Question

所以我有一个基于我的数据库的选择器。一旦人们从该选择器中选择了某些内容，请选择它并按下一步。他们会看到这个页面：

（此处示例中选择train02）

Selected something from the DB

此代码为：

<form method="POST" action="add_to_summary.php?user_id=<?php echo $_SESSION['user_id']?>">
            <div id="customer_list_table">
                <table>
                    <tr>
                        <th>Train id</th>
                        <th>Train Name</th>
                        <th>Tare Weight</th>
                        <th>Numbers of bogies</th>
                        <th>Numbers of axles</th>
                    </tr>
                    <?php 
                        foreach($_POST["checkbox"] as $key=>$val){
                        $data = $database->getAllAssoc_id($val);
                        foreach($data as $data1) { ?>
                            <tr>
                                <input type='hidden' name='user_id[<?php echo $_GET['user_id']?>]' value='<?php echo $_GET['user_id']?>'>
                                <input type='hidden' name='train_id[<?php echo $data1['train_id']?>]' value='<?php echo $data1['train_id']?>'>
                                <td><?= $data1['train_id'] ?></td>
                                <td><?= $data1['train_name'] ?></td>
                                <td><?= $data1['tare_weight'] ?></td>
                                <td><?= $data1['number_of_bogies'] ?></td>
                                <td><?= $data1['number_of_axles'] ?></td>
                            </tr><?php
                        }
                    } ?>
                </table>
                </div>
            <input name="Add to list" type="submit" id="add_to_list" value="Add to list">
        </form>

当他们按添加到列表时，插件会像这样发送到我的数据库：

function summary_add($id) {
            $sql = "INSERT INTO user_train_information "
        . "(train_id, user_id)"
        . "VALUES (:train_id, :user_id) ";
            $sth = $this->pdo->prepare($sql);
            $sth->bindParam(':train_id', $id, PDO::PARAM_STR);
            $sth->bindParam(':user_id', $id, PDO::PARAM_STR);
            $sth->execute();
    }

add_to_summary.php上的代码：

<?php
        foreach($_POST['user_id'] as $id) {
            $test = $database->summary_add($id);
        }
?>

现在，信息将被插入我的数据库中。但是我选择的那个加倍。像这样：

what it does

当我将$_POST['user_id']更改为$_POST['train_id']时，它会输入 14 2次。

当我执行foreach($_POST as $id)时，我得到2个数组到字符串转换错误。

我如何解决这个问题？

我希望表格正确插入，以便user_id列 1 ，train_id列 14 的 编辑：
对于选择器/选择框：form pass selected to next page

编辑：
它现在正在运作！

改变了什么：

更改为隐藏字段：

<input type='hidden' name='user_id' value='<?php echo $_GET['user_id']?>'>
<input type='hidden' name='train_id' value='<?php echo $data1['train_id']?>'>

功能：

function summary_add($train_id, $user_id) {
            $sql = "INSERT INTO user_train_information "
        . "(train_id, user_id)"
        . "VALUES (:train_id, :user_id) ";
            $sth = $this->pdo->prepare($sql);
            $sth->bindParam(':train_id', $train_id, PDO::PARAM_STR);
            $sth->bindParam(':user_id', $user_id, PDO::PARAM_STR);
            $sth->execute();
    }

我插入的地方：

<?php
            $test = $database->summary_add($_POST['train_id'], $_POST['user_id']);
            print_r($_POST);
?>

Answer 1

function summary_add($train_id, $user_id) {
            $sql = "INSERT INTO user_train_information "
        . "(train_id, user_id)"
        . "VALUES (:train_id, :user_id) ";
            $sth = $this->pdo->prepare($sql);
            $sth->bindParam(':train_id', $train_id, PDO::PARAM_STR);
            $sth->bindParam(':user_id', $user_id, PDO::PARAM_STR);
            $sth->execute();
    }

add_to_summary.php上的代码：

<?php
            $test = $database->summary_add($_POST['train_id'][0], $_POST['user_id'][0]);
?>

摘要：您必须将train_id和user_id放在summary_add函数中。

Insert上的数组到字符串转换

1 个答案: