Question

我正在使用W3 Schools示例：http://www.w3schools.com/PHP/php_ajax_database.asp来显示一个数据库数据表，只需从一个选择框中选择。如何修改它以包含另一个选择框，例如我想要匹配名称和年龄的记录？

Answer 1

你可以尝试这个（下面的一段代码没有经过测试，但它应该可以工作）

HTML网页

<html>
<head>
<script type="text/javascript" src="selectuser.js"></script>
</head>
<body>

<form>
Select a User:
<select name="users" id="users">
<option value="1">Peter Griffin</option>
<option value="2">Lois Griffin</option>
<option value="3">Glenn Quagmire</option>
<option value="4">Joseph Swanson</option>
</select>

Select an age:
<select name="age" id="age">
<option value="39">39</option>
<option value="40">40</option>
<option value="41">41</option>
<option value="42">42</option>
</select>

<input type='button' value='Refresh Data' onclick="showUser(document.getElementById('users').value,document.getElementById('age').value)">
</form>
<br />
<div id="txtHint"><b>Person info will be listed here.</b></div>

</body>
</html>

Java脚本代码为：

var xmlhttp;

function showUser(str,age)
{
xmlhttp=GetXmlHttpObject();
if (xmlhttp==null)
  {
  alert ("Browser does not support HTTP Request");
  return;
  }
var url="getuser.php";
url=url+"?q="+str+"&a="+age;
url=url+"&sid="+Math.random();
xmlhttp.onreadystatechange=stateChanged;
xmlhttp.open("GET",url,true);
xmlhttp.send(null);
}

function stateChanged()
{
if (xmlhttp.readyState==4)
{
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}

function GetXmlHttpObject()
{
if (window.XMLHttpRequest)
  {
  // code for IE7+, Firefox, Chrome, Opera, Safari
  return new XMLHttpRequest();
  }
if (window.ActiveXObject)
  {
  // code for IE6, IE5
  return new ActiveXObject("Microsoft.XMLHTTP");
  }
return null;
}

PHP页面

<?php
$q=$_GET["q"];
$a=$_GET["a"];

$con = mysql_connect('localhost', 'peter', 'abc123');
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("ajax_demo", $con);

$sql="SELECT * FROM user WHERE id = '".$q."' and age='".$a."'";

$result = mysql_query($sql);

echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Age</th>
<th>Hometown</th>
<th>Job</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['FirstName'] . "</td>";
  echo "<td>" . $row['LastName'] . "</td>";
  echo "<td>" . $row['Age'] . "</td>";
  echo "<td>" . $row['Hometown'] . "</td>";
  echo "<td>" . $row['Job'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?>

Answer 2

更改

<select name="users" onchange="showUser()" id="name_select">

ADD年龄从1下降到100

<select name="age" onchange="showUser()" id="age_select">
  <option value="1">1</option>
  <option value="2">2</option>
                   |
  <option value="100">100</option>
</select>

更改

function showUser(str)
{
xmlhttp=GetXmlHttpObject();
if (xmlhttp==null)
  {
  alert ("Browser does not support HTTP Request");
  return;
  }
name= document.getElementByID('name_select').value
age= document.getElementByID('age_select').value
var url="getuser.php";
url=url+"?q="+str+"&age="+age;
url=url+"&sid="+Math.random();
xmlhttp.onreadystatechange=stateChanged;
xmlhttp.open("GET",url,true);
xmlhttp.send(null);
}

改变php

$age=$_GET["age"];
$sql="SELECT * FROM user WHERE id = '".$q."' and age=".$age;

AJAX：使用多个选择框中的值显示MySQL数据

2 个答案: