这就是我一直在尝试做的事情:
- 我有一个下拉列表，用于选择学期。
- 选择学期后，在下一个下拉列表中，该学期的相关类(class)应显示为选项。
- 每次我换学期时都会发生这种情况。
数据库结构如下:
- 数据库 - scifac
- 表-course_info
- 列 - courseID:varchar(主键)、courseName varchar、学期 int。
我尝试通过教程使用 ajax 来执行此操作，但它不起作用。
这是我的 html 页面
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<body>
<script language="javascript" type="text/javascript">
<!--
//Browser Support Code
function ajaxFunction()
{
var ajaxRequest; // The variable that makes Ajax possible!
try
{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
}
catch (e)
{
// Internet Explorer Browsers
try
{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e)
{
try
{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}
catch (e)
{
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
// Create a function that will receive data
// sent from the server and will update
// div section in the same page.
ajaxRequest.onreadystatechange = function()
{
if(ajaxRequest.readyState == 4)
{
var ajaxDisplay = document.getElementById('ajaxDiv');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
// Now get the value from user and pass it to
// server script.
//--------------------------------------------
/*var age = document.getElementById('age').value;
var wpm = document.getElementById('wpm').value;
var sex = document.getElementById('sex').value;
var queryString = "?age=" + age ;
queryString += "&wpm=" + wpm + "&sex=" + sex;
ajaxRequest.open("GET", "ajax-example.php" + queryString, true);
ajaxRequest.send(null); */
var sem=document.getElementById('year').value;
var queryString="?sem="+sem;
ajaxRequest.open("GET","ajax-example.php" + queryString, true);
ajaxRequest.send(null);
}
//-->
</script>
<form name='frmform'>
<select id='year' onchange='ajaxFunction()'>
<option value="1">Year I Sem I</option>
<option value="2">Year I Sem II</option>
</select>
<!--<input type='button' onclick='ajaxFunction()' value='Query MySQL'/>-->
<select id="ajaxdiv">
</select>
</form>
<!--<div id='ajaxDiv'>Your result will display here</div>-->
</body>
</html>
我的php页面(ajax-example.php)如下。
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$dbname = "scifac";
//Connect to MySQL Server
mysql_connect($dbhost, $dbuser, $dbpass);
//Select Database
mysql_select_db($dbname) or die(mysql_error());
// Retrieve data from Query String
$sem = $_GET['year'];
$sem=(int)$sem;
$query = "SELECT courseID FROM course_info WHERE semester = '$sem'";
//Execute query
$qry_result = mysql_query($query) or die(mysql_error());
// Insert a new row in the table for each person returned
while($row = mysql_fetch_array($qry_result)){
$display_string.="<select>";
$display_string.="<option>$row[courseID]</option>";
}
echo "Query: " . $query . "<br />";
$display_string .= "</select>";
echo $display_string;
?>
非常感谢您在这方面给予的善意考虑。
最佳答案
您可以在 ajax-example.php 中尝试此操作:
基本上，您需要更改 $_GET['sem'] 来代替 $_GET['year']
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$dbname = "scifac";
//Connect to MySQL Server
mysql_connect($dbhost, $dbuser, $dbpass);
//Select Database
mysql_select_db($dbname) or die(mysql_error());
// Retrieve data from Query String
$sem = $_GET['sem'];
//$sem=(int)$sem;
$query = "SELECT courseID FROM course_info WHERE semester = '".$sem."'";
//Execute query
$qry_result = mysql_query($query) or die(mysql_error());
$display_string.="<select>";
// Insert a new row in the table for each person returned
while($row = mysql_fetch_array($qry_result)){
$display_string.="<option value='".$row[courseID]."'>".$row[courseID]."</option>";
}
$display_string .= "</select>";
echo $display_string;
?>
关于php - 使用mysql数据库在php中动态构建下拉列表，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/24730464/