如何显示在数据库中按名称搜索成员

Question

PHP代码后，共享同一个地址的所有成员迄今

<?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); 
?> 

我需要的代码以匹配该名或姓搜索数据库搜索查询，然后从该查询中找到的成员的地址用于搜索数据库中的地址与第一个查询返回的任何成员相同的地址。我花了大约3个小时，并没有得到任何地方，所以一些帮助将不胜感激。

Answer 1

试试这个第一个查询结果后

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);