Question

我想使用jquery在表中显示mysql的数据。我在下面创建了一个php表，但我想使用jquery instate在#mytable上的表单上显示表。我希望我的代码不会太混乱

> echo "<table class=\"optable\">
<tr>
<th>Device ID</th>
<th>Name</th>
<th>Type</th>
<th>Make & Model</th>
<th>Memory</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
  echo "<tr>";
  echo "<td>" . $row['device_id'] . "</td>";
  echo "<td>" . $row['device_name'] . "</td>";
  echo "<td>" . $row['device_type'] . "</td>";
  echo "<td>" . $row['make_model'] . "</td>";
  echo "<td>" . $row['memory'] . "</td>";
  echo "</tr>";
}
echo "</table>";

形式

<html>
   <script src="http://code.jquery.com/jquery-1.10.2.js"   type="text/javascript"></script>
   <script src="./action/scripts/script1.js" type="text/javascript">      </script> 
      <div>         
          <input id="name" type="text" name="name" onkeyup="updatefrm2('name')"  />      
       </div>
       <div id="mytable"></div> 
</html>

这是我要修改的jquery脚本以显示表

script1.js

function updatefrm2($name) {
    var name = $("#" + $name).val();
    if ($.trim(name) !='') {
    $.post('http://exemple.com/action/subs/name6.php', {name: name},     function(data) {

    $('#member_id').val(data['id']);
    $('#member_id2').val(data['id']);
    $('#member_id_clog').val(data['id']);
    $('#member_id').val(data['id']);
    $('.price2').val(data['id']);

    });
 }    
}

这是我用来从mysql收集dada的php代码

name6.php

<?php

if (!empty($_POST['name'])) {
    require '../db/connect6.php';
    $safe_name = mysql_real_escape_string(trim($_POST['name']));
    $query = mysql_query("
      SELECT * FROM `oz2ts_users`
      WHERE  `oz2ts_users`.`id` = '". $safe_name ."'
  ");

    if (mysql_num_rows($query) > 0) {
      $row = mysql_fetch_assoc($query);
      json($row);
   } else {
      json(null);
  }
 }

function json ($array) {
    header("Content-Type: application/json");
    echo json_encode($array);
}

?>

Answer 1

我认为你的 name6.php 代码应该是：

<?php

if (!empty($_POST['name'])) {
        require '../db/connect6.php';
        $safe_name = mysql_real_escape_string(trim($_POST['name']));
        $query = mysql_query("
              SELECT * FROM `oz2ts_users`
              WHERE  `oz2ts_users`.`id` = '". $safe_name ."'
        ");

        if (mysql_num_rows($query) > 0) {
            $row = mysql_fetch_assoc($query);
            echo "<table class=\"optable\">
                <tr>
                <th>Device ID</th>
                <th>Name</th>
                <th>Type</th>
                <th>Make & Model</th>
                <th>Memory</th>
            </tr>";

            while($row = mysqli_fetch_array($result)) {
                  echo "<tr>";
                  echo "<td>" . $row['device_id'] . "</td>";
                  echo "<td>" . $row['device_name'] . "</td>";
                  echo "<td>" . $row['device_type'] . "</td>";
                  echo "<td>" . $row['make_model'] . "</td>";
                  echo "<td>" . $row['memory'] . "</td>";
                  echo "</tr>";
            }

            echo "</table>";
       } else {
          echo 'null';
      }
 }

script1.js 应为：

function updatefrm2($name) {
    var name = $("#" + $name).val();

    if ($.trim(name) !='') {
        $.post("http://exemple.com/action/subs/name6.php", {name: name}, function(data) {

            $("#mytable").html(data);

        });
    }    
}

希望这有帮助。

Answer 2

使用append。

function updatefrm2($name) {
    var name = $("#" + $name).val();
    if ($.trim(name) !='') {
    $.post('http://exemple.com/action/subs/name6.php', {name: name},     function(data) {

    $('#yourtableid').append('<tr>\
                               <td>'+data.device_id+'</td>\
                               <td>'+data.device_name+'</td>\
                               <td>'+data.device_type+'</td>\
                               <td>'+data.make_model+'</td>\
                               <td>'+data.memory+'</td>\
                              </tr>');
    });
 }    
}

如何创建jquery表？

2 个答案: