Question

每次在表单上更改用户ID（onchange =“loadtable”）时，我需要帮助使jquery函数自动加载表单上的表（在id =“notes_table”上）。所以我需要一个jquery代码函数。

**表格**

<!DOCTYPE html>
<html>
<script src="http://code.jquery.com/jquery-1.10.2.js"   type="text/javascript"></script>
<script src="./action/scripts/global2.js" type="text/javascript"></script> 

    <body>
        User Id:<input type="text" name="user_id" id="user" onchange="loadtable">

        <p>Note List</p>
        <div id="notes_table">Lode Notes table here</div>
    </body>
</html>

** query.php **

<!DOCTYPE html>
<html>
<head>
   <style>
     table {
     width: 100%;
     border-collapse: collapse;
    }

    table, td, th {
    border: 1px solid black;
    padding: 5px;
    }

    th {text-align: left;}
  </style>
</head>
<body>

   <?php

     $q = $_POST['user_id'];

     include '../db/connect.php';
     mysqli_select_db($con,"db3");
     $sql="SELECT * FROM oz2ts_call_logs WHERE member_id = '".$q."'";
     $result = mysqli_query($con,$sql);

    echo "<table class=\"optable\">
   <tr>
    <th>Device ID</th>
    <th>Title</th>
    <th>Note</th>
    <th>Status</th>
    <th>Case Number</th>
   </tr>";
  while($row = mysqli_fetch_array($result)) {
  echo "<tr>";
  echo "<td>" . $row['device_id'] . "</td>";
  echo "<td>" . $row['title'] . "</td>";
  echo "<td>" . $row['description'] . "</td>";
  echo "<td>" . $row['status'] . "</td>";
  echo "<td>" . $row['case_number'] . "</td>";
  echo "</tr>";
  }
  echo "</table>";
  mysqli_close($con);
  ?>
  </body>
 </html>

在过滤器上使用jquery自动加载表

0 个答案: