php - 按姓名搜索成员后如何显示数据库中共享同一地址的所有成员

Question

到目前为止的 PHP 代码

<?php include 'topbit.php';
    if (isset($_POST['address_submitted'])) {
    $given_name = $_POST['address'];
    $find_address_sql = "SELECT * FROM `members` WHERE `LastName` LIKE 
    '%$given_name%' OR `FirstName` LIKE '%$given_name%' ORDER BY `LastName` 
    ASC";
    $find_address_query = mysqli_query($dbconnect, $find_address_sql);
    $find_address_rs = mysqli_fetch_assoc($find_address_query);
    $count = mysqli_num_rows($find_address_query);
?>

我需要代码通过与搜索查询匹配的名字或姓氏来搜索数据库，然后从该查询找到的成员的地址用于在数据库中搜索地址与由返回的任何成员相同的成员第一个查询。我已经花了大约 3 个小时来解决这个问题，但没有取得任何进展，因此我们将不胜感激。

Answer 1

在获得第一个查询结果后尝试此操作

foreach( $find_address_rs as $key =>$value)
    {
      if(isset($addresses))
      {
        $addresses.=','.$value['address'];
      }
     else
    { 
       $addresses=$value['address'];
     }
       }


     $result_members = "SELECT * FROM `members` WHERE `address` in ('".$addresses."')";

$members  = mysqli_query($dbconnect, $result_members);
$members_rs = mysqli_fetch_assoc($members);