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 Fiddle
最佳答案
您只需在id:
后留一个空格
$(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>