Question

我正在尝试用基于用户ID的mysql数据填充表。当我使用少量用户ID作为选项时，它可以工作。但是，如果我使用并输入用户ID，它会显示包含数秒的表格并将其关闭。

<html>
<script src="http://code.jquery.com/jquery-1.10.2.js"    type="text/javascript"></script>
<head>
<script>
function showUser(str) {
if (str == "") {
    document.getElementById("txtHint").innerHTML = "";
    return;
} else {
    if (window.XMLHttpRequest) {
        // code for IE7+, Firefox, Chrome, Opera, Safari
        xmlhttp = new XMLHttpRequest();
    } else {
        // code for IE6, IE5
        xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
    }
    xmlhttp.onreadystatechange = function() {
        if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
            document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
        }
    }
    xmlhttp.open("GET","../action/subs/getcalls.php/?q="+str,true);
    xmlhttp.send();
    }
}
</script>
</head>
<body>

<form>
 <select name="users" onchange="showUser(this.value)">
   <option value="">Select a person:</option>
   <option value="788">user 788</option>
   <option value="786">User 786</option>
   <option value="787">User 787</option>
   <option value="789">User 789</option>
 </select>

</form>
<br>
<div id="txtHint"><b></b></div>

</body>
</html>

**如果我更换

    <form>
  <select name="users" onchange="showUser(this.value)">
    <option value="">Select a person:</option>
    <option value="788">user 788</option>
    <option value="786">User 786</option>
    <option value="787">User 787</option>
    <option value="789">User 789</option>
  </select>

 </form>

通过

    <form>
      <input name="users" onchange="showUser(this.value)"/>

    </form>

该表格将显示几秒钟并关闭

这里是getcall.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


$q = intval($_GET['q']);


include '../db/connect.php';

mysqli_select_db($con,"ajax_demo");
$sql="SELECT * FROM oz2ts_call_logs WHERE member_id = '".$q."'";
$result = mysqli_query($con,$sql);

echo "<table>
 <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>

Answer 1

“守则”对我有用，只有在您点击“输入”时才会应用onchange，或许您只想将<input name="users" onchange="showUser(this.value)"/>更改为<input name="users" onkeyup="showUser(this.value)"/>

同样<script src="http://code.jquery.com/jquery-1.10.2.js" type="text/javascript"></script>属于头

Ajax动态数据表

1 个答案: