Php 如果（isset（$\u POST['Submit']）未获取状态值

从这个ajax函数中，我首先填充了第一个下拉列表first_state.php *下面是完整代码的一部分*

这里我调用函数并获取第一个下拉列表的值 第二个下拉列表在传递varable以获取_district.php时填充

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我如何以一种形式接受_post['district']和_post['state']，上面的代码运行良好，但我一无所知。问题是我有两个php文件**

在函数文件中用$\u get替换$\u post

您好，哪里有一个叫做txtHint的东西。有一个名为form_pin的表单。你把来自ajax的响应放在哪里？实际上我不知道如何把来自ajax的响应放在哪里有JS错误吗？我猜你的showUser JS函数中有一些错误。事实上，当我选择state下拉列表时，它会从mysql中显示district。但是我想在单个isset中输入state和district值，这样我就可以在mysql中基于它插入值。从上面看，有两个php文件，一个有ajax和state下拉列表，另一个填充district

function showUser(str)
{
if (str=="")
  {
  document.getElementById("txtHint").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("txtHint").innerHTML=xmlhttp.responseText;
    }
  }
xmlhttp.open("GET","get_district.php?q="+str,true);
xmlhttp.send();
}

<form class="form2" name="form_pin" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<select name="state"  onchange="showUser(this.value)">
<option value="">Select State</option>
while ($row = mysql_fetch_array($result)){                          
echo "<option value='".$row['State']."'>".$row['State']."</option>";}
</form>

if(isset($_POST['Submit']))
{
$state = $_POST["state"];
$district = $_POST["district"];
echo $state; echo $district;        
}

<?php
include_once "connect_db.inc";
$q = $_GET['q'];
//echo $q;
$con = mysqli_connect($dbhost, $dbusername, $dbuserpassword);
if (!$con)
  {
  die('Could not connect: ' . mysqli_error($con));
  }

mysqli_select_db($con,$default_dbname);
$sql="SELECT DISTRICT FROM state_district WHERE STATE = '".$q."'";
//$sql="SELECT * FROM state_district";

$result = mysqli_query($con,$sql);
?>
<select name="district">
<?PHP
while($row = mysqli_fetch_array($result))
  {
     echo "<option value='".$row['DISTRICT']."'>".$row['DISTRICT']."</option>";
  }
?>
</select>
<?PHP
mysqli_close($con);
?> 

**You are not requesting the below code** 
------------------

<select name="district">
<?PHP
while($row = mysqli_fetch_array($result))
  {
     echo "<option value='".$row['DISTRICT']."'>".$row['DISTRICT']."</option>";
  }
?>
</select>
-----------------------------

**to do so add ID with <select id="txtHint"></select> option in first_step.php**     
--------------------------
<form class="form2" name="form_pin" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<select name="state"  onchange="showUser(this.value)">
<option value="">Select State</option>
while ($row = mysql_fetch_array($result)){                          
echo "<option value='".$row['State']."'>".$row['State']."</option>";}
**// add below**
<select id="txtHint"></select>
</form>