如何在页面重新加载后保持复选框切换div的状态？ [英] How do you maintain the state of a checkbox-toggled div after page reload?

问题描述

我选中了一个复选框时出现div，并在取消选中时隐藏。如果表单有错误，页面会重新加载，即使在提交前选中了复选框，div也会显示为隐藏状态。我可以添加对onLoad事件的调用，但有没有更清晰的方法来确保在页面重新加载后，div是根据复选框的状态呈现的？

I have a div that appears when a checkbox is checked, and hides when unchecked. If the form has errors, the page reloads and the div appears hidden even if the checkbox was checked before submission. I can add a call to the onLoad event, but is there a cleaner way to ensure that, after page reload, the div is rendered based on the status of the checkbox?

Css：

#maintenance-window { display: none; }

jQuery：

$("#check-hasMaintenance").click(function() {
    $(this).is(":checked") ? $('#maintenance-window').show("fast") : $('#maintenance-window').hide("fast")
});

HTML：

<input type="checkbox" value="1" id="check-hasMaintenance">
<div id="maintenance-window">
    Stuff
</div>

推荐答案

选项1

当页面加载时，确保基于复选框显示div：

Ensure div is shown based on check box when the page loads:

$(document).ready(function() {
     // Modified for readability - inline if is fine.
     if($("#check-hasMaintenance").is(":checked"))
         $('#maintenance-window').show("fast")
     else
         $('#maintenance-window').hide("fast");
});

简化隐藏/显示代码：

$("#check-hasMaintenance").click(function() {
    $('#maintenance-window').toggle('fast');
});

选项2

像这样的简写版也应该有效：

A short-hand version like this should work as well:

$(document).ready(function() {
    $('#maintenance-window').toggle($("#check-hasMaintenance").is(":checked"));
});

$("#check-hasMaintenance").click(function() {
    $('#maintenance-window').toggle('fast');
});

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