Question

请耐心等待我，我还不熟悉这种语言。我有一张表格，列出申请人的记录，如申请人编号，姓名和身份。我想更新申请人的状态“雇用”＃39;或者＆＃39;失败＆＃39;在使用PHP表单的特定行上。但是，我不确定如何在提交时获取其行上的特定提交名称。或者，如果你有一个解决方法，我会很感激。非常感谢你的帮助。

<!DOCTYPE html>
<html>
      <h2>Applicant Records</h2>
<?php

$mysql_hostname = "localhost";
$mysql_user = "root";
$mysql_password ="";
$mysql_database = "applicantrecord";

// Create connection
$conn = new mysqli($mysql_hostname, $mysql_user, $mysql_password, $mysql_database);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sqli = "SELECT id, firstname, lastname, status FROM applicant";
$result = $conn->query($sqli);


if ($result->num_rows > 0) { ?>
    <table class="table">
    <thead>
        <tr>
            <th>Applicant No.</th>
            <th>Lastname</th>
            <th>Firstname</th>
            <th>Status</th>
            <th></th>
        </tr>
    </thead>
<?php
    // output data of each row

    echo "<tbody>";
    while($row = $result->fetch_assoc()) 
    { ?>
            <tr>
                <td>
                    <?php echo $row["id"]; 
                        $appid = $row["id"];
                    ?>  
                </td>
                <td>
                    <?php echo $row["lastname"]; ?>
                </td>
                <td>
                    <?php echo $row["firstname"]; ?>
                </td>
                <td>
                    <?php echo $row["status"]; ?>
                </td>
                <td>
                </td>
                <td>

                        <div>
                        <form action="" role="form" method="post" name="form<?php echo $appid; ?>">
                        <select name="applicant_status">
                        <option value="Hired">Hire</option>
                        <option value="Failed">Fail</option>
                        </select>

                        </p>
                        <button type="submit" class="btn btn-default" name = "submit<?php echo $appid; ?>" data-dismiss="modal">Submit</button>                     
                        </form>
                        <?php

                        if(isset($_POST["submit"])){
                            $dbhost = 'localhost';
                            $dbuser = 'root';
                            $dbpass = '';
                            $newappid = $appid;
                            $newapptstatus = $_POST['applicant_status'];
                            $connect = mysql_connect($dbhost, $dbuser, $dbpass);
                        if(! $connect ) {
                            die('Could not connect: ' . mysql_error());
                            }

                        $sql_sub = "UPDATE applicant ". "SET status = '$newappstatus'".  
                                    "WHERE id = '$newappid'" ;
                                    mysql_select_db('applicantrecord');
                        $retval = mysql_query( $sql_sub, $connect );

                        if(! $retval ) {
                        die('Could not update data: ' . mysql_error());
                            echo "<script type= 'text/javascript'>alert('An error occured! Applicant status update failed!');</script>";
                        }

                        echo "<script type= 'text/javascript'>alert('Applicant status updated successfully!');</script>";
                        mysql_close($connect);

            }                   
                        ?>
                        </div>

                </td>
            </tr>
    <?php }
    echo "</tbody>";
    echo "</table>";
} else {
    echo "0 results";
}
$conn->close();
?>

</html>

Answer 1

在您检查$_POST['submit']已设置的if语句中，索引'submit'不存在。因此isset($_POST['submit'])计算为false，并且永远不会执行更新表的查询。

变量$appid正在随着添加的每一行而更改，因此当页面完成加载并且某个行上的提交按钮被推送时，$appid不一定包含正确的行行号。

要解决此问题，您可以在表单中使用隐藏的输入：

<input name="id" value="<?php echo $appid ?>" type="hidden">

然后，您可以将isset($_POST['submit'])替换为isset($_POST['id'])并设置$newappid = $_POST['id']以获取要更改的行号。

如何使用php表单更新表中的特定行？

1 个答案: