Question

我有一个名为payment_status的数据库，其中包含以下列：

-id

-user_id

-firstname

-payment_name

-payment_id

-payment_amount

我正在尝试构建这个，所以如果payment_id = = 1，那么该表将显示该表列中具有该id的所有用户。

如果用户的payment_id为2或3，则同样适用，但2为'部分付费'，3为'付费'。

我的数据库中有payment_id为1的用户，但没有显示任何用户。我只收到了其他陈述。

是否有人能够看到我做错了什么不让任何用户显示？

<?php
//Payment Section

    $con = mysqli_connect("localhost", "root", "", "db");
    $run = mysqli_query($con,"SELECT * FROM payment_status ORDER BY id DESC");
    $numrows = mysqli_num_rows($run);
    $payment_id = $row['payment_id'];

        if($payment_id == 3 AND $numrows > 0) {
                while($row = mysqli_fetch_assoc($run)){
                    $paid_id = $row['user_id'];
                    $paid_name = $row['firstname'];
                }
            } else {
            echo "No Payments made";
        }
        if($payment_id == 2 AND $numrows > 0) {
                    while($row = mysqli_fetch_assoc($run)){
                        $partially_paid_id = $row['user_id'];
                        $partially_paid_name = $row['firstname'];
                        $partially_paid_amount = $row['payment_amount'];
                    }
                } else {
            echo "No Partial Payments made";
        }
        if($payment_id == 1 AND $numrows > 0) {
                    while($row = mysqli_fetch_assoc($run)){
                        $owes_id = $row['user_id'];
                        $owes_name = $row['firstname'];
                    }
            } else {
            echo "No one owes";
        }       
?>

表

<table class="paymentTable" id="dragTable">
        <tr>
            <th class="thPayment">Paid</th>
            <th class="thPayment">Partially Paid</th>
            <th class="thPayment">Owes</th>
        </tr>
        <tr>
            <td class="tdPayment" id="paid">
                            <div>
            <?php
                if ($paid_name == true) {
                    echo $paid_name;
                } else {
                    echo "No one has paid";
                }
            ?>
                            </div>
            </td>
            <td class="tdPayment" id="partially_paid">
            <div>
            <?php 
                if ($partially_paid__name == true) {
                    echo $partially_paid__name . " - " . $partially_paid_amount;
                } else {
                    echo "No one has made a partial payment";
                }
            ?>  
            </div>
            </td>
            <td class="tdPayment" id="owes">
            <div>
            <?php
                if ($owes_name == true) {
                    echo $owes_name;
                } else {
                    echo "Everyone has paid something";
                }
            ?>  
            </div>
            </td>
        </tr>
    </table>

Answer 1

您需要执行以下操作： -

<?php
//Payment Section

    $con = mysqli_connect("localhost", "root", "", "db");
    $run = mysqli_query($con,"SELECT * FROM payment_status ORDER BY id DESC");
    $numrows = mysqli_num_rows($run);
        if( $numrows > 0) {
            while($row = mysqli_fetch_assoc($run)){
                $payment_id = $row['payment_id'];
                    if($payment_id == 3){
                        $paid_id = $row['user_id'];
                        $paid_name = $row['firstname'];
                    }else {
                        echo "No Payments made";
                    }
                    if($payment_id == 2){
                        $partially_paid_id = $row['user_id'];
                        $partially_paid_name = $row['firstname'];
                        $partially_paid_amount = $row['payment_amount'];
                    }else {
                         echo "No Partial Payments made";
                    }
                    if($payment_id == 1){
                         $owes_id = $row['user_id'];
                        $owes_name = $row['firstname'];
                    }else {
                         echo "No one owes";
                    }
            }
        }
?>

num rows if语句不从我的数据库输出数据

1 个答案: