UPDATE ...这是我从教程中实现的代码,在网络中的chrome开发工具中我可以在标题中看到变量正在发送,并且在预览中我可以看到下拉菜单,但是它没有插入到加载的网页ajax get方法
<script type="text/javascript">
$(document).ready(function() {
$('#selectEvidence').change(function(){
alert($(this).val());
});
});
function evidencesearch(str)
{
if (str=="")
{
document.getElementById("case").innerHTML="";
return;
}
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("case").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","searchfunction.php?variable="+str,true);
xmlhttp.send();
}
</script>
<?php
$variable = $_GET['variable']; //used for second drop down menu
//echo "test test test $variable";
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$db = 'fid';
$conn = mysql_connect($dbhost,$dbuser,$dbpass);
if (!$conn)
die('Could not connect: ' . mysql_error());
mysql_select_db($db);
echo '<label class="input" for="case" type="input">Specify: </label><select id="case" name="case"><option=value"null"></option>'; //Insert to loaded page
$resource = mysql_query("SELECT $variable FROM `evidence`");
if($resource && mysql_num_rows($resource)) {
while ($row = mysql_fetch_assoc($resource)){
echo '<option value="'.$row[$variable].'">'.$row[$variable].'</option></select>';//Insert to loaded page
}
}
mysql_close($conn)
?>
为什么不使用jQuery? – kabuto178 2013-03-23 21:14:16
我确实看过jquery,但是我已经尝试过在页面上显示下拉菜单,这与通过类而不是id调用div有关。我只是有一个问题,我的sql语句现在打破了,谢谢! – James 2013-03-23 21:24:12