Jquery根据输入检查值隐藏div

Question

我正在尝试根据输入检查值隐藏 div。如果该项目已被检查，它也应该工作，到目前为止我的代码不起作用。正如我想隐藏的 div 总是显示的那样？我创建了一个JS Fiddle 来复制问题

代码如下：

var test = $("input[value='home']");

if (test == true) {
  $(".title").hide();
} else {
  $(".title").show();
}

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div role="radiogroup">
  <input type="radio" name="gender" value="home" checked> Male
  <br>
  <input type="radio" name="gender" value="female"> Female
  <br>
  <input type="radio" name="gender" value="other"> Other
  <br />
  <br />
  <div class='title'>
    This is the title not on home value
  </div>
</div>

任何输入都会很棒。

Answer 1

您只能使用 CSS 来实现：

input[name="gender"][value="home"]:checked ~ div.title {
  display: none;
}

<input type="radio" name="gender" value="home" checked> Male
<br>
<input type="radio" name="gender" value="female"> Female
<br>
<input type="radio" name="gender" value="other"> Other
<br />
<br />
<div class='title'>
    This is the title not on home value
</div>

Answer 2

你需要为radio组添加一个事件监听，然后获取checked选项的值。此外，您可能希望在页面加载时运行一次，这就是最后一行的 .change() 所做的。

//run this anytime a radio input changes
$('input[name="gender"]').on('change', function() {
    //get the currently selected value
    let test = $('input[name="gender"]:checked').val();

    //check the current value
    if (test == 'home') {
        $(".title").hide();
    } else {
        $(".title").show();
    }
}).change(); //trigger this on page load

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.0.0/jquery.min.js"></script>
<div role="radiogroup">
    <input type="radio" name="gender" value="home" checked> Male
    <br>
    <input type="radio" name="gender" value="female"> Female
    <br>
    <input type="radio" name="gender" value="other"> Other
    <br />
    <br />
    <div class='title'>
        This is the title not on home value
    </div>
</div>

Answer 3

If Radio button checked show div 的副本

这是为您的案例改编的代码：

$('input[type="radio"]').click(function(){
    if($(this).attr("value")=="home"){
        $(".title").hide('slow');
    }
    else{
       $(".title").show('slow');
    }  
});
$('input[type="radio"]').trigger('click');

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div role="radiogroup">
    <input type="radio" name="gender" value="home" checked> Male
    <br>
    <input type="radio" name="gender" value="female"> Female
    <br>
    <input type="radio" name="gender" value="other"> Other
    <br />
    <br />
    <div class='title'>
        This is the title not on home value
    </div>
</div>

Answer 4

试试这个。 首先在准备功能检查输入是否被检查。然后检查输入更改功能。

$(document).ready(function() {
    var test = $("input[name='gender']");
    var home = $("input[value='home']")[0];
    var title = $(".title");
    if(home.checked) {
        title.hide();
    }

    test.on('change', function(e) {
        if(e.target.value === 'home') {
        title.hide();
      } else {
        title.show();
      }
    })
});

Answer 5

检查是否有任何输入值为checked：

// hide show title on load
if ($("input[value='home']:checked").length) {  // can use :checked selector to see if input with value of home is checked (see link above)
  $(".title").hide();
} else {
  $(".title").show();
}

$("input").on('change', function() {             // hide or show title on change of input (probably use a class here instead of a bare input selector)
  if (this.value === "home" && this.checked) {
    $(".title").hide();                          // hide title if value is home and checkbox is checked
  } else {
    $(".title").show();                          // otherwise show title
  }
})

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div role="radiogroup">
  <input type="radio" name="gender" value="home" checked> Male
  <br>
  <input type="radio" name="gender" value="female"> Female
  <br>
  <input type="radio" name="gender" value="other"> Other
  <br />
  <br />
  <div class='title'>
    This is the title not on home value
  </div>