将变量传递给 jQuery 属性包含选择器

Question

是否可以将变量传递给 jQuery 属性包含选择器？

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div data-vp='{
  id: vp-485858383,
  customTwo: "test"
}'>
asdf
</div>

$(function() {
    var testString = "vp-485858383";
  $('[data-vp*="id: ${testString}"]').css('color', 'red');
});

我也试过：

$('[data-vp*="id:' +  testString + '"]').css('color', 'red');

JS Fiddle

Answer 1

你只需要 id:

之后的 space

$(function() {
    var testString = "vp-485858383";
  $('[data-vp*="id: ' +  testString + '"]').css('color', 'red');
});

演示

$(function() {
 var testString = "vp-485858383";
  $('[data-vp*="id: ' +  testString + '"]').css('color', 'red');
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div data-vp='{
      id: vp-485858383,
      customTwo: "test"
}'>
asdf
</div>

将变量传递给 jQuery 属性包含选择器

Pass variable into jQuery attribute-contains selector

javascript

jquery

custom-data-attribute