AJAX：显示MySQL数据与来自多个选择框值 [英] AJAX: Display MySQL data with value from multiple select boxes

问题描述

我使用的是W3学校的例子： http://www.w3schools.com/PHP /php_ajax_database.asp 显示给从单一选择框中选择数据库中的数据表。我怎么能修改此包含另一个选择框，例如说我想匹配的记录姓名和年龄？

I'm using the W3 Schools example: http://www.w3schools.com/PHP/php_ajax_database.asp to display a table of database data given the choice from a single select box. How can I modify this to include another select box, say for example I wanted records matching a name and an age?

推荐答案

修改

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

ADD年龄下拉从1到100

ADD age drop down from 1 to 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>

修改

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);
}

变化在PHP

change in php

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

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