我知道这可能是一个小问题,但我花了很多时间并没有找到任何解决方案。让我简要介绍一下我的问题.. 我有这个表格,我正在尝试为州,城市和城市制作一个下拉列表。通过ajax调用的区域,这是ajax调用的代码。
function getsstate(category_name) {
var strURL="state.php?sstate="+category_name;
var req = getXMLHTTP();
if (req) {
req.onreadystatechange = function() {
if (req.readyState == 4) {
// only if "OK"
if (req.status == 200) {
document.getElementById('sstatediv').innerHTML=req.responseText;
} else {
alert("There was a problem while using XMLHTTP:\n" + req.statusText);
}
}
}
req.open("GET", strURL, true);
req.send(null);
}
}
function getccity(subcategory_name) {
var strURL="City.php?ccity="+subcategory_name;
var req = getXMLHTTP();
if (req) {
req.onreadystatechange = function() {
if (req.readyState == 4) {
// only if "OK"
if (req.status == 200) {
document.getElementById('ccitydiv').innerHTML=req.responseText;
} else {
alert("There was a problem while using XMLHTTP:\n" + req.statusText);
}
}
}
req.open("GET", strURL, true);
req.send(null);
}
}
</script>
它的工作正常..见图像
现在我正在尝试将这些值插入到我的数据库中,但只有状态名称才会进入数据库。 我的表格是:
<form name="register" action="register_tuto.php" id="formID" method="post">
<select name="state" onChange="getsstate(this.value)">
<option value="">Select State</option>
<?php
//database connection here
$q=mysqli_query($con,"select * from state where category_status='1'");
while($n=mysqli_fetch_array($q)){
echo "<option value=$n[category]>$n[category]</option>";
}
?>
</select>
<tr>
<td align="left" valign="top" style="font-size:13px;">City <span class="redtxt_01">*</span>
<div id="sstatediv"><select name="ccity" >
<option>Select City First</option>
</select></div> </td>
</tr>
<tr>
<td align="left" valign="top" style="font-size:13px;">Area <span class="redtxt_01">*</span>
</td>
<div id="ccitydiv"><select name="aarea">
<option>Select Area First</option>
</select></div>
</td>
</tr>
作为测试,我试图在register_tuto.php页面上打印这些值:
<?php
echo $a=$_POST['state'];
echo $b=$_POST['ccity'];
echo $c=$_POST['aarea'];
?>
这里只有国家价值回归......任何想法或帮助都会被高兴地接受。