Question

我正在尝试为我的网站创建Ajax实时搜索。我目前遇到一些麻烦，认为该代码无法正常工作。有人可以帮我:) 我尝试过的查询没有响应的第一个ajax 第二，当我尝试使用search.php时，http请求出错 idk我努力解决了这个问题

我正在使用php 5.6 AJAX jQuery 2.1.3 Bootstrap 4

live.php

<?php include "header.php"; //ajax and jquery script loaded are here?>  
<body>
    <div class="container">
        <label for="ajaxs">Search</label>
        <input type="text" name="search_text" id="search_text" placeholder="Search Keyword" class="form-control" /><br>
        <div id="result"></div>
        <br>
    </div>
</body>
</html>
<script>
$(document).ready(function(){

 load_data();

 function load_data(query)
 {
  $.ajax({
   url:"search.php",
   method:"POST",
   data:{query:query},
   success:function(data)
   {
    $('#result').html(data);
   }
  });
 }
 $('#search_text').keyup(function(){
  var search = $(this).val();
  if(search != '')
  {
   load_data(search);
  }
  else
  {
   load_data();
  }
 });
});
</script>

search.php

<?php
require "koneksi2.php"; //mysql connection are here
$output = '';
if(isset($_POST["query"]))
{
    $search = mysq_real_escape_string($_POST["query"],$koneksi);
    $query = "
    SELECT * FROM bukutamu 
    WHERE id LIKE '%".$search."%'
    OR dari LIKE '%".$search."%' 
    OR agama LIKE '%".$search."%' 
    OR email LIKE '%".$search."%' 
    OR komentar LIKE '%".$search."%'
    ";
}
else
{
    $query = "SELECT * FROM bukutamu ORDER BY id";
}
$result = mysq_query($query,$koneksi);
if(mysq_num_rows($result) > 0)
{
    $output .= '
    <div class="table-responsive">
        <table class="table table-bordered table-hover table-light:hover">
            <tr class="thead bg-primary text-white">
                <th>No</th>
                <th>Dari</th>
                <th>Agama</th>
                <th>Email</th>
                <th>Komentar</th>
                <th>Tanggal</th>
            </tr>';
    $no = 0;
    while($row = mysql_fetch_array($result))
    {
        $no++;
        $output .= '
        <tr>
        <td>'.$no.'</td>
        <td>'.$row["dari"].'</td>
        <td>'.$row["agama"].'</td>
        <td>'.$row["email"].'</td>
        <td>'.$row["komentar"].'</td>
        <td>'.$row["tglsimpan"].'</td>
        </tr>
        ';
    }
    echo $output;
}
else
{
    echo 'Data Not Found';
}

?>

使用PHP MYSQL进行AJAX实时搜索

0 个答案: