方法。但是,在不使用 jQuery 的情况下,如何禁用与 localStorage.buttonColor 同名的按钮?

我也可以在for (var button in themeButtons)循环中完成所有这些吗?如果我能做到这一点,它可能是一个更干净的解决方案。

4

1 回答 1

7

如果只有一个按钮:

document.querySelector('button[name="' + localStorage.buttonColor + '"]').disabled = true;

或者:

var el = document.getElementsByName(localStorage.buttonColor);
if (el) {
    el[0].disabled = true;
}

如果有多个元素:

var all = document.querySelectorAll('button[name="' + localStorage.buttonColor + '"]');
for (var i = 0, len = all.length; i<len; i++){
    all[i].disabled = true;
}

如果有多个按钮,并且您需要启用不共享名称的按钮localStorage.buttonColor

var buttons = document.getElementsByTagName('button'),
    buttonsQSA = document.querySelectorAll('button');

// iterate over whichever collection you prefer to use
for (var i = 0, len = buttonsQSA.length; i<len; i++){
    buttonsQSA[i].disabled = buttonsQSA[i].name == localStorage.buttonColor;
}
于 2013-10-11T10:25:23.490 回答