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Php 如何为特定用户显示内容

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Php 如何为特定用户显示内容,php,Php,我想显示添加的计费组织，仅针对特定用户（由其ID确定）。可能吗 <?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

我想显示添加的计费组织，仅针对特定用户（由其ID确定）。可能吗

<?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));
?>

您可以试试这个。。您必须先获取$user\u id，然后对所选用户执行查询


<?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));
?>



是-大多数事情都是可能的，包括找到这个问题的答案