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javascript - Questions about data obtained by ajax

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Release： 2023-03-01 19:24:01

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There is ul in a.html, and now there are three li in ul. Now I get the corresponding data through ajax every time I click on one li.
Then there are three a tags in b.html. How can I achieve this? Click the a tag in b.html to jump to the corresponding li in a.html and get the corresponding data? ? ? javascript - Questions about data obtained by ajax

Reply content:

The following is pseudo code for reference only

b.html

<code><a href="a.shtml?li=0>跳转到LI_A并取得数据</a></code>

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a.html

<code><li class="myli">li_a</li>
<li class="myli">li_b</li>
<li class="myli">li_c</li>

$('.myli').click(function () {
    loadData($(this).index());
});

// 进入页面判断参数，然后点击对应LI标签
window.onload = function () {
    var index = getQuery('li')
    $('.myli').get(index).click();
}</code>

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<code>这个demo你看下：
前台：
<html>
    <meta charset="utf-8">

    <head>
        <script type="text/javascript" src="jquery.min.js"></script>
        <script type="text/javascript">
$(function()
{
    $(document).on('mouseout', '[name="state"]', function()
    {
        var html;
        var partialState = $(this).val();
        $.getJSON("getStates.php",
        {
            partialState: partialState
        }, function(states)
        {
            $('input').val(states[0]);
            $.each(states, function(i, value)
            {
                html += value;
                $("#results").html(html);
            });
        });

    });
});

        </script>
    </head>

    <body>
        <input type="text" name="state" autocomplete="off" />
        <br>
        <div id="results"> </div>
    </body>

</html>

后台：
<?php
  error_reporting(E_ALL);
  ini_set('display_errors', 1);

  $con = mysqli_connect("localhost", "root", "")
  or die("Failed to connect to the server: " . mysql_error());

  mysqli_select_db($con, "dedecms")
  or die("Failed to connect to the database: " . mysql_error());

  $partialStates = strtoupper($_GET['partialState']);

  if(!$partialStates)
  {
     echo "###";
  }
  else
  {
     $states = mysqli_query($con,"select typename  from dede_arctype where typename like '%$partialStates%'") or die(mysql_error());
     $sources = array();
   while($row = mysqli_fetch_array($states)) {
       $sources[] = $row['typename'];
   }
   header('Content-Type: application/json');
   echo json_encode($sources);
  }
</code>

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