将变量传递到jQuery属性包含选择器

Question

是否可以将变量传递到jQuery属性包含选择器中？

<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');
});

我也尝试过：

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

JS小提琴

Answer 1

您只需在id:后id:一个空格id:

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

演示版

 $(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>