从2种不同形式获取数据并继续同一表行

Question

请帮助我完成此操作，因为它已经使我发疯。

我是整个过程的新手，所以对您来说似乎容易的事对我来说可能是一场噩梦，不，谷歌没有帮助:(

因此，我有一个名为member的mysql表，其结构如下：

• mem_id
• 用户名
• 密码
• 名字
• 姓
• titlu（标题）
• 描述（描述）
• 欢乐（整数）
• 通讯

我正在使用execute.php解析用户详细信息，如下所示：

<?php
session_start();
include('db.php');
$username=$_POST['username'];

$result  =  mysqli_query($db,"SELECT  *  FROM  member  WHERE  username='$username'");
$num_rows  =  mysqli_num_rows($result);

if  ($num_rows)  {
header("location:  index.php?remarks=failed");
}
else
{
$firstname=$_POST['firstname'];
$lastname=$_POST['lastname'];
$username=$_POST['username'];
$password=$_POST['password'];
mysqli_query($db,"INSERT  INTO  member(firstname,  lastname,  username,  password)VALUES('$firstname',  '$lastname',  '$username',  '$password')");
header("location:  index.php?remarks=success");
}
?>

现在我有另一种形式，可以插入礼物详细信息，并且必须继续填充mysql中的同一行。

我尝试了以下方法，但是没有运气：

<?php
session_start();
include('db.php');
$username=$_POST['username'];

$result  =  mysqli_query($db,"SELECT  *  FROM  member  WHERE  username='$username'");
$num_rows  =  mysqli_num_rows($result);

if  ($num_rows)  {
header("location:  index.php?remarks=failed");
}
else
{
$titlu = $_POST['titlu'];
$descriere = $_POST['descriere'];
$joy = $_POST['joy'];
$comm = $_POST['comm'];
$sql = "UPDATE member 
            SET titlu = '".mysql_real_escape_string($_POST[titlu])."'
            SET descriere = '".mysql_real_escape_string($_POST[descriere])."'
            SET joy = '".mysql_real_escape_string($_POST[joy])."'
            SET comm = '".mysql_real_escape_string($_POST[comm])."'
            WHERE username='".mysql_real_escape_string($_POST['username'])."'";
header("location:  welcome.php?remarks=success");
}
?>

非常感谢您的支持！

Answer 1

将您的第二个文件更改为以下代码。 仅当不存在具有给定用户名的用户时，它才会重定向到index.php?remark=failed

 <?php
    session_start();
    include('db.php');
    $username=$_POST['username'];

    $result  =  mysqli_query($db,"SELECT  *  FROM  member  WHERE  username='$username'");
    $num_rows  =  mysqli_num_rows($result);

    if  (!$num_rows)  {
    header("location:  index.php?remarks=failed");
    }
    else
    {
    $titlu = $_POST['titlu'];
    $descriere = $_POST['descriere'];
    $joy = $_POST['joy'];
    $comm = $_POST['comm'];
    $sql = "UPDATE member 
                SET titlu = '".mysql_real_escape_string($_POST[titlu])."',
                descriere = '".mysql_real_escape_string($_POST[descriere])."',
               joy = '".mysql_real_escape_string($_POST[joy])."',
                comm = '".mysql_real_escape_string($_POST[comm])."'
                WHERE username='".mysql_real_escape_string($_POST['username'])."'";
mysqli_query($sql);
    header("location:  welcome.php?remarks=success");
    }
    ?>

Answer 2

我设法通过使用完成它：

<?php
    session_start();
    include('db.php');
    include('session.php');

    $res  =  mysqli_query($db,"SELECT * FROM member where mem_id=$loggedin_id");    
    $num_rows  =  mysqli_num_rows($res);

    if  (!$num_rows)  {
        header("location:  welcome.php?remarks=failed");
    }
    else
    {
    $titlu = $_POST['titlu'];
    $descriere = $_POST['descriere'];
    $joy = $_POST['joy'];
    $comm = $_POST['comm'];

mysqli_query($db,"UPDATE member 
               SET titlu = '$titlu',
               descriere = ' $descriere ',
               joy = '$joy',
               comm = '$comm'
               where mem_id=$loggedin_id");
    header("location:  welcome.php?remarks=success");
    }
    ?>

但是现在，我需要下一步的帮助：

我需要以html表的登录用户身份从数据库中获取所有数据，但我的数据库除外。 我正在尝试以下操作，但没有带来任何回报：

<? 
$result = mysql_query($db,"SELECT * FROM member")

or die(mysql_error());

echo "<table border='1' cellpadding='10'>";

echo "<tr> <th>ID</th> <th>First Name</th> <th>Last Name</th> <th></th> <th></th></tr>";

while($row = mysql_fetch_array( $result )) {

echo "<tr>";

echo '<td>' . $row['mem_id'] . '</td>';

echo '<td>' . $row['firstname'] . '</td>';

echo '<td>' . $row['lastname'] . '</td>';

echo "</tr>";

}



// close table>

echo "</table>";

?>

有任何想法吗？

非常感谢你！

Answer 3

您可以使用mysqli_insert_id获取最后的ID插入。 然后使用ID更新。