특정 사용자의 콘텐츠를 표시하는 방법

Question

특정 사용자 (ID로 결정)에 대해서만 추가 된 결제 조직을 표시하려고합니다. 가능한가?

<?php 
$query3 = "SELECT * FROM account WHERE user_id = '$user_id' AND account_id = '$account_id'"; 
$result3 = mysqli_query($link, $query3) or die(mysqli_error($link)); 
$row2 = mysqli_fetch_array($result1); 
$user_id = $_SESSION['user_id']; 
$billing_name_id = $_POST['billing_name_id']; 
$billing_reference_num = $_POST['billing_reference_num']; 
$transaction_amount = $_POST['transaction_amount']; 
$account_number = $_POST['account_number']; 
$account_id = $_POST['account_id']; 
$query4 = "INSERT INTO billinghistory (billing_name_id, billing_reference_num, transaction_amount, account_id) 
     VALUES('$billing_name_id','$billing_reference_num', '$transaction_amount','$account_id')"; 
$result4 = mysqli_query($link, $query4) or die(mysqli_error($link)); 
?> 

고맙습니다!

Answer 1

시도해 볼 수 있습니다. 먼저 $ user_id를 가져와 선택한 사용자에 대한 쿼리를 실행해야합니다.

<?php 
$user_id = $_SESSION['user_id']; 
$account_id = $_POST['account_id']; 
$query3 = "SELECT * FROM account WHERE user_id = '$user_id' AND account_id = '$account_id'"; 
$result3 = mysqli_query($link, $query3) or die(mysqli_error($link)); 
$row2 = mysqli_fetch_array($result1); 

$billing_name_id = $_POST['billing_name_id']; 
$billing_reference_num = $_POST['billing_reference_num']; 
$transaction_amount = $_POST['transaction_amount']; 
$account_number = $_POST['account_number']; 

$query4 = "INSERT INTO billinghistory (billing_name_id, billing_reference_num, transaction_amount, account_id) 
     VALUES('$billing_name_id','$billing_reference_num', '$transaction_amount','$account_id')"; 
$result4 = mysqli_query($link, $query4) or die(mysqli_error($link)); 
?>