PHP -选择下拉菜单后直接显示来自MySQL的消息

它使用Ajax很简单，我已经在下面检查了

db_file.php

<?php
$db_name="pet"; // Database name
$tbl_name="lover"; // Table name
$pet_id=$_POST['pet_id']                        
$sql="SELECT details FROM $tbl_name WHERE id = $pet_id";
$result=mysql_query($sql);
while($rows=mysql_fetch_array($result)){
$id=$rows['id'];
$details=$rows['details'];
echo $details;//this output is transeffered to ur actual page
}
?>

current_file.php

//不要忘记包含Jquery文件…

<script>
$("#pet_id").change(function()
    {
        var value = $("#pet_id option:selected").text();
        $.ajax({ 
            url: "current_file.php",
            data: {value: value},
            type: "POST",
            success: function(output) {
                    alert(output);//Here u wil get your output, You can append this where you want to display
                   }
            }
        })
});
</script>
<select name="pet_id" id ="pet_id" >
<option value="">Select PET:</options>
<option value="1">DOG</options>
<option value="2">BIRD</options>
<option value="3">FISH</options>

HTML

<select name="pet_id" onchange="redirect(this.value)">
    <option value="">Select PET:</options>
    <option value="1">DOG</options>
    <option value="2">BIRD</options>
    <option value="3">FISH</options>
</select>

function redirect(selectedOption)
{
  if(selectedOption)
  {
     window.location = 'yourpagename.php?opt='+selectedOption;
  }
  else
  {
     window.location = 'yourpagename.php';
  }
}

if(isset($_GET['opt']))
{
  $opt = $_GET['opt'];  // do not forget to sanitize this
  $sql="SELECT details FROM $tbl_name WHERE id = $opt";
  $result=mysql_query($sql);
  while($rows=mysql_fetch_array($result)){
  $id=$rows['id'];
  $details=$rows['details'];
  // do something with $details; in your case echo it out      
  }
}

PS:停止使用mysql_*函数。为什么?查找在这里

正确的代码是:

<form action="$_SERVER['PHP_SELF']" method="POST">
<select name="pet_id" id="pet_id" onChange="this.form.submit()">
    <option value="">Select PET:</options>
<option value="1">DOG</options>
<option value="2">BIRD</options>
<option value="3">FISH</options>
</select>
</form>

则显示数据库

$pet_id=$_POST['pet_id']; 
if (!empty($pet_id)){
    $db_name="pet"; // Database name
    $tbl_name="lover"; // Table name

    $sql="SELECT details FROM $tbl_name WHERE id = $pet_id";
    $result=mysql_query($sql);
    while($rows=mysql_fetch_array($result)){
        $id=$rows['id'];
        $details=$rows['details'];
       // close while loop 
     }
}//close if statement