在使用 javascript 加载时将“checked”添加到单选按钮

Add "checked" to radio button on load with javascript

提问人:Robert Banks 提问时间:11/16/2023 最后编辑:Stephen PRobert Banks 更新时间:11/16/2023 访问量:50

问:

基于 Bootstrap 5 的暗模式切换 javascript (https://getbootstrap.com/docs/5.3/customize/color-modes/#javascript) 我正在尝试将单选按钮设置为在页面加载时以及更改颜色模式时“选中”。

使用以下 javascript:

    <script>
      (() => {
        'use strict'

        const getStoredTheme = () => localStorage.getItem('theme')
        const setStoredTheme = theme => localStorage.setItem('theme', theme)

        const getPreferredTheme = () => {
          const storedTheme = getStoredTheme()
          if (storedTheme) {
            return storedTheme
          }

          return window.matchMedia('(prefers-color-scheme: dark)').matches ? 'dark' : 'light'
        }

        const setTheme = theme => {
          if (theme === 'auto' && window.matchMedia('(prefers-color-scheme: dark)').matches) {
            document.documentElement.setAttribute('data-bs-theme', 'dark')
          } else {
            document.documentElement.setAttribute('data-bs-theme', theme)
          }
        }

        setTheme(getPreferredTheme())

        const showActiveTheme = (theme, focus = false) => {
          const themeSwitcher = document.querySelector('#bd-theme')

          if (!themeSwitcher) {
            return
          }

          const btnToActive = document.querySelector(`[data-bs-theme-value="${theme}"]`)

          document.querySelectorAll('[data-bs-theme-value]').forEach(element => {
            element.classList.remove('checked')
          })

          btnToActive.classList.add('checked')

          if (focus) {
            themeSwitcher.focus()
          }
        }

        window.matchMedia('(prefers-color-scheme: dark)').addEventListener('change', () => {
          const storedTheme = getStoredTheme()
          if (storedTheme !== 'light' && storedTheme !== 'dark') {
            setTheme(getPreferredTheme())
          }
        })

        window.addEventListener('DOMContentLoaded', () => {
          showActiveTheme(getPreferredTheme())

          document.querySelectorAll('[data-bs-theme-value]')
            .forEach(toggle => {
              toggle.addEventListener('click', () => {
                const theme = toggle.getAttribute('data-bs-theme-value')
                setStoredTheme(theme)
                setTheme(theme)
                showActiveTheme(theme, true)
              })
            })
        })
      })()
    </script>

我正在尝试控制按钮组:

  <div class="btn-group" id="bd-theme" role="group">
    <input autocomplete="off" checked class="btn-check" data-bs-theme-value="light" id="btnradio1" name="btnradio" type="radio">
    <label class="btn btn-outline-dark" for="btnradio1">
      Light
    </label>
    <input autocomplete="off" class="btn-check" data-bs-theme-value="dark" id="btnradio2" name="btnradio" type="radio">
    <label class="btn btn-outline-dark" for="btnradio2">
      Dark
    </label>
    <input autocomplete="off" class="btn-check" data-bs-theme-value="auto" id="btnradio3" name="btnradio" type="radio">
    <label class="btn btn-outline-dark" for="btnradio3">
      Auto
    </label>
  </div>

默认情况下,“Light”模式是“选中”的,但如果已经存储了不同的“主题”,则应“选中”其他单选按钮。

我认为特别需要更改的代码是:

        const showActiveTheme = (theme, focus = false) => {
          const themeSwitcher = document.querySelector('#bd-theme')

          if (!themeSwitcher) {
            return
          }

          const btnToActive = document.querySelector(`[data-bs-theme-value="${theme}"]`)

          document.querySelectorAll('[data-bs-theme-value]').forEach(element => {
            element.classList.remove('checked')
          })

          btnToActive.classList.add('checked')

          if (focus) {
            themeSwitcher.focus()
          }
        }
javascript twitter-bootstrap 引导-5

评论

0赞 Daniel Beck 11/16/2023
您的代码看起来已经为所选主题设置了选中的值(在行中)。您只需要在页面加载时调用 showActiveTheme,并将从本地存储中检索到的当前主题传递给它。btnToActive.classList.add('checked')
0赞 Robert Banks 11/16/2023
感谢您@DanielBeck回复。该代码片段似乎将“checked”作为新值添加到“class”属性中,而不是作为新属性,例如 .<input autocomplete="off" checked class="btn-check" ...
0赞 Daniel Beck 11/16/2023
啊,射击你是绝对正确的,我的错误!我看到了“检查”这个词,并掩盖了其余的:(

答:

1赞 dave 11/16/2023 #1

您确定要切换吗?似乎只是您关心的 DOMNode 上的 checked 属性吗?意思是,你只需要设置 ,而不是classListcheckedelement.checked = trueelement.classList.add('checked')

document.getElementById('makeDark').addEventListener('click', () => {
    document.getElementById('dark').checked = true;
});

document.getElementById('makeDarkBroken').addEventListener('click', () => {
    document.getElementById('dark').classList.toggle('checked');
});
.checked {
  width: 20px;
  height: 20px;
}
<label><input id="light" name="demo" value="light" type="radio" checked /> Light</label>
<label><input id="dark" name="demo" value="dark" type="radio" /> Dark</label>

<button id="makeDark">Make Dark</button>
<button id="makeDarkBroken">Make Dark Broken</button>

评论

0赞 Robert Banks 11/16/2023
完美 - 这似乎已经解决了它!谢谢你的帮助,@dave。我唯一的问题是,当单击“浅色”或“深色”时,我是否应该在检查网页时看到源代码中的“已检查”输入属性更改?
1赞 dave 11/16/2023
如果你想在检查元素或其他东西时看到它,你不会这样做(但它不会真正影响任何事情 - 属性是浏览器“关心”的),然后添加和.checkeddocument.getElementById('dark').setAttribute('checked', 'checked');document.getElementById('dark').removeAttribute('checked');