javascript-搜索后Ajax重置页面恢复正常

我有一个实时搜索,它从数据库中提取所有相关信息,并使用Ajax将其显示在与搜索相关的表中,因此它不会刷新页面,因此在每次搜索后它都重置为零,直到它恢复收到另一个输入,但是我希望它显示其正常运行(所有信息).

删除输入后希望它看起来像什么：http://prntscr.com/hnmvhr

index.PHP

    <!DOCTYPE html>
<html>
    <head>
        <title>Webslesson Tutorial | Autocomplete TextBox using Bootstrap Typehead with Ajax PHP</title>
        <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
        <script src="https://cdnjs.cloudflare.com/ajax/libs/bootstrap-3-typeahead/4.0.2/bootstrap3-typeahead.min.js"></script>
        <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" />
    </head>
    <body>
        <br /><br />
        <div class="container" style="width:600px;">
        <h2 align="center">Ajax live data search using Jquery PHP MysqL</h2>
        <div class="form-group">
            <div class="input-group">
                <span class="input-group-addon">Search</span>
                <input type="text" name="search_text" id="search_text" placeholder="Search by Customer Details" class="form-control" />
            </div>
        </div>
        <br />
        <div id="result">
            <div class='table-responsive'>
                <table class='table table-bordered'>
                <tr>
                    <th>Customer name</th>
                    <th>Address</th>
                    <th>City</th>
                    <th>Potal code</th>
                    <th>Country</th>
                </tr>
                <?PHP
                    include('db.PHP');
                    $customers = DB::query('SELECT * FROM live');
                    foreach($customers as $p){
                      echo '<tr>
                              <td>'.$p["name"].'</td>
                              <td>'.$p["address"].'</td>
                              <td>'.$p["city"].'</td>
                              <td>'.$p["postCode"].'</td>
                              <td>'.$p["country"].'</td>
                            </tr>';
                    }

                    ?>
            </div>
        </div>
    </body>
</html>
<script>
    $('#search_text').keyup(function(){
      var txt = $(this).val();
      if(txt != ''){
        $.ajax({
          url: "fetch.PHP",
          method: "POST",
          data:{search:txt},
          dataType: "text",
          success:function(data){
            $('#result').html(data);
          }
        });
      }else{
        $('#result').html('');
    });
</script>

fetch.PHP

<?PHP
$connect = MysqLi_connect("localhost", "root", "", "ajax");
$output = '';
$sql = "SELECT * FROM live WHERE name LIKE '%".$_POST['search']."%'";
$result = MysqLi_query($connect, $sql);

if(MysqLi_num_rows($result) > 0){
  $output .= "<h4 align='center'>Search result</h4>";
  $output .= "<div class='table-responsive'>
                <table class='table table-bordered'>
                  <tr>
                    <th>Customer name</th>
                    <th>Address</th>
                    <th>City</th>
                    <th>Potal code</th>
                    <th>Country</th>
                  </tr>";

 while($row = MysqLi_fetch_array($result)){
   $output .= '
               <tr>
                <td>'.$row["name"].'</td>
                <td>'.$row["address"].'</td>
                <td>'.$row["city"].'</td>
                <td>'.$row["postCode"].'</td>
                <td>'.$row["country"].'</td>
               </tr>
              ';
 }
 echo $output;
}else{
  echo "There are no customers.";
}

?>

谢谢,
伊森

解决方法:

您可以将原始数据集保存到变量中,如果输入是“”,则可以将变量中的内容还原为以下内容,而不是将html内容设置为“”：

var originalData = $('#result').html();
$('#search_text').keyup(function(){
  var txt = $(this).val();
  if(txt != ''){
    $.ajax({
      url: "fetch.PHP",
      method: "POST",
      data:{search:txt},
      dataType: "text",
      success:function(data){
        $('#result').html(data);
      }
    });
  } else {
    $('#result').html(originalData);
  }
});

javascript-搜索后Ajax重置页面恢复正常

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