php - 通过名称搜索成员后如何显示数据库中共享相同地址的所有成员

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);
?>

我需要代码来按与搜索查询匹配的名字或姓氏搜索数据库，然后从该查询中找到的成员具有用于搜索数据库的成员的地址，这些成员的地址与第一个查询返回的任何成员的地址相同。我为此花了大约3个小时，却一无所获，因此不胜感激。

最佳答案

在您的第一个查询结果之后尝试一下

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);