Question

我正在尝试这个HTML代码：

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

和脚本：

function setThemeColor(buttonName) {
    localStorage.themeColor = buttonName;
    document.getElementsByTagName('html')[0].className = buttonName
    var themeButtons = document.querySelectorAll(".theme");
    for (var button in themeButtons) {
        themeButtons[button].disabled = false;
    }
    // this.disabled = false;
    // element.setAttribute("disabled", "disabled");
}

我在设置调用该函数的按钮的禁用状态时遇到问题。有人能告诉我怎么做到这一点。我尝试了两件事，但似乎都没有效果。

Answer 1

传递对按钮的引用，而不只是名称：

<强> HTML

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

<强> JS

function setThemeColor(button) {
    localStorage.themeColor = button.name;
    document.getElementsByTagName('html')[0].className = button.name;
    var themeButtons = document.querySelectorAll(".theme");
    for (var button in themeButtons) {
        themeButtons[button].setAttribute("disabled", "disabled");
    }
    button.setAttribute("disabled", "disabled");
}

Answer 2

document.getElementById("buttonid1").disabled=false;

如何在函数内处理调用函数的元素？

2 个答案: