Question

我想使用jQuery将所有CSS设置为null。我知道你可以做到

$("[class^=col-]").css('width', '');
$("[class^=col-]").css('right', '');
$("[class^=col-]").css('margin', '');
$("[class^=col-]").css('position', '');

但是可以用一个命令来完成，例如：

$("[class^=col-]").css('');

编辑：我在上面的帖子中看到了答案。

通过删除样式属性，它可以完美运行$("[class^=col-]").removeAttr("style")

Answer 1

您需要使用这些属性的默认值来覆盖您设置的值

# Copyright 2017, Google, Inc.
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
#
#    http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.

# [START app_yaml]
handlers:
    # Should probably be at the beginning of the list
    # so it's not clobbered by a catchall route
    - url: /datadog
      script: gae_datadog.datadog.app

env_variables:
    DATADOG_API_KEY: 'MY_KEY'

runtime: nodejs
env: flex
# [END app_yaml]

$(document).ready(function() {
  $("[class^=col-]").css({
    'width': 'auto',
    'position': 'inherit',
    'margin': '0',
    'right': '0'
  });
});

.col-md-12 {
  width: 10px;
  position: fixed;
  margin: 10px;
  right: 10px;
}

Answer 2

使用对象符号指定要删除的all the properties：

$("[class^=col-]").css({
    width: "",
    right: "",
    margin: "",
    position: ""
});

这里是一个例子：

$(document).ready(function() {
  // remove color and background from style "attribute"
  // leave everything else unchanged
  $("[class^=col-]").css({
    'color': '',
    'background': ''
  });
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="col-md-12" style="color: red; background: orange; position: relative;">some test</div>

jQuery删除多种样式

2 个答案: