Question

我有三张桌子，我想检查它们是否包含特定元素，例如一个值为＆＃39; Previous＆＃39;的按钮。我通过使用jquery函数find并编写函数来解决它，但我需要在没有jquery的情况下解决这个问题。这可能吗？

＆＃13;

var t1 = document.getElementById("table_one");
var t2 = document.getElementById("table_two");
var t3 = document.getElementById("table_three");

has_prev_button(t1);
has_prev_button(t2);
has_prev_button(t3);

function has_prev_button(element)
{
  var has_prev_button = false;
  var check = $(element).find("input[type=button]");
  
  for (i=0; i<=check.length-1; i++)
  {
    if (check[i].getAttribute("value") == "Previous")
    {
      has_prev_button = true;
    }
  }

  if (has_prev_button)
  {
    document.write("<p>The selected table has a Previous button</p>");
  }
  else
  {
    document.write("<strong><p style='color:red'>The selected table has NO Previous button</p></strong>");
  }
}

＆＃13;

table {
  margin-bottom:40px;
  border: 1px solid black;
}

td {
  border: 2px solid #D8D8D8;
  width: 70px;
}

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table id="table_one">
  <tr>
    <td>Foo</td>
    <td>Bar</td>
  <tr>
  <tr>
    <td><input type="button" value="Previous"></td>
    <td><input type="button" value="Next"></td>
  </tr>
</table>

<table id="table_two">
  <tr>
    <td>Foo</td>
    <td>Bar</td>
  <tr>
  <tr>
    <td></td>
    <td><input type="button" value="Next"></td>
  </tr>
</table>

<table id="table_three">
  <tr>
    <td>Foo</td>
    <td>Bar</td>
  <tr>
  <tr>
    <td><input type="button" value="Previous"></td>
    <td><input type="button" value="Next"></td>
  </tr>
</table>

＆＃13;

Answer 1

使用element.querySelectorAll：

＆＃13;

var t1 = document.getElementById("table_one");
var t2 = document.getElementById("table_two");
var t3 = document.getElementById("table_three");

has_prev_button(t1);
has_prev_button(t2);
has_prev_button(t3);

function has_prev_button(element)
{
  var has_prev_button = false;
  var check = element.querySelectorAll("input[type=button]");
  
  for (i=0; i<=check.length-1; i++)
  {
    if (check[i].value == "Previous")
    {
      has_prev_button = true;
    }
  }

  if (has_prev_button)
  {
    document.write("<p>The selected table has a Previous button</p>");
  }
  else
  {
    document.write("<strong><p style='color:red'>The selected table has NO Previous button</p></strong>");
  }
}

＆＃13;

table {
  margin-bottom:40px;
  border: 1px solid black;
}

td {
  border: 2px solid #D8D8D8;
  width: 70px;
}

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table id="table_one">
  <tr>
    <td>Foo</td>
    <td>Bar</td>
  <tr>
  <tr>
    <td><input type="button" value="Previous"></td>
    <td><input type="button" value="Next"></td>
  </tr>
</table>

<table id="table_two">
  <tr>
    <td>Foo</td>
    <td>Bar</td>
  <tr>
  <tr>
    <td></td>
    <td><input type="button" value="Next"></td>
  </tr>
</table>

<table id="table_three">
  <tr>
    <td>Foo</td>
    <td>Bar</td>
  <tr>
  <tr>
    <td><input type="button" value="Previous"></td>
    <td><input type="button" value="Next"></td>
  </tr>
</table>

＆＃13;

Answer 2

只需使用 document.querySelectorAll 以及隔开的类。

例如，如果我想在第三部分中找到按钮并更改其背景颜色：

jQuery

$('.my_sections').eq(2).find('.my_button').css('background', 'pink')

香草JS

document.querySelectorAll('.my_sections .my_button')[2].style.background = 'pink'

类似地，如果我想检查my_button类中有多少按钮：

document.querySelectorAll('.my_sections .my_button').length

jquery find（）等效于javascript

2 个答案: