Question

我有一个html网页，其中有一个表格，可以从我的数据库中的表中获取数据。每行都有一个按钮，点击后可以使用javascript动态添加到网页上的另一个表中。动态表有一个提交按钮，单击该按钮时，表中的数据应添加到我的数据库中的表中。我的问题是表中的内容没有添加到数据库中，但每次按下提交按钮时，都会添加一个具有唯一ID的新行。还有一个输入文本框，也可以正确添加到我的数据库中。这让我相信我的数据库连接正在工作，但由于某种原因，表中的数据不会被添加。如何使动态表中的数据进入数据库？这是我的代码。代码已经更新了建议。

    <?php
        $sql = "SELECT *
                FROM inventoryCars";
        $result = mysqli_query($connection, $sql);
?>                              
<div class="inventory">
    <div class ="cars">
    <table>
        <?php           
            if (mysqli_num_rows($result) > 0)
            {
            while($row = mysqli_fetch_assoc($result)) 
            {
                $RegNo = $row["regNo"];
                $CarMake = $row["carMake"];
                $CarModel = $row["carModel"];
        ?>  
        <tr>
            <input type="hidden" name="regNo" value="<?php echo $row['regNo'] ?>">
            <input type="hidden" name="carMake" value="<?php echo $row['carMake'] ?>">
            <input type="hidden" name="carModel" value="<?php echo $row['carModel'] ?>">
            <td><?php echo $row["regNo"] ?></td>
            <td><?php echo $row["carMake"] ?></td>
            <td><?php echo $row["carModel"] ?></td>
            <td><button onclick="addRowToTable('<?php echo $RegNo ?>','<?php echo $CarMake ?>','<?php echo $CarModel ?>')">+</button></td>
        </tr>
        <script type ="text/javascript">
            function addRowToTable(regNo, carMake, carModel) {
                var table = document.getElementById("newhireTBL");
                var row = table.insertRow(0);
                var cell1 = row.insertCell(0);
                var cell2 = row.insertCell(1);
                var cell3 = row.insertCell(2);
                cell1.innerHTML = regNo;
                cell2.innerHTML = carMake;
                cell3.innterHTML = carModel;
                }
        </script>
        <?php
            }
            } 
        ?>
    </table>
    </div>
</div>              

    <div class="order">
    <div class ="newhire">
    <table id="newhireTBL">
    </table>

    <form action = "addNewHire.php" method = "post">
            <input type="hidden" name="regNo" value="<?php echo $row['regNo'] ?>">
            <input type="hidden" name="carMake" value="<?php echo $row['carMake'] ?>">
            <input type="hidden" name="carModel" value="<?php echo $row['carModel'] ?>">

            <input id="submit" type="submit" name="submit" value="Submit">
    </form> 
    </div>
</div>

addNewHire.php

<?php session_start(); ?>
<?php require_once("connectToSQL.php"); ?>
<?php
echo "<pre>";var_dump($_POST);echo "</pre>";die;
if (isset($_POST['submit'])){
    $RegNo = $_POST['regNo'];
    $CarMake = $_POST['carMake'];
    $CarModel = $_POST['carModel'];

    $_SESSION["regNo"] = $RegNo;
    $_SESSION["carMake"] = $CarMake;
    $_SESSION["carModel"] = $CarModel;

$sql = "INSERT INTO newHire(regNo, carMake, carModel)
        VALUES('$RegNo', '$CarMake', '$CarModel')";

if (mysqli_query($connection, $sql)) {
header("location:order_success.php");
echo "Order has been made";
} else {
header("location:order_fail");
echo "Order fail";

}
}
?>

}
?>

测试显示了这个

array(4) { 
["regNo"]=>
string(0) "" 
["carMake"]=> 
string(0) "" 
["carModel"]=>
string(0) "" 
["submit"]=> 
string(6) "Submit" 
}

Answer 1

表单和表格标签的嵌套无效，这可能会阻止提交表单数据。尝试为＆＃34; newhireTBL＆＃34;移动结束标记。表位于表格的开始标记（表格外）的正下方，因此表格和表格标签不会重叠。

例如：

<table...>
</table>

<form...>
</form>

编辑：这是一个基本示例，其中包含一些可以帮助您解决问题的虚拟数据。祝你好运！

 <?php
    $rows = array(
        array(
            "regNo" => "abc123",
            "carMake" => "toyota",
            "carModel" => "camry",
        ),
        array(
            "regNo" => "def456",
            "carMake" => "ford",
            "carModel" => "laser",
        ),        
    );
?>    

<?php foreach ($rows as $row): ?>
    <?php echo $row["regNo"] ?>
    <?php echo $row["carMake"] ?>
    <?php echo $row["carModel"] ?>
    <button onclick="addRow('<?php echo $row["regNo"] ?>','<?php echo $row["carMake"] ?>','<?php echo $row["carModel"] ?>')">+</button>
    <br/>
<?php endforeach ?>

<form action = "addNewHire.php" method = "post">
    <input id=regNo type="text" name="regNo" value="">
    <input id=carMake type="text" name="carMake" value="">
    <input id=carModel type="text" name="carModel" value="">
    <input id="submit" type="submit" name="submit" value="Submit">
</form>

<script type ="text/javascript">
    function addRow(regNo, carMake, carModel) {
        document.getElementById('regNo').value = regNo;
        document.getElementById('carMake').value = carMake;
        document.getElementById('carModel').value = carModel;
    }
</script>

将数据从动态表提交到数据库 - PHP

1 个答案: