从2种不同的表格中获取数据，并继续使用相同的表格行

Question

请帮我完成这件事，因为它已经让我发疯了。

我是新来的这整个过程有啥似乎很容易对你可能是我的噩梦，没有，谷歌并没有帮助:(

所以，我有一个MySQL表命名为成员具有以下结构：

• mem_id
• 用户名
• 密码
• 姓名
• 姓氏
• titlu（标题）
• descriere（介绍）
• 喜悦（整数）
• 通讯

我解析使用execute.php看起来像这样用户的详细信息：

<?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"); 
} 
?> 

现在我有另一种插入gi ft细节，并且必须继续填充mysql中的同一行。

我试过以下，但没有运气：

<?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下面的代码。它会重定向到index.php?remark=failed只有在没有用户与给定用户名

<?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

可以使用mysqli_insert_id得到最后ID插入存在。然后用id更新。

Answer 3

我设法得到它通过做：

<?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"); 
    } 
    ?> 

但现在，我需要下一个步骤的帮助：

我需要从数据库中获取的所有数据，除了我的，如一个登录用户，在一个HTML表格。我正在尝试以下但没有带回：

<? 
$result = mysql_query($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>"; 

?> 

任何想法？

非常感谢！