我有一個顯示用戶名的下拉菜單。我如何使用Ajax來顯示他們的信息？

Question

我的下拉列表顯示我從數據庫中獲得的用戶名。我想實現一個displayInfo函數，以便當有人選擇一個用戶時，它會自動在下面顯示他/她的信息。

如何在有人選擇他們的名字時顯示用戶信息？

這是我的下拉代碼：

<?php 
//connect             
$conn = mysqli_connect("localhost","user","123abc"); 
mysqli_select_db($conn, "users"); 

//query 
$sql= mysqli_query($conn, "SELECT person_id,first_name FROM users"); 

echo "<select name='dropdown' onchange='displayInfo' id='dropdown'>"; 
while ($row = mysqli_fetch_array($sql))  
{ 
    //display friends' first names on dropdown 
    if($row['person_id'] == $row['first_name']) { 
     echo "<option value='" . $row['person_id'] . "' selected>" . $row['list_name'] . "</option>"; 
    } else { 
     echo "<option value='" . $row['person_id'] . "'>" . $row['first_name'] . "</option>"; 
    } 

} 
echo "</select>"; 

Answer 1

調用上的選擇一個Javascript功能選擇像改寫這個

<?php 
//connect             
$conn = mysqli_connect("localhost","user","123abc"); 
mysqli_select_db($conn, "users"); 

//query 
$sql= mysqli_query($conn, "SELECT person_id,first_name FROM users"); 

echo "<select name='dropdown' onchange='displayInfo' id='dropdown'>"; 
while ($row = mysqli_fetch_array($sql))  
{ 
    //display friends' first names on dropdown 
    if($row['person_id'] == $row['first_name']) { 
     **echo "<option onchange='myFunction(this);' value='" . $row['person_id'] . "' selected>" . $row['list_name'] . "</option>";** 
    } else { 
     echo "<option onchange='myFunction(this);' value='" . $row['person_id'] . "'>" . $row['first_name'] . "</option>"; 
    } 

} 
echo "</select>"; 

<script> 
function myFunction(control){ 

$(control).val() //to access value or any other functionality 
} 
</script>