从2种不同形式获取数据并继续同一表行

Question

Please help me get this done since it's driving me crazy already.

I'm new to this whole process so what seems easy for you might be a nightmare for me, and no, google didn't help :(

So, i'm having one mysql table named member with the following structure:

• mem_id
• username
• password
• firstname
• lastname
• titlu (title)
• descriere (description)
• joy (integer)
• comm

I'm parsing user details using execute.php looking like this:

<?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");
}
?>

Now i have another form that inserts gift details and must continue filling the same row in mysql.

I've tried the following but no luck:

<?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");
}
?>

Thank you very much for your support!

Answer 1

Change your second file to the below code. it will redirect to index.php?remark=failed only if no user exist with the given username

 <?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

I managed to get it done by using:

<?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");
    }
    ?>

But now, i need help with the next step:

I need to get all the data from the database,except mine, as a logged in user, in a html table. I'm trying the following but brings back nothing:

<? 
$result = mysql_query($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>";

?>

Any ideas?

Thank you very much!

Answer 3

you can use mysqli_insert_id get last id insert. Then update with id.