PDO 在 foreach 循环中更新行值

Question

我创建了一个表格，在一个 foreach 循环中显示表格中的所有记录。然后我在选择中包含了一个 SQL 更新并更新了一个数组值。当我选择不同的值时，循环会更新表中的所有记录，而不仅仅是选择要更新的记录。我整天都在为此苦苦挣扎，请帮助我。

如果 foreach 是错误的方式，我非常感谢您对如何显示可以显示记录并允许更新每个数组的表的建议。

<?php
            foreach ($stmt as $key => $row)
            {

                $newcustomerid = htmlentities($row['customer_id']);

                echo '<tr><td>' . htmlentities($row['customer_company']) . '</td>';
                echo '<td> id:' . $newcustomerid . 'key: ' . $key .'</td>';
                echo '<td>' . htmlentities($row['customer_email']) . '</td>';
                echo '<td>' . htmlentities($row['customer_phone']) . '</td>';
                echo '<td>' . htmlentities($row['customer_country']) . '</td>';
                echo '<td>' . htmlentities($row['customer_city']) . '</td>';
                echo '<td>' . htmlentities($row['customer_segment']) . '</td>';
                echo '<td>' . htmlentities($row['customer_updated']) . '</td>';
                $customer_status = htmlentities($row['customer_status']);

                if(isset($_POST['Submitbb'])){ //check if form was submitted
                $input = $_POST['Submitbb']; //get input text

                    $stmtUpdateStatus = $conn->prepare("UPDATE user_customers SET `customer_status` = :customer_status WHERE `customer_id` = :customer_id");
                    $stmtUpdateStatus->execute(array(':customer_status' => $input, ':customer_id' => $row['customer_id']));
                }

                echo '<td>
                    <form action="" method="post">
                        <select name="Submitbb" onchange="this.form.submit();">
                            <option> - ' . $customer_status . ' - </option>
                            <option>Susisiekti</option>
                            <option>Priminimas 1</option>
                            <option>Priminimas 2</option>
                            <option>Paskambinti</option>
                            <option>Netinkamas klientas</option>
                            </select>
                    </form>
                </td>';


                echo '<td>' . 'Išsaugoti' . '</td></tr>';

            }



        ?>

谢谢。

Answer 1

您可以使用隐藏输入来捕获您希望更新的数据库中的行

 echo '<td>
                <form action="" method="post">
                    <input name="customer_id" type="hidden" value="'.$newcustomerid.'"/>  <!-- hidden -->
                    <select name="Submitbb" onchange="this.form.submit();">
                        <option> - ' . $customer_status . ' - </option>
                        <option>Susisiekti</option>
                        <option>Priminimas 1</option>
                        <option>Priminimas 2</option>
                        <option>Paskambinti</option>
                        <option>Netinkamas klientas</option>
                        </select>
                </form>
            </td>';

然后在表单发布的时候进行比较

            if(isset($_POST['Submitbb'])){ //check if form was submitted
            $input = $_POST['Submitbb']; //get input text
            $customer_id =  $_POST['customer_id']; // get the posted customer ID
            if($row['customer_id'] == $customer_id){

                    $stmtUpdateStatus = $conn->prepare("UPDATE user_customers SET `customer_status` = :customer_status WHERE `customer_id` = :customer_id");
                    $stmtUpdateStatus->execute(array(':customer_status' => $input, ':customer_id' => $customer_id)); 
                }
            }