[How to display a certain image, depending on a variable.]

Ok, then. So first, get the data from the database. Then, using the code I've posted above...

 

<?php
// I'm assuming you've already made the db connection and have executed the query...
// Substitute the row name

while($row = mysql_fetch_array($result)) {
echo '<img src="/images/' . $row['lang'] . 'flag.gif">';
}
// blah...
?>

[Ideas Anyone?]

http://www.phpbb.com/

 

... or simply search it on Google...

[How to display a certain image, depending on a variable.]

Yes, there is a way. First, name all of your image files the same way (ex: LANGflag.gif) and they should all be in the same folder. Then when you display it, just use

 

<?php
$lang = $_GET['lang'] // (or however you get your user's language)
echo '<img src="/images/' . $lang . 'flag.gif">';
?>

[Ideas Anyone?]

Heh, I'd look into phpBB's lang source code. They have it in arrays, but you'll have to download additional language packs.

[session destroy woes.]

Did you put session_start() at the very beginning of both files?

[how do i get a variable form the url and use it in a link - noob question]

<?php
$idr = $_GET['idr'];
$table = $_GET['table'];

echo '<a href="printquote.php?idr=' . $idr . '&table=' . $table . '">link</a>';
?>

[session destroy woes.]

First of all, I suggest you put session_start() at the beginning of login.php. Then, in logout.php, you have to start the session first, then destroy it.

[how do i get a variable form the url and use it in a link - noob question]

Yes, it's a $_GET[''] function.

 

<?php
$idr = $_GET['idr'];
$table = $_GET['table'];

echo '<a href="printquote.php?idr=' . $idr . '&table=' . $table . ">link</a>';
?>

[Mysql Query Question.]

$query = "SELECT rank FROM `ranks` WHERE id IN(" . implode(',', $rankid) . ")";
$result = mysql_query($query) OR DIE (mysql_error());

 

Putting it in variables is sometimes easier.

[[SOLVED] Stuck on Strange array error!]

It'll be better if you posted the relevant part(s) of the code here, as I am not going to click those links. However, have you looked at array_merge() ?

[Mysql Query Question.]

Try this:

$result = mysql_query("SELECT rank FROM `ranks` WHERE id IN (" . implode (',',$rankid) . ")");

[File downloads on secure server]

What exactly were the errors that you got?

[[SOLVED] insert content script]

We're here to help, not to make the entire thing for you. Try to get started on it first, and we'll be glad to help you along the way.

[managing "hit points" in a simple game]

Put

<?php
error_reporting(E_ALL);
?>

 

at the very top of the page. See if any error message comes up.

[[SOLVED] Loops and arrays... problem]

That worked! Thank you all so much for helping me out with this.

[[SOLVED] Loops and arrays... problem]

I don't really get what you mean by the question, but the field "msg_id" is the same in both tables.

 

I'll try that out.

[[SOLVED] Sessions Test Not Working]

Nevermind. You started the session already ($_SESSION['username'] = $username). I overlooked that before. Sorry.

 

Solution:

check_login.php:

<?php

session_start();

mysql_connect("localhost", "root", "xxxxxxxx");
mysql_select_db("xxxxxxxxxxxx");

$username = $_POST['username'];
$password = $_POST['password'];

$sql = "SELECT * FROM users WHERE username = '$username' AND password = '$password'";
$result = mysql_query($sql); 

if (mysql_num_rows($result) == 0) {

echo "Username and password combination not found.";

} else {
    while($row = mysql_fetch_array($result)) {
        $_SESSION['username'] = $row['username'];
    }
header("members.php");

}  

?>

[[SOLVED] Loops and arrays... problem]

try to change this to

DELETE FROM table_name

WHERE column_name = some_value

 

Since I'm dealing with multiple tables, that wouldn't work.

[[SOLVED] Sessions Test Not Working]

You didn't start $_SESSION['username'] in check_login.php...

[[SOLVED] Loops and arrays... problem]

That just deletes ALL of the records. I believe I need the limit in there.

[[SOLVED] Loops and arrays... problem]

Echo the SQL query and paste it here.

 

DELETE privmsgs, privmsgs_spec FROM privmsgs, privmsgs_spec WHERE privmsgs.msg_id IN (4,5) LIMIT 2

[[SOLVED] A closing echo in include() file?]

File 1:

<?php
// Blah Blah
include ('file2.php');
?>

 

File 2: (the include file)

<?php
echo '<td></td>';
echo '<td>';
?>

 

You don't really have to use PHP in the include file...

 

<td></td>
<td>

[[SOLVED] Loops and arrays... problem]

@ roopurt18:

I got this error:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'LIMIT 2' at line 8

 

My query now looks like this:

<?php
$query = "DELETE
                  $privmsgs_tbl_name,
                  $privmsgs_spec_tbl_name
             FROM
                 $privmsgs_tbl_name,
                 $privmsgs_spec_tbl_name
             WHERE
                 $privmsgs_tbl_name.msg_id IN (" . implode(', ', $_POST['pm_delete']) . ") LIMIT " . count($_POST['pm_delete']);
?>

 

Any ideas?

 

@ Hypnos: I'll try it out.

[Foreach Required for $_POST Array?]

Try $_POST['check']['1']

or maybe

$_POST['check'][1]

[[SOLVED] Loops and arrays... problem]

Will this work with deleting rows from multiple tables that I have now? They all have the same ID as their common row.