我做了一个while循环,显示了Ploegen数据库中的所有用户,我让它输出了应该被隐藏的文本,但我暂时对其进行了测试。无论如何,它只需要更新1行,它在页面上显示的文本和ID是正确的,但是更新始终是ID 1,而我更改了很多选项值,而valueid(名称)始终相同。

我不知道如何完成您尝试的这项工作,希望您能解释一下。

我不需要反馈,因为代码还不错,只需要解决问题即可。

<div class="content-module-main">
                        <?php if(isset($_POST['submit'])){
                        $idtje = $_POST['valueid'];
                        $valueid = $_POST['ploeg'];
                        $idtje = mysqli_real_escape_string($connect, $idtje);
                        $valueid = mysqli_real_escape_string($connect, $valueid);
    //$go = "UPDATE `ploegen` SET `ploeg`='$id' where `id`='$valueid'";
    //$go2 = mysqli_query($connect, $go);
                        echo "<di" . "v class=\"confirmation-box round\">Ploegentest var_dump($valueid); var_dump($idtje);</div>";
                        }
                        ?>
                        <table>
                            <thead>
                                <tr>
                                    <th>Ploeg</th>
                                    <th>Naam</th>
                                    <th>Tel</th>
                                    <th>Gas of Elektra</th>
                                    <th>Wijzig</th>
                                </tr>

                            </thead>
                            <tbody>

                                <tr>
                                    <form method="post" action="home_ploegen_b.php">
                                    <?php
                                    $query = "SELECT naam, telefoonnummer, id, ploeg, gasofelektra FROM ploegen ORDER by naam DESC LIMIT 100";
                                    if ($result = mysqli_query($connect, $query)) {
                                    while ($get = mysqli_fetch_assoc($result)) {
                                    echo"<tr><td><center>" . $get['ploeg'] . "</center></td>";
                                    echo"<td>" . $get['naam'] . "</td>";
                                    echo"<td>" . $get['telefoonnummer'] . "</td>";
                                    echo"<td>" . $get['gasofelektra'] . "</td>";
                                    echo"<td><center><input type=\"text\" name=\"valueid\" value=\"" . $get['naam'] . "\" /><select name=\"ploeg\">
                                    <option value=\"1\">Ploeg 1</option>
                                    <option value=\"2\">Ploeg 2</option>
                                    <option value=\"3\">Ploeg 3</option>
                                    <option value=\"4\">Ploeg 4</option>
                                    <option value=\"5\">Ploeg 5</option>
                                    <option value=\"6\">Ploeg 6</option>
                                    <option value=\"7\">Ploeg 7</option>
                                    <option value=\"8\">Ploeg 8</option>
                                    <option value=\"9\">Ploeg 9</option>
                                    <option value=\"10\">Ploeg 10</option>
                                    </select><input type=\"submit\" name=\"submit\" value=\"UPDATE\"></center></td></tr>";

                                                                          }
                                    }
                                        ?>                  </td></form>

最佳答案

我认为这是您遇到的问题。

<input type=\"text\" name=\"valueid\" value=\"" . $get['naam'] . "\" />
<select name=\"ploeg\".


您的文本字段和选择字段名称在所有行中都相同。请在每个字段中输入动态名称,或将名称用作数组

参考链接:

Array Name

Dynamic Naming

这是使用数组命名的代码

while ($get = mysqli_fetch_assoc($result)) {
    echo"<tr><td><center>" . $get['ploeg'] . "</center></td>";
    echo"<td>" . $get['naam'] . "</td>";
    echo"<td>" . $get['telefoonnummer'] . "</td>";
    echo"<td>" . $get['gasofelektra'] . "</td>";
    echo"<td><center><input type=\"text\" name=\"valueid[]\" value=\"" . $get['naam'] . "\" />
    <select name=\"ploeg[]\">
    <option value=\"1\">Ploeg 1</option>
    <option value=\"2\">Ploeg 2</option>
    <option value=\"3\">Ploeg 3</option>
    <option value=\"4\">Ploeg 4</option>
    <option value=\"5\">Ploeg 5</option>
    <option value=\"6\">Ploeg 6</option>
    <option value=\"7\">Ploeg 7</option>
    <option value=\"8\">Ploeg 8</option>
    <option value=\"9\">Ploeg 9</option>
    <option value=\"10\">Ploeg 10</option>
    </select>
    <input type=\"submit\" name=\"submit\" value=\"UPDATE\"></center></td></tr>";
}


这是动态命名的

$row = 1;
while ($get = mysqli_fetch_assoc($result)) {
    echo"<tr><td><center>" . $get['ploeg'] . "</center></td>";
    echo"<td>" . $get['naam'] . "</td>";
    echo"<td>" . $get['telefoonnummer'] . "</td>";
    echo"<td>" . $get['gasofelektra'] . "</td>";
    echo"<td><center><input type=\"text\" name=\"valueid$row\" value=\"" . $get['naam'] . "\" />
    <select name=\"ploeg$row\">
    <option value=\"1\">Ploeg 1</option>
    <option value=\"2\">Ploeg 2</option>
    <option value=\"3\">Ploeg 3</option>
    <option value=\"4\">Ploeg 4</option>
    <option value=\"5\">Ploeg 5</option>
    <option value=\"6\">Ploeg 6</option>
    <option value=\"7\">Ploeg 7</option>
    <option value=\"8\">Ploeg 8</option>
    <option value=\"9\">Ploeg 9</option>
    <option value=\"10\">Ploeg 10</option>
    </select>
    <input type=\"submit\" name=\"submit\" value=\"UPDATE\"></center></td></tr>";
    $row ++;
}

10-04 11:10
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