Question

我想在用户从下拉列表中选择一个选项后显示我的数据库中的详细信息。类似的东西：

SELECT `details` FROM $tbl_name WHERE id= value of selected option


<?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'];

    // close while loop 
    }

 ?>
<select name="pet_id">
    <option value="">Select PET:</options>
    <option value="1">DOG</options>
    <option value="2">BIRD</options>
    <option value="3">FISH</options>

    <a> then it will display details from database</a>

Answer 1

简单使用Ajax，我在下面做了检查

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>

Answer 2

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>

JAVASCRIPT

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

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_*个功能。为什么？找出HERE

Answer 3

这是正确的代码：

<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>

然后它将显示数据库中的详细信息

PHP

$pet_id=$_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

PHP - 在选择下拉列表后直接显示来自MySQL的消息

3 个答案: