Id receiver php

Hello, I have this code that takes the id and send it to the showFullArticle.php:


$rs_result = mysql_query ($sql,$mysql_connection);
while ($row = mysql_fetch_assoc($rs_result)) {
?>
<div><a href="showFullArticle.php?id=<?php echo $row["id"];?>" target="Article">
<?php echo $row['title']; ?></a>
</div>
<div class="description"><?php echo $row["description"];?></div>
<?php
};
?>


I put that code in the showFullArticle.php file:


<?php
$id = $_GET["id"];
echo $id;
$mysql_connection = mysql_connect("host", "user", "password") or die ('Eror: ' .mysql_error());
$database_myconnection = "database";
$query = "select * from tablename where id = '".$id."'";
if (mysql_query($query)) {
echo "

'".$title."'

";
}
mysql_close($mysql_connection);
?>


but something is wrong. mysql data are not displayed. Can you please help me? Thank you in advance!

You connect to the database there, but you're not selecting one.


$mysql_connection = mysql_connect("host", "user", "password") or die ('Eror: ' .mysql_error());
$database_myconnection = "database";
mysql_select_db($database_myconnection,$mysql_connection) or die('Failed to select database: 'mysql_error());


Also, echo out $query and see if that's getting filled with $id.

Aside...

1. you can simplify this

$query = "select * from tablename where id = '".$id."'";
to this
$query = "select * from tablename where id = '$id'";


2. Do you want to display the title and the article?

$query = "select * from tablename where id = '$id'";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
echo $row['title'] . "<br>";
echo nl2br($row['article']);

Great! It worked perfectly! Thank you very much Nightfire and litebearer!

Great! It worked perfectly! Thank you very much Nightfire and litebearer!