我有 3 页 Index.php findasset.php 和 findid.php。我有 2 个下拉菜单，最后一个值将回显到页面的另一部分。我正在使用 ajax 来查询其他下拉列表，它正在部分工作。
除了 findid 页面上的 device_category_name='$cId' 之外，它大部分是动态的并且可以工作，应该用 $category 替换，但我想将代码显示为工作模型。我认为我的问题最初是在 findasset 页面 $category=isset($_GET['category']);
当我尝试在 findid 上回显变量时，它回显的是“1”而不是单词
索引页面有一个从 mysql 数据库中提取的下拉列表，它工作得很好。我已经尽可能地标记了代码。这是部分 工作示例。如果您选择 Category-Drawing，那么其中任何一个资产都可以工作，但这是因为在 findid 页面上，查询是部分硬编码的，我不希望它被硬编码。
我知道，我非常接近解决这个问题，但我被卡住了。你能帮帮我吗？
索引.php
function getXMLHTTP() { //function to return the xml http object
var xmlhttp=false;
try{
xmlhttp=new XMLHttpRequest();
}
catch(e) {
try{
xmlhttp= new ActiveXObject("Microsoft.XMLHTTP");
}
catch(e){
try{
xmlhttp = new ActiveXObject("Msxml2.XMLHTTP");
}
catch(e1){
xmlhttp=false;
}
}
}
return xmlhttp;
}
function getcategory(category) {
var strURL="findasset.php?category="+category;
var req = getXMLHTTP();
if (req) {
req.onreadystatechange = function() {
if (req.readyState == 4) {
// only if "OK"
if (req.status == 200) {
document.getElementById('assetdiv').innerHTML=req.responseText;
} else {
alert("There was a problem while using XMLHTTP:\n" + req.statusText);
}
}
}
req.open("GET", strURL, true);
req.send(null);
}
}
function getid(category,asset) {
var strURL="findid.php?category="+category+"&asset="+asset;
var req = getXMLHTTP();
if (req) {
req.onreadystatechange = function() {
if (req.readyState == 4) {
// only if "OK"
if (req.status == 200) {
document.getElementById('iddiv').innerHTML=req.responseText;
} else {
alert("There was a problem while using XMLHTTP:\n" + req.statusText);
}
}
}
req.open("GET", strURL, true);
req.send(null);
}
}
</script>
</head>
<body>
<form method="post" action="" name="form1">
<table width="60%" border="0" cellspacing="0" cellpadding="0">
<tr>
<td width="150">Category</td>
<td width="150"><select name="category" onChange="getcategory(this.value)">
<?
require "config.php";// connection to database
$query = "SELECT DISTINCT device_category_name FROM fgen_structures ORDER BY device_category_name ASC";
$result = mysql_query($query);
while ($myrow = mysql_fetch_array($result))
{
echo "<option value='$myrow[device_category_name]'>$myrow[device_category_name]</option>";
}
?>
</select></td>
</tr>
<tr style="">
<td>Asset</td>
<td ><div id="assetdiv"><select name="asset" >
<option>Select Category First</option>
</select></div></td>
</tr>
<tr style="">
<td>ID</td>
<td ><div id="iddiv"></div></td>
</tr>
<tr>
<td> </td>
<td> </td>
</tr>
<tr>
<td> </td>
<td> </td>
</tr>
</table>
</form>
</body>
</html>
findasset.php
$category= isset($_GET['category']);// Could be the Start of the PROBLEM
$cate=$_GET['category'];
require "config.php";// connection to database
$query="SELECT * FROM fgen_structures WHERE device_category_name='$cate'";
$result=mysql_query($query);
?>
<select name="asset" onchange="getid(<?=$category;?>,this.value)">
<option>Select State</option>
<? while($row=mysql_fetch_array($result)) { ?>
<option value=<? echo $row['device_type_name'];?>><? echo $row['device_type_name'];?></option>
<? } ?>
</select>
findid.php
<?
$category=isset($_GET['category']); // This is where I think the problem is as well!!!!
$asset=isset($_GET['asset']);
$cate=$_GET['category'];
$assets=$_GET['asset'];
$cId='Drawing'; //If Hard Coded works
require "config.php";// connection to database
$query="SELECT * FROM fgen_structures WHERE device_category_name='$cId' AND device_type_name='$assets'"; // Currently hardcoded with $cid and it works but I need it dynamic with $cate or $category
$result=mysql_query($query);
while($row=mysql_fetch_array($result)) {
echo $row['fgen_structure_id'];
//echo $category; // This displays a 1 ??
} ?>